6.3 Trigonometric Substitutions

One of the fundamental formulas in geometry is for the area \(A\) of a circle of radius r: \(A = \pi r^2\). The calculus-based proof of that formula uses a definite integral evaluated by means of a trigonometric substitution, as will now be demonstrated.

Use the circle of radius \(r >0\) centered at the origin \((0,0)\) in the \(xy\)-plane, whose equation is \(x^2 + y^2 = r^2\) (see Figure 6.3.1(a) below).

tikzpicture[>=latex,every node/.style=font=] [black!60,line width=1pt,->] (-2.2,0) -- (2.5,0) node[right] x; [black!60,line width=1pt,->] (0,-1.8) -- (0,2.2) node[above] y; [linecolor,line width=1.5pt] (0,0) circle (1.5); [dashed] (0,0) -- (60:1.5) node[black,midway,above] r; [-latex,dashed] (0:0.6) arc (0:60:0.6); (60:1.5) circle (2.5pt); at (35:0.4) ; [below right] at (0:1.5) r; [below left] at (0:0) 0; [above right] at (60:1.5) (x,y)=(r \, ,r \, ); [right] at (-2.2,1.8) x^2 + y^2 = r^2; tikzpicture — Figure 6.3.1 Circle of radius \(r\)

tikzpicture[>=latex,every node/.style=font=] [fill=fillcolor] (0:1.5) arc (0:180:1.5); [black!60,line width=1pt,->] (-2.2,0) -- (2.5,0) node[right] x; [black!60,line width=1pt,->] (0,0) -- (0,2.2) node[above] y; [linecolor,line width=1.5pt] (0:1.5) arc (0:180:1.5); [below] at (1.5,-0.1) r; [below] at (-1.5,-0.1) -r; [above right] at (1.5,0.1) =0; [above left] at (-1.5,0.1) = ; (0:1.5) circle (2.5pt); (180:1.5) circle (2.5pt); [below] at (0,-0.1) 0; [above right] at (60:1.5) y= r^2 - x^2; tikzpicture — Figure 6.3.1 Circle of radius \(r\)

By symmetry about the \(x\)-axis, the area \(A\) of the circle is twice the area of its upper hemisphere (see Figure 6.3.1(b) above), which is the area under the curve \(y =\sqrt{r^2 - x^2}\):

\[ A ~=~ 2 \, \int_{-r}^{r} \sqrt{r^2 - x^2}~\dx \]

To evaluate this integral, recall from trigonometry that any point \((x,y)\) on the circle can be written as \((x,y)=(r\cos\,\theta,r\sin\,\theta)\), where \(0 \le \theta < 2\pi\) (in radians) is the angle shown in Figure 6.3.1(a). Figure 6.3.1(b) shows that as \(x\) goes from \(x=-r\) to \(x=r\), the angle \(\theta\) goes from \(\theta = \pi\) to \(\theta = 0\). Now substitute \(x=r\,\cos\,\theta\) and \(\dx=-r\,\sin\,\theta\,\dtheta\) into the integral and change the limits of integration from \(x=-r\) and \(x=r\) to \(\theta =\pi\) and \(\theta=0\), respectively:

\begin{align*} A ~&=~ 2 \,\int_{\pi}^{0} \sqrt{r^2 - r^2\,\cos^2 \,\theta}~(-r\,\sin\,\theta)~\dtheta ~=~ -2 \,\int_{0}^{\pi} \sqrt{r^2\,(1 - \cos^2 \,\theta)}~(-r\,\sin\,\theta)~\dtheta\\[6pt] &=~ 2 \,\int_{0}^{\pi} r\,\sqrt{\sin^2\,\theta}~r\,\sin\,\theta~\dtheta ~=~ 2r^2 \,\,\int_{0}^{\pi} \sin^2\,\theta~\dtheta ~=~ \cancel{2}r^2 \,\int_{0}^{\pi} \frac{1 - \cos\,2\theta}{\cancel{2}}~\dtheta\\[6pt] &=~ r^2 \,\left(\theta ~-~ \frac{1}{2}\,\sin\,2\theta\right)~\Biggr|_0^{\pi} ~=~ r^2 \,\left(\pi ~-~ \frac{1}{2}\,\sin\,2\pi ~-~ \left(0 ~-~ \frac{1}{2}\,\sin\,0\right)\right)\\[6pt] &=~ \pi r^2 \quad\checkmark \end{align*}

