6.5 Miscellaneous Integration Methods

The integration methods presented so far are considered “standard,” meaning every calculus student should know them. This section will discuss a few additional methods, some more common than others. One such method is the Leibniz integral rule for “differentiation under the integral sign.”^[1] This powerful and useful method is best explained with a simple example.

Recall from Section 5.4 that

\begin{equation} \int e^{\alpha x}\;\dx ~=~ \tfrac{1}{\alpha}\,e^{\alpha x} ~+~ C \tag{6.14}\end{equation}

for any constant $\alpha \ne 0$. This antiderivative involves the variable $x$ and the constant $\alpha$. The idea behind the Leibniz rule is to reverse those roles: view $\alpha$ as the variable and $x$ as a constant. The antiderivative is then seen as a function of $\alpha$, and so its derivative can be taken with respect to $\alpha$. The big leap that this method makes is to move the differentiation operation inside the integral:^[2]

\[ \frac{d}{\dalpha} \int e^{\alpha x}\;\dx ~=~ \int \frac{d}{\dalpha}\,(e^{\alpha x})~\dx ~=~ \int x\,e^{\alpha x}\;\dx \]

However, differentiating the right side of formula (6.14) shows that

\[ \frac{d}{\dalpha} \int e^{\alpha x}\;\dx ~=~ \frac{d}{\dalpha} \left(\tfrac{1}{\alpha}\,e^{\alpha x} ~+~ C\right) ~=~ \frac{\alpha\,\left(x\,e^{\alpha x}\right) ~-~ 1\,\cdot\,e^{\alpha x}}{\alpha^2} ~=~ \tfrac{1}{\alpha}\,x\,e^{\alpha x} ~-~ \tfrac{1}{\alpha^2}\,e^{\alpha x} \]

Thus,

\[ \int x\,e^{\alpha x}\;\dx ~=~ \tfrac{1}{\alpha}\,x\,e^{\alpha x} ~-~ \tfrac{1}{\alpha^2}\,e^{\alpha x} ~+~ C \]

which can be verified via integration by parts with the tabular method:

$tikzpicture[>=latex] [left] at (0,0.2) u; [right] at (1,0.2) ; (-0.6,-0.15) -- (2.3,-0.15); [left] at (0,-0.5) x; [right] at (1,-0.5) e^ x\, ; [left] at (0,-1.2) 1; [right] at (1,-1.2) 1 \,e^ x; [->] (0,-0.5) -- (1,-1.2); [right] at (2.4,-1.2) (+); [->] (3.4,-1.2) -- (4.4,-1.2) node[right] + (x)\,( 1 \,e^ x); [left] at (0,-1.9) 0; [right] at (1,-1.9) 1 ^2\,e^ x; [->] (0,-1.2) -- (1,-1.9); [right] at (2.4,-1.9) (--); [->] (3.4,-1.9) -- (4.4,-1.9) node[right] - (1)\,( 1 ^2\,e^ x); [->] (-1.5,-1.9) node[left] STOP -- (-0.5,-1.9); tikzpicture$

\[ \int x\,e^{\alpha x}\;\dx ~=~ \tfrac{1}{\alpha}\,x\,e^{\alpha x} ~-~ \tfrac{1}{\alpha^2}\,e^{\alpha x} ~+~ C\quad\checkmark \]

What was actually done in the above example? A known integral,

\[ \int e^{\alpha x}\;\dx ~=~ \tfrac{1}{\alpha}\,e^{\alpha x} ~+~ C ~, \]

was differentiated with respect to $\alpha$ via the Leibniz rule to produce a new integral,

\[ \int x\,e^{\alpha x}\;\dx ~=~ \tfrac{1}{\alpha}\,x\,e^{\alpha x} ~-~ \tfrac{1}{\alpha^2}\,e^{\alpha x} ~+~ C ~, \]

with the constant $\alpha$ treated temporarily—only during the differentiation—as a variable. In general, that is how the Leibniz rule is used. Typically this means if you want to evaluate a certain integral with the Leibniz rule, then you “work backwards” to figure out which integral you need to differentiate with respect to some constant (e.g. $\alpha$) in the integrand.

Example 6.24

Use the Leibniz rule to evaluate $~\displaystyle\int \frac{\dx}{(1 + x^2)^2}~$.

Solution: By formula (5.4) in Section 5.4,

\[ \int\,\frac{\dx}{a^2 + x^2} ~=~ \tfrac{1}{a}\,\tan^{-1}\left(\tfrac{x}{a}\right) ~+~ C \]

for any constant $a > 0$. So differentiate both sides with respect to $a$:

