Trig Substitution

Trig substitution reduces certain integrals to integrals of trig functions. The idea is to match the given integral against one of the following trig identities:

$$ \eqalign { 1 - (\sin \theta)^2 &= (\cos \theta)^2\cr 1 + (\tan \theta)^2 &= (\sec \theta)^2\cr (\sec \theta)^2 - 1 &= (\tan \theta)^2\cr } $$

If you don't obtain one of the identities above after substituting, you've probably used the wrong substitution.


Example.

$$\int (4 - x^2)^{3/2}\,dx = \int \left(4 - 4 (\sin \theta)^2\right)^{3/2} (2 \cos \theta)\,d\theta =$$

$$\left[x = 2 \sin \theta,\quad dx = 2 \cos \theta\,d\theta\right]$$

$$\int \left(4 (\cos \theta)^2\right)^{3/2} (2 \cos \theta)\,d\theta = 16 \int (\cos \theta)^4\,d\theta = 16 \int \left((\cos \theta)^2\right)^2\,d\theta =$$

$$16 \int \left(\dfrac{1}{2} (1 + \cos 2\theta)\right)^2\,d\theta = 4 \int \left(1 + 2 \cos 2\theta + (\cos 2\theta)^2\right)\,d\theta =$$

$$4 \int \left(1 + 2 \cos 2\theta + \left(\dfrac{1}{2} (1 + \cos 4\theta)\right)\right)\,d\theta = 4 \left(\theta + \sin 2\theta + \dfrac{1}{2} (\theta + \dfrac{1}{4} \sin 4\theta)\right) + C =$$

$$6\theta + 4 \sin 2\theta + \dfrac{1}{2} \sin 4\theta + C.$$

To "match" the "4" in "$4 - x^2$ ", I had to use $x = 2 \sin \theta$ (since $2^2 = 4$ ). I used the double angle formula to reduce the even powers of cosine.

To put the x's back, I need to express everything in terms of trig functions of $\theta$ (as opposed to $2 \theta$ or $4 \theta$ ). I use the double angle formulas for sine:

$$\sin 2\theta = 2 \sin \theta \cos \theta, \quad \sin 4\theta = 2 \sin 2\theta \cos 2\theta = 2(2 \sin \theta \cos \theta)\left(2 (\cos \theta)^2 - 1\right) = 4\sin \theta \cos \theta \left(2 (\cos \theta)^2 - 1\right).$$

Therefore,

$$\int (4 - x^2)^{3/2}\,dx = 6\theta + 8 \sin \theta \cos \theta + 2 \sin \theta \cos \theta \left(2 (\cos \theta)^2 - 1\right) = 6\theta + 6 \sin \theta \cos \theta + 4 (\cos \theta)^3 \sin \theta.$$

Now draw a right triangle which shows the substitution.

$$\hbox{\epsfysize=1in \epsffile{trigsub1.eps}}$$

The triangle shows $\sin \theta =
   \dfrac{x}{2}$ , and by Pythagoras the third side is $\sqrt{4 - x^2}$ . Therefore,

$$\int (1 - x^2)^{3/2}\,dx = 6 \arcsin \dfrac{x}{2} + \dfrac{3}{2} x \sqrt{4 - x^2} + \dfrac{1}{4} x (4 - x^2)^{3/2} + C.\quad\halmos$$


Example. Compute $\displaystyle \int \dfrac{dx}{\sqrt{25 + x^2}}$ .

$25 + x^2$ looks like $1 + (\tan
   \theta)^2$ , so let $x = 5\tan \theta$ . Then $dx = 5(\sec
   \theta)^2\,d\theta$ , so

$$\int \dfrac{dx}{\sqrt{25 + x^2}} = \int \dfrac{5(\sec \theta)^2\,d\theta} {\sqrt{25 + 25(\tan \theta)^2}} = \int \dfrac{5(\sec \theta)^2\,d\theta}{\sqrt{25(\sec \theta)^2}} = \int \dfrac{5(\sec \theta)^2\,d\theta}{5\sec \theta} = \int \sec \theta\,d\theta =$$

$$\ln |\sec \theta + \tan \theta| + C = \ln \left|\dfrac{\sqrt{25 + x^2}}{5} + \dfrac{x}{5}\right| + C.$$

$$\hbox{\epsfysize=1in \epsffile{trigsub2.eps}}\quad\halmos$$


Example. Compute $\displaystyle \int \dfrac{x\,dx}{\sqrt{25 + x^2}}$ .

This could be done using $x = 5\tan
   \theta$ . But it's easier to do a u-substitution:

$$\int \dfrac{x\,dx}{\sqrt{25 + x^2}} = \int \dfrac{x\cdot \dfrac{du}{2x}}{\sqrt{u}} = \dfrac{1}{2} \int \dfrac{du}{\sqrt{u}} = \sqrt{u} + C = \sqrt{25 + x^2} + C.$$

$$\left[u = 25 + x^2, \quad du = 2x\,dx, \quad dx = \dfrac{du}{2x}\right]\quad\halmos$$


Example. Compute $\displaystyle \int \sqrt{x^2 - 4}\,dx$ .

$x^2 - 4$ looks like $(\sec
   \theta)^2 - 1$ , so let $x = 2\sec \theta$ . Then $dx = 2\sec
   \theta \tan \theta\,d\theta$ , and

$$\int \sqrt{x^2 - 4}\,dx = \int \sqrt{4(\sec \theta)^2 - 4}(2\sec \theta \tan \theta\,d\theta) = \int \sqrt{4(\tan \theta)^2}(2\sec \theta \tan \theta\,d\theta) =$$

$$\int (2\tan \theta)(2\sec \theta \tan \theta\,d\theta) = 4\int \sec \theta (\tan \theta)^2\,d\theta = 4\int \sec \theta\left((\sec \theta)^2 - 1\right)\,d\theta =$$

$$4 \int (\sec \theta)^3\,d\theta - 4\int \sec \theta\,d\theta = 2 \sec \theta \tan \theta + 2\ln |\sec \theta + \tan \theta| - 4\ln |\sec \theta + \tan \theta| + C =$$

$$2 \sec \theta \tan \theta - 2\ln |\sec \theta + \tan \theta| = \dfrac{1}{2}x\sqrt{x^2 - 4} - 2\ln \left|\dfrac{x}{2} + \dfrac{\sqrt{x^2 - 4}}{2}\right| + C.$$

$$\hbox{\epsfysize=1in \epsffile{trigsub3.eps}}\quad\halmos$$


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