For an indefinite integral of the general form \(\int \sqrt{a^2 - u^2}\;\du\), the same calculation as above with the substitutions \(u=a \cos\,\theta\) and \(\du=-a \sin\,\theta\,\dtheta\) yields

\begin{align*} \int \sqrt{a^2 - u^2}~\du ~&=~ \int \sqrt{a^2 - a^2\,\cos^2 \,\theta}~(-a\,\sin\,\theta)~\dtheta ~=~ -a^2 \,\int \sin^2\,\theta~\dtheta\\[6pt] &=~ -a^2 \,\int \frac{1 - \cos\,2\theta}{2}~\dtheta ~=~ -\frac{a^2}{2}\,\theta ~+~ \frac{a^2\,\sin\,2\theta}{4} ~+~ C ~, \end{align*}

which is still in terms of \(\theta\). To put this back in terms of \(u\), use \(\theta = \cos^{-1} (\frac{u}{a})\), the double-angle formula \(\sin\,2\theta = 2\,\sin\,\theta\,\cos\,\theta\), and \(\sqrt{a^2 - u^2} = \sqrt{a^2\,\sin^2\,\theta} = a\,\sin\,\theta\). Then

\begin{align*} \int \sqrt{a^2 - u^2}~\du ~&=~ -\frac{a^2}{2}\,\cos^{-1} \left(\frac{u}{a}\right) ~+~ \frac{2a^2\,\sin\,\theta\,\cos\,\theta}{4}\\[6pt] &=~ -\frac{a^2}{2}\,\cos^{-1} \left(\frac{u}{a}\right) ~+~ \frac{(a\,\cos\,\theta)\,(a\,\sin\,\theta)}{2} ~+~ C \end{align*}

which results in the following formula:

Theorem 6.2

\begin{equation} \int \sqrt{a^2 - u^2}~\du ~=~ -\frac{a^2}{2}\,\cos^{-1} \left(\frac{u}{a}\right) ~+~ \frac{1}{2}\,u\,\sqrt{a^2 - u^2} ~+~ C \tag{6.6}\end{equation}

It is left as an exercise to show that the substitution \(u=a \sin\,\theta\) gives:

Theorem 6.3

\begin{equation} \int \sqrt{a^2 - u^2}~\du ~=~ \frac{a^2}{2}\,\sin^{-1} \left(\frac{u}{a}\right) ~+~ \frac{1}{2}\,u\,\sqrt{a^2 - u^2} ~+~ C \tag{6.7}\end{equation}

That these two seemingly different antiderivatives are equivalent follows immediately from the identity \(\sin^{-1} x \,+\, \cos^{-1} x \,=\, \frac{\pi}{2}\) for all \(-1 \le x \le 1\), which shows that the antiderivatives differ by the constant \(\frac{\pi a^2}{4}\) (absorbed in the generic constant \(C\)):

\begin{align*} \frac{a^2}{2}\,\sin^{-1} \left(\frac{u}{a}\right) ~+~ \frac{1}{2}\,u\,\sqrt{a^2 - u^2} ~+~ C ~&=~ \frac{a^2}{2}\,\left(\frac{\pi}{2} ~-~ \cos^{-1} \left(\frac{u}{a}\right)\right) ~+~ \frac{1}{2}\,u\,\sqrt{a^2 - u^2} ~+~ C\\[6pt] &=~ -\frac{a^2}{2}\,\cos^{-1} \left(\frac{u}{a}\right) ~+~ \frac{1}{2}\,u\,\sqrt{a^2 - u^2} ~+~ \underbracket[0.3pt]{C +\frac{\pi a^2}{4}}_{C} \end{align*}

Thus, either substitution—\(u=a \cos\,\theta\) or \(u=a \sin\,\theta\)—can be used when evaluating the integral \(\int \sqrt{a^2 - u^2}\,\du\). The latter choice is sometimes preferred, to avoid the negative sign in \(\du\) and the resulting formula.