\begin{align*} \frac{d}{\da}\,\int\,\frac{\dx}{a^2 + x^2} ~&=~ \frac{d}{\da}\,\left(\tfrac{1}{a}\,\tan^{-1}\left(\tfrac{x}{a}\right) ~+~ C\right)\\[6pt] \int\,\frac{d}{\da}\,\left(\frac{1}{a^2 + x^2}\right)~\dx ~&=~ -\tfrac{1}{a^2}\,\tan^{-1}\left(\tfrac{x}{a}\right) ~+~ \tfrac{1}{a}\,\cdot\,\frac{1}{1 + \left(\tfrac{x}{a}\right)^2}\,\cdot\,-\tfrac{x}{a^2}\\[6pt] \int -\frac{2a}{(a^2 + x^2)^2}\,\dx ~&=~ -\tfrac{1}{a^2}\,\tan^{-1}\left(\tfrac{x}{a}\right) ~-~ \frac{x}{a\,(a^2 + x^2)}\\[6pt] \int \frac{\dx}{(a^2 + x^2)^2} ~&=~ \tfrac{1}{2a^3}\,\tan^{-1}\left(\tfrac{x}{a}\right) ~+~ \frac{x}{2a^2\,(a^2 + x^2)} ~+~ C \end{align*}

That general formula is useful in itself. In particular, for $a=1$,

\[ \int \frac{\dx}{(1 + x^2)^2} ~=~ \tfrac{1}{2}\,\tan^{-1} x ~+~ \frac{x}{2\,(1 + x^2)} ~+~ C ~, \]

which agrees with the result from Example 6.17 in Section 6.3.
Notice that there was no generic constant (e.g. $a$ or $\alpha$) in the statement of the problem. When that happens, you will need to figure out where the constant should be in order to use the Leibniz rule.

You can also use differentiation under the integral sign to evaluate definite integrals.

Example 6.25

Show that $~\displaystyle\int_0^{\infty} e^{-x^2} \,\dx ~=~ \tfrac{1}{2}\sqrt{\pi}~$.

Solution: Let $I = \int_0^{\infty} e^{-x^2} \,\dx$. The integral is convergent, since by Exercise 11 in Section 4.4, for all $x$

\[ e^{x^2} ~\ge~ 1 ~+~ x^2 \quad\Rightarrow\quad 0 ~\le~ e^{-x^2} ~\le~ \frac{1}{1 + x^2} \]

implies $I$ is convergent by the Comparison Test, since $\int_0^{\infty} \frac{1}{1 + x^2}\,\dx$ is convergent (and equals $\tfrac{1}{2}\pi$) by Example 5.32 in Section 5.5. For $\alpha \ge 0$, define

\[ \phi(\alpha) ~=~ \int_0^{\infty} \,\frac{\alpha\,e^{-\alpha^2 x^2}}{1 + x^2} \,\dx ~. \]

Then clearly $\phi(0) = 0$, and differentiating under the integral sign shows

\[ \phi'(\alpha) ~=~ \int_0^{\infty} \,\frac{-2\alpha^2 e^{-\alpha^2 x^2} + e^{-\alpha^2 x^2}}{1 + x^2}~\dx \qquad\Rightarrow\qquad \phi'(0) ~=~ \int_0^{\infty} \frac{\dx}{1 + x^2} ~=~ \tfrac{1}{2}\pi ~. \]

The substitution $y = \alpha x$, so that $\dy = \alpha\,\dx$, shows $\phi(\alpha)$ can be written as

\[ \phi(\alpha) ~=~ \int_0^{\infty} \,\frac{e^{-y^2}}{1 + \left(\tfrac{y}{\alpha}\right)^2} \,\dy \qquad\Rightarrow\qquad 0 ~\le~ \lim_{\alpha \to \infty}~ \phi(\alpha) ~\le~ I ~<~ \infty ~. \]

Also, for $\alpha > 0$,

\begin{align*} \frac{d}{\dalpha}\,\left(\frac{1}{\alpha}\,e^{-\alpha^2}\,\phi(\alpha)\right) ~&=~ \frac{d}{\dalpha}\,\int_0^{\infty} \,\frac{e^{-\alpha^2 (1+x^2)}}{1 + x^2} \,\dx ~=~ \int_0^{\infty} \,\frac{-2\alpha\,(1+x^2)\, e^{-\alpha^2 (1+x^2)}}{1 + x^2}~\dx\\[6pt] &=~ -2\alpha\, e^{-\alpha^2}\,\int_0^{\infty} e^{-\alpha^2 x^2}~\dx \quad\text{, now substitute $u = \alpha x$ and $\du = \alpha \dx$ to get}\\[6pt] &=~ -2\alpha\, e^{-\alpha^2}\,\frac{1}{\alpha}\,\int_0^{\infty} e^{-u^2} \,\du ~=~ -2\, e^{-\alpha^2}\,I \quad\text{, and so integrating both sides yields}\\[6pt] \int_0^{\infty} \frac{d}{\dalpha}\,\left(\frac{1}{\alpha}\,e^{-\alpha^2}\,\phi(\alpha)\right)~\dalpha ~&=~ -2I\,\int_0^{\infty} e^{-\alpha^2} \,\dalpha ~=~ -2I^2 ~. \end{align*}

However, by the Fundamental Theorem of Calculus.