Example 6.16

Evaluate \(~\displaystyle\int \sqrt{9-4x^2}~\dx~\).

Solution: The integrand is of the form \(\sqrt{a^2 - u^2}\) with \(a=3\) and \(u=2x\), so that \(\du = 2\dx\). Then \(\dx = \frac{1}{2}\du\) and so:

\begin{align*} \int \sqrt{9-4x^2}~\dx ~&=~ \frac{1}{2}\,\int \sqrt{a^2 - u^2}~\du\\[6pt] &=~ \frac{1}{2}\,\left(\frac{a^2}{2}\,\sin^{-1} \left(\frac{u}{a}\right) ~+~ \frac{1}{2}\,u\,\sqrt{a^2 - u^2}\;\right) ~+~ C\\[6pt] &=~ \frac{9}{4}\,\sin^{-1} \left(\frac{2x}{3}\right) ~+~ \frac{1}{2}\,x\,\sqrt{9 - 4x^2} ~+~ C \end{align*}

In general, when other methods fail, use the table below as a guide for certain types of integrals, making use of the specified substitution and trigonometric identity:

\(\vphantom{\displaystyle\int}\)Integral contains	Substitution	Identity
\(\sqrt{a^2 \,-\, u^2}\vphantom{\displaystyle\int^{x^2}}\)	\(u ~=~ a\,\sin\,\theta\)	\(1 ~-~ \sin^2 \theta ~=~ \cos^2 \theta\)
\(\sqrt{a^2 \,+\, u^2}\vphantom{\displaystyle\int^{x^2}}\)	\(u ~=~ a\,\tan\,\theta\)	\(1 ~+~ \tan^2 \theta ~=~ \sec^2 \theta\)
\(\sqrt{u^2 \,-\, a^2}\vphantom{\displaystyle\int^{x^2}}\)	\(u ~=~ a\,\sec\,\theta\)	\(\sec^2 \theta ~-~ 1 ~=~ \tan^2 \theta\)

For example, the substitution \(u = a\,\tan\,\theta\) leads to the following formula:

Theorem 6.4

\begin{equation} \int \sqrt{a^2 + u^2}~\du ~=~ \frac{1}{2}\,u\,\sqrt{a^2 + u^2} ~+~ \frac{a^2}{2}\;\ln\;\Abs{u + \sqrt{a^2 + u^2}\,} ~+~ C \tag{6.8}\end{equation}

Similarly, the substitution \(u = a\,\sec\,\theta\) yields this formula:

Theorem 6.5

\begin{equation} \int \sqrt{u^2 - a^2}~\du ~=~ \frac{1}{2}\,u\,\sqrt{u^2 - a^2} ~-~ \frac{a^2}{2}\;\ln\;\Abs{u + \sqrt{u^2 - a^2}\,} ~+~ C \tag{6.9}\end{equation}

The proof of each formula requires this result from Example 6.7 in Section 6.1:

Theorem 6.6

\begin{equation} \int \sec^3 \theta~\dtheta ~=~ \frac{1}{2}\,\left(\sec\,\theta\;\tan\,\theta ~+~ \ln\;\abs{\,\sec\,\theta \;+\; \tan\,\theta\,}\right) ~+~ C \tag{6.10}\end{equation}

The above substitutions can be used even if no square roots are present.

Example 6.17

Evaluate \(~\displaystyle\int \frac{\dx}{(1 + x^2)^2}~\).