\begin{align*} \int_0^{\infty} \frac{d}{\dalpha}\,\left(\frac{1}{\alpha}\,e^{-\alpha^2}\,\phi(\alpha)\right)~\dalpha ~&=~ \frac{1}{\alpha}\,e^{-\alpha^2}\,\phi(\alpha)~\Biggr|_0^{\infty} ~=~ \left(\lim_{\alpha \to \infty}~\frac{\phi(\alpha)}{\alpha \,e^{\alpha^2}}\right) ~-~ \left(\lim_{\alpha \to 0}~\frac{\phi(\alpha)}{\alpha\,e^{\alpha^2}}\right)\\[6pt] &=~ 0 ~-~ \left(\lim_{\alpha \to 0}~\frac{\phi(\alpha)}{\alpha\,e^{\alpha^2}}\right) ~\to~ \frac{0}{0} \quad\text{, so by L'H\^{o}pital's Rule}\\[6pt] &=~ -\lim_{\alpha \to 0}~\frac{\phi'(\alpha)}{e^{\alpha^2} + 2\alpha^2\,e^{\alpha^2}} ~=~ -\frac{\phi'(0)}{1+0} ~=~ -\tfrac{1}{2}\pi ~. \end{align*}

Thus,

\[ -2I^2 ~=~ -\tfrac{1}{2}\pi \qquad\Rightarrow\qquad I ~=~ \tfrac{1}{2}\sqrt{\pi} \]

which is the desired result.

One immediate consequence of Example 6.25 is that

\[ \int_{-\infty}^{\infty} e^{-x^2} \,\dx ~=~ \sqrt{\pi} \]

since $e^{-x^2}$ is an even function. The following example shows another consequence, as well as how useful substitutions can be in writing integrals in a different form.

Example 6.26

Show that the Gamma function $\Gamma\,(t)$ can be written as

\[ \Gamma\,(t) ~=~ 2\,\int_0^{\infty} y^{2t-1} \, e^{-y^2} ~\dy \quad\text{for all $t > 0$,} \]

and that $\Gamma\,\left(\tfrac{1}{2}\right) ~=~ \sqrt{\pi}$.

Solution: Let $x = y^2$, so that $\dx = 2y\;\dy$. Then $x=0~\Rightarrow~y=0~$ and $x=\infty~\Rightarrow~y=\infty$, so

\[ \Gamma\,(t) ~=~ \int_0^{\infty} x^{t-1} \, e^{-x} ~\dx ~=~ \int_0^{\infty} (y^2)^{t-1}\,e^{-y^2}~2y~\dy\\[6pt] ~=~ 2\,\int_0^{\infty} y^{2t-1} \, e^{-y^2} ~\dy ~. \]

In this form, with the help of Example 6.25 it is now easy to evaluate $\Gamma\,\left(\tfrac{1}{2}\right)$:

\[ \Gamma\,\left(\tfrac{1}{2}\right) ~=~ 2\,\int_0^{\infty} y^{1-1} \, e^{-y^2} ~\dy ~=~ 2\,\int_0^{\infty} e^{-y^2}~\dy ~=~ 2\,\left(\tfrac{1}{2}\sqrt{\pi}\right) ~=~ \sqrt{\pi} \]

A function closely related to the Gamma function is the Beta function $B(x,y)$, defined by:

\begin{equation} B(x,y) ~=~ \int_0^1 t^{x-1}\,(1-t)^{y-1}\,\dt \qquad\text{for all $x > 0$ and $y > 0$} \tag{6.15}\end{equation}

It can be shown that^[3]

\begin{equation} B(x,y) ~=~ \frac{\Gamma\,(x)\;\Gamma\,(y)}{\Gamma\,(x+y)} \qquad\text{for all $x > 0$ and $y > 0$.} \tag{6.16}\end{equation}

Example 6.27

Show that the Beta function $B(x,y)$ can be written as

\[ B(x,y) ~=~ \int_0^{\infty} \frac{u^{x-1}}{(1+u)^{x+y}}~\du ~. \]

Solution: Let $u=\frac{t}{1-t}$, so that $t=\frac{u}{1+u}$, $1-t=\frac{1}{1+u}$, and $\dt = \frac{\du}{(1+u)^2}$. Then $t=0~\Rightarrow~u=0$ and $t=1~\Rightarrow~u=\infty$, so

\[ B(x,y) ~=~ \int_0^1 t^{x-1}\,(1-t)^{y-1}\,\dt ~=~ \int_0^{\infty} \left(\frac{u}{1+u}\right)^{x-1}\;\left(\frac{1}{1+u}\right)^{y-1} \frac{\du}{(1+u)^2} ~=~ \int_0^{\infty} \frac{u^{x-1}}{(1+u)^{x+y}}~\du ~. \]

Another application of substitutions in integrals is in the evaluation of fractional derivatives. Recall from Section 1.6 that the zero-th derivative of a function is just the function itself, and that derivatives of order $n$ are well-defined for integer values $n \ge 1$. It turns out that derivatives of fractional orders (e.g. $n=\frac{1}{2}$) can be defined, with the Riemann-Louiville definition being the most common:

Definition 6.1

For all $0 < \alpha < 1$, the fractional derivative of order $\alpha$ of a function $f(x)$ is

\begin{equation} \frac{d^{\alpha}}{\dx^{\alpha}}\,f(x) ~=~ \frac{1}{\Gamma\,(1-\alpha)}\;\ddx\,\int_0^x \frac{f(t)}{(x-t)^{\alpha}}\,\dt ~. \tag{6.17}\end{equation}

Example 6.28

Calculate $~\dfrac{d^{1/2}}{\dx^{1/2}}\,(x)~$.