Solution: Notice that this integral cannot be evaluated by using the Power Formula with the substitution \(u=1+x^2\) (why?). Integration by parts does not look promising, either. So try a trigonometric substitution. The integrand contains a term of the form \(a^2 + u^2\) (with \(a=1\) and \(u=x\)), so use the substitution \(x=\tan\,\theta\). Then \(\dx = \sec^2 \theta\,\dtheta\) and so

\begin{align*} \int \frac{\dx}{(1 + x^2)^2} ~&=~ \int \frac{\sec^2 \theta~\dtheta}{(1 + \tan^2 \theta)^2}\\[6pt] &=~ \int \frac{\sec^2 \theta\,\dtheta}{(\sec^2 \theta)^2}\\[6pt] &=~ \int \frac{\dtheta}{\sec^2 \theta^2}\\[6pt] &=~ \int \cos^2 \theta~\dtheta\\[6pt] &=~ \int \frac{1 + \cos\,2\theta}{2}\,\dtheta\\[6pt] &=~ \frac{\theta}{2} ~+~ \frac{1}{4}\,\sin\,2\theta ~+~ C\\[6pt] &=~ \frac{\theta}{2} ~+~ \frac{1}{2}\,\sin\,\theta\;\cos\,\theta ~+~ C \end{align*}

by the trigonometric double-angle identity \(\sin\,2\theta = 2\,\sin\,\theta\;\cos\,\theta\). [r] tikzpicture[every node/.style=font=] [linecolor,line width=1.5pt] (0,0) -- (2.5,1.5) node[black,midway,above,sloped] 1 + x^2 -- (2.5,0) node[black,midway,right] x -- cycle; [below] at (1.25,0) 1; [linecolor] (2.3,0) -- (2.3,0.2) -- (2.5,0.2); at (0.75,0.2) ; [linecolor] (0:1) arc (0:30.9:1); tikzpicture The simplest way to get expressions for \(\sin\,\theta\) and \(\cos\,\theta\) in terms of \(x\) is to draw a right triangle with an angle \(\theta\) such that \(\tan\,\theta = x = \frac{x}{1}\), as in the drawing on the right. The hypotenuse must then be \(\sqrt{1+x^2}\) (by the Pythagorean Theorem), which makes it easy to read off the values of \(\sin\,\theta\) and \(\cos\,\theta\):

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

Since \(\theta = \tan^{-1} x\), putting the integral back in terms of \(x\) yields:

\begin{align*} \int \frac{\dx}{(1 + x^2)^2} ~&=~ \frac{1}{2}\,\tan^{-1} x ~+~ \frac{1}{2}\,\frac{x}{\sqrt{1+x^2}}\;\frac{1}{\sqrt{1+x^2}} ~+~ C\\[6pt] &=~ \frac{1}{2}\,\tan^{-1} x ~+~ \frac{x}{2\,(1 + x^2)} ~+~ C \end{align*}

Note: An alternative method for getting \(\sin\,\theta\) and \(\cos\,\theta\) in terms of \(x\) would be to put \(\tan\,\theta = x\) in the identity \(\sec^2 \theta = 1 + \tan^2 \theta\) to solve for \(\cos\,\theta\), then use the identity \(\sin^2 \theta = 1 - \cos^2 \theta\) to solve for \(\sin\,\theta\).

By completing the square, quadratic expressions in \(x\) can be put in one of the forms \(a^2 \pm u^2\) or \(u^2 - a^2\), enabling the use of the corresponding trigonometric substitution.

Example 6.18

Evaluate \(~\displaystyle\int \frac{\dx}{(4x^2 + 8x - 5)^{3/2}}~\).

Solution: This integral cannot be evaluated by using the Power Formula, so try a trigonometric substitution. Complete the square on the expression \(4x^2 + 8x - 5\):

\[ 4x^2 ~+~ 8x ~-~ 5 ~=~ 4\,(x^2 + 2x) ~-~ 5 ~=~ 4\,(x^2 + 2x + 1) ~-~ 5 ~-~ 4 ~=~ 4\,(x + 1)^2 ~-~ 9 \]