Solution: Here $\alpha = \frac{1}{2}$ and $f(x)=x$, so that

\[ \frac{d^{1/2}}{\dx^{1/2}}\,(x) ~=~ \frac{1}{\Gamma\,(1-1/2)}\;\ddx\,\int_0^x \frac{t}{(x-t)^{1/2}}\,\dt ~=~ \frac{1}{\sqrt{\pi}}\;\ddx\,\int_0^x \frac{t~\dt}{\sqrt{x-t}} \]

since $\Gamma\,\left(\tfrac{1}{2}\right) ~=~ \sqrt{\pi}$ by Example 6.26. Use the substitution $u=\sqrt{x-t}$, so that $t=x-u^2$ and $\dt=-2u\,\du$. Then $t=0~\Rightarrow~u=\sqrt{x}~$ and $t=x~\Rightarrow~u=0$, so

\begin{align*} \frac{d^{1/2}}{\dx^{1/2}}\,(x) ~&=~ \frac{1}{\sqrt{\pi}}\;\ddx\,\int_{\sqrt{x}}^0 \frac{(x-u^2)\,(-2u~\du)}{u} ~=~ \frac{2}{\sqrt{\pi}}\;\ddx\,\int_0^{\sqrt{x}} (x ~-~ u^2)~\du\\[6pt] &=~ \frac{2}{\sqrt{\pi}}\;\ddx\,\left(xu ~-~ \tfrac{1}{3}\,u^3\right)~\Biggr|_{u=0}^{u=\sqrt{x}} ~=~ \frac{2}{\sqrt{\pi}}\;\ddx\,\left(x^{3/2} ~-~ \tfrac{1}{3}\,x^{3/2}\right)\\[6pt] &=~ \frac{2}{\sqrt{\pi}}\;\ddx\,\left(\tfrac{2}{3}\,x^{3/2}\right) ~=~ \frac{2}{\sqrt{\pi}}\,\sqrt{x} \end{align*}

Part of the motivation for creating fractional derivatives was to find if it were possible to take two “half” derivatives to form a “whole” derivative:

\[ \frac{d^{1/2}}{\dx^{1/2}}\,\left(\frac{d^{1/2}}{\dx^{1/2}}\,f(x)\right) ~=~ \ddx\,f(x) \]

It is left as an exercise to show that the above relation does hold for the function $f(x) = x$. Derivatives with fractional order $0 < \alpha < 1$ and integer order $n \ge 1$ can be combined by taking the derivative of integer order first:^[4]

\[ \frac{d^{n+\alpha}}{\dx^{n+\alpha}}\,f(x) ~=~ \frac{d^{\alpha}}{\dx^{\alpha}}\,\left(\frac{d^{n}}{\dx^{n}}\,f(x)\right) \]

Recall from Section 6.3 that the trigonometric substitution $x=r\,\cos\,\theta$—or its sister substitution $x=r\,\sin\,\theta$—was motivated by trying to find the area of a circle of radius $r$. To simplify matters, let $r=1$ so that points on the unit circle can be identified with the angle $\theta$ via that substitution, with $\theta$ as shown in Figure 6.5.1(a) below.

tikzpicture[>=latex,every node/.style=font=] [dashed] (0,0) -- (60:1.5) node[black,midway,above] 1; [->,dashed] (0:0.6) arc (0:60:0.6); [black!60,line width=1pt,->] (-2.2,0) -- (2.5,0) node[right] x; [black!60,line width=1pt,->] (0,-2.2) -- (0,2.2) node[above] y; [linecolor,line width=1.5pt] (0,0) circle (1.5); (60:1.5) circle (2.5pt); at (35:0.4) ; [below right] at (0:1.5) 1; [below left] at (0:0) 0; [above right] at (60:1.5) (x,y)=( \, , \, ); [fill=white] (0,-2.42) circle (1.5pt); tikzpicture — Figure 6.5.1 Points on the unit circle $x^2 + y^2 = 1$

tikzpicture[>=latex,every node/.style=font=] [black!60,line width=1pt,->] (-2.5,0) -- (2.8,0) node[right] x; [black!60,line width=1pt,->] (0,-2.2) -- (0,2.2) node[above] y; [name path=cir,linecolor,line width=1.5pt] (0,0) circle (1.5); [dashed] (-1.5,0) -- (1.8,0); [name path=s1,dashed] (1.5,1.5) node[right] slope = 12 -- (-1.5,0) -- (1.5,-1.5) node[right] slope = - 12; [dashed] (-1.5,0) -- (0.7,2.2) node[right] slope = 1; [dashed] (-1.5,0) -- (0.7,-2.2) node[right] slope = -1; [name path=s2,dashed] (-0.5,2) node[left] slope = 2 -- (-1.5,0) -- (-0.5,-2) node[left] slope = -2 ; [below right] at (1.5,-0.1) 1; [below left] at (-1.5,-0.1) -1; [above right] at (1.5,0.1) slope = 0; (0:1.5) circle (2.5pt); [name intersections=of=cir and s1,by=a,b,c] (a) circle (2.5pt) (b) circle (2.5pt) (c) circle (2.5pt); (0,1.5) circle (2.5pt); (0,-1.5) circle (2.5pt); [name intersections=of=cir and s2,by=d,e,f] (d) circle (2.5pt) (e) circle (2.5pt) (f) circle (2.5pt); [fill=white] (-1.5,0) circle (1.5pt); [below left] at (0,0) 0; tikzpicture — Figure 6.5.1 Points on the unit circle $x^2 + y^2 = 1$

Figure 6.5.1(b) shows a different identification of points on the unit circle—by slope. This will be the basis for a half-angle substitution for evaluating certain integrals.

Let $A$ be the point $(-1,0)$, then for any other point $P$ on the unit circle draw a line from $A$ through $P$ until it intersects the line $x=1$, as shown in Figure 6.5.2 below:

tikzpicture[>=latex,every node/.style=font=] [->,shift=(-3,0)] (0:1.3) arc (0:30:1.3); [dashed] (0,0) -- (60:3); [->] (0:0.7) arc (0:60:0.7); [black!60,line width=1pt,->] (-3,0) -- (3.5,0) node[right] x; [name path=yaxis,black!60,line width=1pt,->] (0,0) -- (0,4) node[above] y; [->,black!40] (3,0) -- (3,4) node[black,pos=0.7,right] x=1; [name path=cir,linecolor,line width=1.5pt] (0:3) arc (0:180:3); [name path=ap,dashed] (-3,0) -- (3,3.46); (60:3) circle (2.5pt); [name intersections=of=cir and ap,by=a,b] (a) circle (2.5pt) (b) circle (2.5pt); [fill=white] (-3,0) circle (1.5pt); [below right] at (0,0) 0; [below] at (3,0) 1; [below left] at (-3,0) A; [above right] at (63:3) P; [above right] at (-2.4,0) /2; [above right] at (0.2,0) ; [|<->|] (-3,-0.4) -- (0,-0.4) node[black,fill=white,midway] 1; [name intersections=of=yaxis and ap,by=T]; [red,<->,line width=1.2pt] (0,0) -- (T) node[black,midway,left] t; (aux1) at ((-3,0)!7pt!90:(T)); (aux2) at ((T)!7pt!-90:(-3,0)); [|<->|] (aux1) -- (aux2) node[midway,sloped,above] 1+t^2; tikzpicture — Figure 6.5.2 Half-angle substitution: $t = \tan\left(\tfrac{\theta}{2}\right) = $ slope of $\overline{AP}$

tikzpicture[>=latex,every node/.style=font=] [->] (0:0.6) arc (0:107:0.6); [->,shift=(-1.5,0)] (0:0.85) arc (0:53.8:0.85); [black!60,line width=1pt,->] (-2.5,0) -- (2.8,0) node[right] x; [name path=yaxis,black!60,line width=1pt,->] (0,0) -- (0,4) node[above] y; [->,black!40] (1.5,0) -- (1.5,4.4) node[black,pos=0.7,right] x=1; [name path=cir,linecolor,line width=1.5pt] (0:1.5) arc (0:180:1.5); [name path=ap,dashed] (1.5,4.1) -- (-1.5,0); [below] at (1.5,0) 1; [below left] at (-1.5,0) A; [|<->|] (-1.5,-0.4) -- (0,-0.4) node[black,fill=white,midway] 1; [name intersections=of=cir and ap,by=d,e] (d) circle (2.5pt) (e) circle (2.5pt); [dashed] (0,0) -- (d); [name intersections=of=yaxis and ap,by=T]; [red,<->,line width=1.2pt] (0,0) -- (T) node[black,pos=0.5,right] t; [fill=white] (-1.5,0) circle (1.5pt); [below right] at (0,0) 0; [above right] at (-1.3,0) /2; [above right] at (0.15,0) ; (aux1) at ((-1.5,0)!11pt!90:(T)); (aux2) at ((T)!11pt!-90:(-1.5,0)); [|<->|] (aux1) -- (aux2) node[midway,sloped,above] 1+t^2; tikzpicture — Figure 6.5.2 Half-angle substitution: $t = \tan\left(\tfrac{\theta}{2}\right) = $ slope of $\overline{AP}$