This expression is now of the form \(u^2 - a^2\) for \(u=2\,(x+1)\) and \(a=3\). Use the substitution \(u=a\,\sec\,\theta\), which means \(2\,(x+1) = 3\,\sec\,\theta\). Then \(2\,\dx = 3\,\sec\,\theta\;\tan\,\theta\;\dtheta\) and so:

\begin{align*} \int \frac{\dx}{(4x^2 + 8x - 5)^{3/2}} ~&=~ \int \frac{\dx}{(4\,(x + 1)^2 ~-~ 9)^{3/2}}\\[6pt] &=~ \frac{3}{2}\,\int \frac{\sec\,\theta\;\tan\,\theta~\dtheta}{(9\,\sec^2 \theta ~-~ 9)^{3/2}} ~=~ \frac{3}{2}\,\int \frac{\sec\,\theta\;\tan\,\theta~\dtheta}{(9\,(\sec^2 \theta ~-~ 1))^{3/2}}\\[6pt] &=~ \frac{1}{18}\,\int \frac{\sec\,\theta\;\tan\,\theta~\dtheta}{\tan^3 \theta} ~=~ \frac{1}{18}\,\int \frac{\sec\,\theta~\dtheta}{\tan^2 \theta} ~=~ \frac{1}{18}\,\int \frac{\cos\,\theta~\dtheta}{\sin^2 \theta}\\[6pt] &=~ \frac{1}{18}\,\int \csc\,\theta\;\cot\,\theta~\dtheta ~=~ -\frac{1}{18}\,\csc\,\theta ~+~ C \end{align*}

[r] tikzpicture[every node/.style=font=] [linecolor,line width=1.5pt] (0,0) -- (2.5,1.5) node[black,midway,above,sloped] 2(x+1) -- (2.5,0) node[black,midway,right] 4x^2 + 8x - 5 -- cycle; [below] at (1.25,0) 3; [linecolor] (2.3,0) -- (2.3,0.2) -- (2.5,0.2); at (0.75,0.2) ; [linecolor] (0:1) arc (0:30.9:1); tikzpicture To get an expression for \(\csc\,\theta\) in terms of \(x\), draw a right triangle with an angle \(\theta\) such that \(\sec\,\theta = \frac{2(x+1)}{3}\), as in the drawing on the right. The side opposite \(\theta\) must then be \(\sqrt{4x^2 + 8x - 5}\) (by the Pythagorean Theorem), and hence:

\[ \csc\,\theta ~=~ \frac{2(x+1)}{\sqrt{4x^2 + 8x - 5}} \]

Putting the integral back in terms of \(x\) yields:

\begin{align*} \int \frac{\dx}{(4x^2 + 8x - 5)^{3/2}} ~&=~ -\frac{1}{18}\,\frac{2(x+1)}{\sqrt{4x^2 + 8x - 5}} ~+~ C\\[6pt] &=~ -\frac{x+1}{9\,\sqrt{4x^2 + 8x - 5}} ~+~ C \end{align*}

Note: Trigonometric identities could have been used to obtain \(\csc\,\theta\) by knowing \(\sec\,\theta\).

The following integrals from Section 5.4 might be helpful for the exercises:

\begin{align} \int\,\tan\,u~\du ~&=~ \ln\;\abs{\sec\,u} ~+~ C\tag{6.11}\\ \int\,\sec\,u~\du ~&=~ \ln\;\abs{\sec\,u ~+~ \tan\,u} ~+~ C\tag{6.12}\\ \int\,\csc\,u~\du ~&=~ -\ln\;\abs{\csc\,u ~+~ \cot\,u} ~+~ C \tag{6.13}\end{align}

Exercises

For Exercises 1-16, evaluate the given integral.

\(\displaystyle\int \sqrt{9 + 4x^2}~\dx\)
\(\displaystyle\int \sqrt{2 - 3x^2}~\dx\)
\(\displaystyle\int \sqrt{4x^2 - 9}~\dx\)
\(\displaystyle\int \sqrt{x^2 + 2x + 10}~\dx\)

4
\(\displaystyle\int \frac{\sqrt{1 - x^2}}{x^2}~\dx\)
\(\displaystyle\int \frac{x^2~\dx}{\sqrt{x^2 - 9}}\vphantom{\displaystyle\int \frac{\sqrt{1 - x^2}}{x^2}}\)
\(\displaystyle\int \frac{\dx}{x\,\sqrt{1 + x^2}}\vphantom{\displaystyle\int \frac{\sqrt{1 - x^2}}{x^2}}\)
\(\displaystyle\int \frac{\dx}{x^2\,\sqrt{a^2 + x^2}}~~~(a > 0)\vphantom{\displaystyle\int \frac{\sqrt{1 - x^2}}{x^2}}\)