From geometry you know that the inscribed angle that the line $\overline{AP}$ makes with the $x$-axis is half the measure of the central angle $\theta$. So the slope of $\overline{AP}$ is the tangent of that angle: $\tan\,\frac{1}{2}\theta = \frac{t}{1} = t$, which is measured along the $y$-axis and can take any real value. Each point on the unit circle—except $A$—can be identified with that slope $t$. Figure 6.5.2 shows only positive slopes—reflect the picture about the $x$-axis for negative slopes. The figure shows that

\[ \sin\,\tfrac{1}{2}\theta ~=~ \frac{t}{\sqrt{1+t^2}} \qquad\text{and}\qquad \cos\,\tfrac{1}{2}\theta ~=~ \frac{1}{\sqrt{1+t^2}} \]

so that by the double-angle identities for sine and cosine,

\[ \sin\,\theta ~=~ 2\,\sin\,\tfrac{1}{2}\theta\,\cos\,\tfrac{1}{2}\theta ~=~ 2\,\frac{t}{\sqrt{1+t^2}}\,\frac{1}{\sqrt{1+t^2}} ~=~ \frac{2t}{1+t^2} \]

and

\[ \cos\,\theta ~=~ \cos^2 \tfrac{1}{2}\theta ~-~ \sin^2 \tfrac{1}{2}\theta ~=~ \frac{1}{1+t^2} ~-~ \frac{t^2}{1+t^2} ~=~ \frac{1-t^2}{1+t^2} ~. \]

Since $\theta = 2\,\tan^{-1} \,t$, then

\[ \dtheta ~=~ d\,\left(2\,\tan^{-1} t\right) ~=~ \frac{2\,\dt}{1+t^2} ~. \]

Below is a summary of the substitution:

Half-angle substitution: The substitution $t = \tan\,\tfrac{1}{2}\theta$ yields:

\[ \sin\,\theta ~=~ \frac{2t}{1+t^2} \qquad\qquad \cos\,\theta ~=~ \frac{1-t^2}{1+t^2} \qquad\qquad \dtheta ~=~ \frac{2\,\dt}{1+t^2} \]

The half-angle substitution thus turns rational functions of $\sin\,\theta$ and $\cos\,\theta$ into rational functions of $t$, which can be integrated using partial fractions or another method.

Example 6.29

Evaluate $~\displaystyle\int \frac{\dtheta}{1 \;+\; \sin\,\theta \;+\; \cos\,\theta} $.

Solution: Using $t = \tan\,\tfrac{1}{2}\theta$, the denominator of the integrand is

\[ 1 ~+~ \sin\,\theta ~+~ \cos\,\theta ~=~ \frac{1+t^2}{1+t^2} ~+~ \frac{2t}{1+t^2} ~+~ \frac{1-t^2}{1+t^2} ~=~ \frac{2t + 2}{1+t^2} \]

so that

\begin{align*} \int \frac{\dtheta}{1 \;+\; \sin\,\theta \;+\; \cos\,\theta} ~&=~ \mathop{\mathlarger{\mathlarger{\int}}} \frac{\frac{2\,\dt}{1+t^2}}{\frac{2t + 2}{1+t^2}} ~=~ \int \frac{\dt}{t+1}\\[6pt] &=~ \ln\,\abs{t+1} ~+~ C\\ &=~ \ln\,\Abs{\tan\,\tfrac{1}{2}\theta \;+\;1} ~+~ C \end{align*}

Example 6.30

Evaluate $~\displaystyle\int \frac{\dtheta}{3\,\sin\,\theta \;+\; 4\,\cos\,\theta}~$.

Solution: Using $t = \tan\,\tfrac{1}{2}\theta$, the integral becomes

\begin{align*} \int \frac{\dtheta}{3\,\sin\,\theta \;+\;4\,\cos\,\theta} ~&=~ \mathop{\mathlarger{\mathlarger{\int}}} \frac{\frac{2\,\dt}{1+t^2}}{3\,\frac{2t}{1+t^2} \;+\; 4\,\frac{1-t^2}{1+t^2}} ~=~ \int \frac{-1}{2t^2 - 3t - 2}\,\dt\\[6pt] &=~ \int \frac{-1}{(2t+1)\,(t-2)}\,\dt ~=~ \int \left(\frac{A}{2t+1} ~+~ \frac{B}{t-2}\right)\,\dt \end{align*}

where

\begin{alignat*}{3} \text{coefficient of $t$}&: \quad & A ~+~ 2B ~&=~ 0 \quad\Rightarrow\quad A ~=~ -2B\\ \text{constant term}&: & -2A ~+~ B ~&=~ -1 \quad\Rightarrow\quad 4B ~+~ B ~=~ -1 \quad\Rightarrow\quad B ~=~ -\frac{1}{5} ~~\text{and}~~ A ~=~ \frac{2}{5} \end{alignat*}

Thus,

\begin{align*} \int \frac{\dtheta}{3\,\sin\,\theta \;+\;4\,\cos\,\theta} ~&=~ \int \left(\frac{\frac{2}{5}}{2t+1} ~+~ \frac{-\frac{1}{5}}{t-2}\right)\,\dt ~=~ \frac{1}{5}\,\ln\,\abs{2t+1} ~-~ \frac{1}{5}\,\ln\,\abs{t-2} ~+~ C\\[4pt] &=~ \frac{1}{5}\,\ln\,\Abs{2\,\tan\,\tfrac{1}{2}\theta \;+\; 1} ~-~ \frac{1}{5}\,\ln\,\Abs{\tan\,\tfrac{1}{2}\theta \;-\; 2} ~+~ C \end{align*}

By the half-angle substitution $t = \tan\,\tfrac{1}{2}\theta$,

\[ \frac{\sin\,\theta}{1 \;+\; \cos\,\theta} ~=~ \frac{\dfrac{2t}{1+t^2}}{\dfrac{1+t^2}{1+t^2} + \dfrac{1-t^2}{1+t^2}} ~=~ \frac{\dfrac{2t}{1+t^2}}{\dfrac{2}{1+t^2}} ~=~ t \]

which yields the useful half-angle identities:^[5]

Theorem 6.7

\begin{equation} \tan\,\tfrac{1}{2}\theta ~=~ \frac{\sin\,\theta}{1 \;+\; \cos\,\theta} ~=~ \frac{1 \;-\; \cos\,\theta}{\sin\,\theta} \tag{6.18}\end{equation}

Example 6.31

Evaluate $~\displaystyle\int \frac{\sin\,\theta}{1 \;+\; \cos\,\theta}\,\dtheta~$.

Solution: Though you could use the half-angle substitution $t = \tan\,\tfrac{1}{2}\theta$, it is easier to use the half-angle identity (6.18) directly, since

\[ \int \frac{\sin\,\theta}{1 \;+\; \cos\,\theta}\,\dtheta ~=~ \int \tan\,\tfrac{1}{2}\theta~\dtheta ~=~ 2\,\ln\,\Abs{\sec\,\tfrac{1}{2}\theta} ~+~ C \]

by formula (6.11) in Section 6.3.

Exercises

For Exercises 1-12, evaluate the given integral.

$\displaystyle\int \frac{1 \;-\; 2\,\cos\,\theta}{\sin\,\theta}\;\dtheta$
$\displaystyle\int \frac{\dtheta}{3 \;-\; 5\,\sin\,\theta}$
$\displaystyle\int \frac{\dtheta}{2 \;-\; \sin\,\theta}$
$\displaystyle\int \frac{\dtheta}{4 \;+\; \sin\,\theta}$

4
$\displaystyle\int \frac{\sin\,\theta}{2 \;-\; \sin\,\theta}\;\dtheta$
$\displaystyle\int \frac{\dtheta}{5 \;-\; 3\,\cos\,\theta}$
$\displaystyle\int \frac{\dtheta}{1 \;+\; \sin\,\theta \;-\; \cos\,\theta}$
$\displaystyle\int \frac{\dtheta}{1 \;-\; \sin\,\theta \;+\; \cos\,\theta}$

4
$\displaystyle\int \frac{\cot\,\theta}{1 \;+\; \sin\,\theta}\;\dtheta$
$\displaystyle\int \frac{1 \;-\; \cos\,\theta}{3\,\sin\,\theta}\;\dtheta$
$\displaystyle\int_{-\infty}^{\infty} e^{-x^2/2}\,\dx$
$\displaystyle\int_{-\infty}^{\infty} x^2 \,e^{-x^6}\,\dx$
Consider the integral $~\displaystyle\int \frac{\sin\,\theta}{1 \;+\; \cos\,\theta}\,\dtheta~$ from Example 6.31.
1. Evaluate the integral using the substitution $u=1 + \cos\,\theta$.
2. Evaluate the integral using the half-angle substitution $t = \tan\,\tfrac{1}{2}\theta$.
3. Show that the answers from parts (a) and (b) are equivalent to the result from Example 6.31.

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Evaluate the integral $~\displaystyle\int \frac{\dtheta}{3\,\sin\,\theta \;+\; 4\,\cos\,\theta}~$ from Example 6.30 by noting that

\begin{align*} \int \frac{\dtheta}{3\,\sin\,\theta \;+\; 4\,\cos\,\theta} ~&=~ \int \frac{\dtheta}{5\,\left(\frac{3}{5}\,\sin\,\theta \;+\; \frac{4}{5}\,\cos\,\theta\right)}\\[5pt] &=~ \int \frac{\dtheta}{5\,\left(\cos\,\phi\;\sin\,\theta \;+\; \sin\,\phi\;\cos\,\theta\right)}\\[5pt] &=~ \int \frac{\dtheta}{5\,\sin\,(\theta + \phi)} ~=~ \frac{1}{5}\,\int \csc\,(\theta + \phi)~\dtheta \end{align*}

by the sine addition formula, where $\phi$ is the angle in the right triangle shown above. Complete the integration and show that your answer is equivalent to the result from Example 6.30.