4
\(\displaystyle\int \frac{x^3~\dx}{\sqrt{x^2 + 4}}\vphantom{\displaystyle\int \frac{x^2~\dx}{\sqrt{a^2 - x^2}}}\)
\(\displaystyle\int \frac{dx}{(4x^2 - 9)^{3/2}}\vphantom{\displaystyle\int \frac{x^2~\dx}{\sqrt{a^2 - x^2}}}\)
\(\displaystyle\int \frac{dx}{(9 + 4x^2)^2}\vphantom{\displaystyle\int \frac{x^2~\dx}{\sqrt{a^2 - x^2}}}\)
\(\displaystyle\int \frac{x^2~\dx}{\sqrt{a^2 - x^2}}~~~(a > 0)\)

4
\(\displaystyle\int \frac{x^3~\dx}{\sqrt{9 - x^2}}\vphantom{\displaystyle\int \frac{\sqrt{4 - x^2}}{x}}\)
\(\displaystyle\int \frac{\sqrt{4 - x^2}}{x}~\dx\)
\(\displaystyle\int \frac{(x - 4)~\dx}{\sqrt{-9x^2 + 36x - 32}}\vphantom{\displaystyle\int \frac{\sqrt{4 - x^2}}{x}}\)
\(\displaystyle\int \frac{dx}{(4x^2 + 16x + 15)^{3/2}}\vphantom{\displaystyle\int \frac{\sqrt{4 - x^2}}{x}}\)
Prove formula (6.7) directly by using the substitution \(u=a \sin\,\theta\). 2
Prove formula (6.8).
Prove formula (6.9).
Show that using the substitution \(u=a \cot\,\theta\) to evaluate the integral \(\int \sqrt{a^2 + u^2}\,\dx~\) leads to an antiderivative equivalent to the one in formula (6.8).
Show that using the substitution \(u=a \csc\,\theta\) to evaluate the integral \(\int \sqrt{u^2 - a^2}\,\dx~\) leads to an antiderivative equivalent to the one in formula (6.9).

The integrals \(~\displaystyle\int \frac{\dx}{\sqrt{x^2 \pm a^2}}~\) can be evaluated without the use of trigonometric substitutions, by using differentials:
1. For \(u^2 = x^2 \pm a^2\), show that
  
  \[ \frac{\dx}{u} ~=~ \frac{d\,(x+u)}{x+u} ~. \]
2. Integrate both sides of the result from part (a).
Note: In general, many integrals involving \(\sqrt{x^2 \pm a^2}\) can be handled with a similar manipulation of differentials, with varying complexity.
According to Newtonian physics the path of a photon grazing the surface of the Sun should be deflected by the Sun’s gravitational field by an angle \(\theta\), given approximately by

\[ \theta ~=~ \ABS{\frac{2\,GMR}{c^2}\,\int_{\infty}^0 \frac{\dy}{(R^2 + y^2)^{3/2}}} \]

where \(c= 2.998 \times 10^8\) m/s is the speed of light, \(G = 6.67 \times 10^{-11}\) N/m²/kg² is the gravitational constant, \(M = 1.99 \times 10^{30}\) kg is the mass of the Sun, and \(R = 6.96 \times 10^{8}\) m is the radius of the Sun. Show that

\[ \theta ~=~ \frac{2\,GM}{c^2 R} ~=~ 4.24 \times 10^{-6}~\text{radians} ~=~ 2.43 \times 10^{-4}~\text{degrees} ~\approx~ 0.875~\text{seconds of arc,} \]

where 1 second of arc \(= 1/3600\) of 1 degree.^[1]

Albert Einstein published this result in 1911, then showed in 1915 that the true angle should be double that amount, due to the curvature of space. Experiments verified Einstein’s prediction. See pp.69-71 in Serway, R.A., C.J. Moses and C.A. Moyer, Modern Physics, Orlando, FL: Harcourt Brace Jovanovich Publishers, 1989. ↩