Show directly from the definition of the Beta function that $B(x,y) = B(y,x)$ for all $x > 0$ and $y > 0$.
Show that the Beta function $B(x,y)$ can be written as

\[ B(x,y) ~=~ \int_0^{\pi/2} 2\,\sin^{2x-1}(\theta)~\cos^{2y-1}(\theta)~\dtheta \qquad\text{for all $x > 0$ and $y > 0$.} \]
Use Exercise 16 and formula (6.16) to show that

\[ \int_0^{\pi/2} \sin^{m}\theta~\cos^{n}\theta~\dtheta ~=~ \frac{\Gamma\,\left(\dfrac{m+1}{2}\right) \; \Gamma\,\left(\dfrac{n+1}{2}\right)}{2\,\Gamma\,\left(\dfrac{m+n}{2} + 1\right)} \qquad\text{for all $m > -1$ and $n > -1$.} \]
Use Exercise 28 from Section 6.1, as well as Exercise 17 above, to show that for $m=1$, $2$, $3$, $\ldots$,

\[ \int_0^{\pi/2} \sin^{2m}\theta~\dtheta ~=~ \frac{\sqrt{\pi}\;\Gamma\,\left(m + \frac{1}{2}\right)}{2\,(m!)} \qquad\text{and}\qquad \int_0^{\pi/2} \sin^{2m+1}\theta~\dtheta ~=~ \frac{\sqrt{\pi}\;(m!)}{2\,\Gamma\,\left(m + \frac{3}{2}\right)} ~. \]

2
Show that $~\displaystyle\int_0^{\infty} \dfrac{\ln\,x}{1 + x^2}\,\dx ~=~ 0$.
Show that $~\displaystyle\int_0^{\infty} \dfrac{x^a}{a^x}\,\dx ~=~ \dfrac{\Gamma\,(a+1)}{(\ln\,a)^{a+1}}~$ for $a > 1$.
Use the result from Example 6.28 to show that

\[ \frac{d^{1/2}}{\dx^{1/2}}\,\left(\frac{d^{1/2}}{\dx^{1/2}}\,(x)\right) ~=~ 1 ~=~ \ddx\,(x) ~. \]

2
Calculate $~\dfrac{d^{1/2}}{\dx^{1/2}}\,(c)~$ for all constants $c$.
Calculate $~\dfrac{d^{1/3}}{\dx^{1/3}}\,(x)~$.
Show that $~\displaystyle\int_0^1 \dfrac{1}{\sqrt{1 - x^n}}\,\dx ~=~ \tfrac{1}{n}\,B\left(\tfrac{1}{n},\tfrac{1}{2}\right)~$ for $n \ge 1$.
Show that the Gamma function $\Gamma\,(t)$ can be written as

\[ \Gamma\,(t) ~=~ p^t\,\int_0^{\infty} u^{t-1} \,e^{-pu}~\du \quad\text{for all $t > 0$ and $p > 0$.} \]
Show that the Gamma function $\Gamma\,(t)$ can be written as

\[ \Gamma\,(t) ~=~ \int_0^1 \left(\ln\,\left(\frac{1}{u}\right)\right)^{t-1}\,\du \quad\text{for all $t > 0$.} \]
Using the result from Exercise 27 in Section 6.1 that

\[ \int e^{ax}\,\cos\,bx~\dx ~=~ \frac{e^{ax}\,(a\,\cos\,bx ~+~ b\,\sin\,bx)}{a^2 + b^2} \]

for all constants $a$ and $b \ne 0$, differentiate under the integral sign to show that for all $\alpha > 0$

\[ \int_0^{\infty} x\,e^{-x} \sin\,\alpha x~\dx ~=~ \frac{2 \alpha}{(1 + \alpha^2)^2} ~. \]

Use the Leibniz rule and formula (6.8) from Section 6.3 to show that for all $a > 0$,

\[ \int \frac{\dx}{\sqrt{a^2 + x^2}} ~=~ \ln\;\Abs{x + \sqrt{a^2 + x^2}\,} ~+~ C ~. \]
Use Example 6.27 to show that the Beta function satisfies the relation

\[ B(x,1-x) ~=~ \int_0^1 \,\frac{t^{-x} \;+\; t^{x-1}}{1 + t}\,\dt \quad\text{for all $0 < x < 1$.} \]

(Hint: First use a substitution to show that $\displaystyle\int_0^{\infty} \dfrac{u^{x-1}}{1 + u}\,\du = \displaystyle\int_0^{\infty} \dfrac{t^{-x}}{1 + t}\,\dt$.)
Show that for all $a > -1$,

\[ \int_0^{\pi/2} \frac{\dtheta}{1 \;+\; a\,\sin^2 \theta} ~=~ \frac{\pi}{2\,\sqrt{1+a}} ~. \]

The renowned physicist Richard Feynman (1918-1988) famously lamented that the technique was no longer being taught. See p.72 in Feynman, R.P., Surely You’re Joking, Mr. Feynman!, New York: Bantam Books, 1986. ↩
It can be proved that this is valid when the derivative of the integrand is a continuous function of $\alpha$, which will always be the case in this book. See pp.121-122 in Sokolnikoff, I.S., Advanced Calculus, New York: McGraw-Hill Book Company, Inc., 1939. ↩
See p.18-19 in Rainville, E.D., Special Functions, New York: Chelsea Publishing Company, 1971. ↩
For more details about fractional derivatives, as well as examples of their applications in physics and engineering, see Oldham, K.B. and J. Spanier, The Fractional Calculus, New York: Academic Press, 1974. ↩
For a different derivation, see pp.79-80 in Corral, M., Trigonometry, http://mecmath.net/trig/, 2009. ↩