# Trigonometric substitution

## $$\sqrt{a^2-x^2}$$.

• Consider the integral $$\int\sqrt{a^2-x^2}\,\mathrm dx$$.
• So far, we have only seen definite integrals with particular boundaries like $\int_0^a\sqrt{a^2-x^2}\,\mathrm dx=\frac14a^2\pi$ in Calculus I, which follows from the geometrical reasoning that the graph $$y=\sqrt{a^2-x^2}$$ over $$[0,a]$$ is the top right quarter of the circle with radius $$a$$ and centre the origin, therefore the area under the graph is one quarter of the area of a disk with radius $$a$$.
• In this section, we'll see that we can get an antiderivative via the substitution $$x=a\sin\theta$$.

## $$\sqrt{a^2-x^2}$$.

• Recall the Substitution rule formula $\int f(u(x))u'(x)\,\mathrm dx=\int f(u)\,\mathrm du.$
• There, we use the substitution $$u=u(x)$$. That is, the new variable $$u$$ is a function of the old variable $$x$$.
• Now we want to use the substitution $$x=x(\theta)=a\sin\theta$$, that is the old variable $$x$$ is a function of the new variable $$\theta$$.
• We also want to make sure that the function $$x(\theta)$$ is invertible, that is one-to-one. We can do this by requiring $$-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$$.
• Correspondingly, here we need to use the so-called inverse substitution: $\int f(x)\,\mathrm dx=\int f(x(\theta))x'(\theta)\,\mathrm d\theta$

## $$\sqrt{a^2-x^2}$$.

• Let's see what happens in the prototypical case $$f(x)=\sqrt{a^2-x^2}$$, $$x(\theta)=a\sin\theta$$. Suppose that $$a>0$$.
• We have $$x'(\theta)=a\cos\theta$$.
• Therefore, the inverse substitution formula gives $\int\sqrt{a^2-x^2}\,\mathrm dx=\int\sqrt{a^2-a^2(\sin\theta)^2}a\cos\theta\,\mathrm d\theta.$
• We can keep going using a trigonometric formula: $=\int\sqrt{a^2(1-(\sin\theta)^2)}a\cos\theta\,\mathrm d\theta=\int\sqrt{a^2(\cos\theta)^2}a\cos\theta\,\mathrm d\theta=\int a|\cos\theta|a\cos\theta\,\mathrm d\theta$
• Note that $$\cos\theta\ge0$$ when $$-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$$. Therefore, we can get rid of the absolute value sign: $=\int a^2(\cos\theta)^2\,\mathrm d\theta=\frac12a^2\int1+\cos2\theta\,\mathrm d\theta=\frac{a^2}{2}\left(\theta+\frac12\sin2\theta\right)=\frac{a^2}{2}\sin^{-1}\frac{\theta}{a}+\frac12x\sqrt{a^2-x^2}.$

## $$\sqrt{a^2-x^2}$$.

• We can use the previous calculation to compute the area of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$
• Note that the curve defined by this equation is not a graph. If we want to solve for $$y$$, we get $$y=\pm b\sqrt{1-\frac{x^2}{a^2}}=\pm\frac{b}{a}\sqrt{a^2-x^2}$$.
• In the textbook, they cut the ellipse in half and calculate the area under the graph $$y=\frac{b}{a}\sqrt{a^2-x^2}$$.
• It will amount to the same to say that the ellipse is the region between the graphs $$y=\frac{b}{a}\sqrt{a^2-x^2}$$ and $$y=-\frac{b}{a}\sqrt{a^2-x^2}$$ over $$[-a,a]$$.
• Therefore, its area can be given by the definite integral $\int_{-a}^a\frac{b}{a}\sqrt{a^2-x^2}-\left(-\sqrt{a^2-x^2}\right)\,\mathrm dx=2\frac{b}{a}\int_{-a}^a\sqrt{a^2-x^2}\,\mathrm dx.$
• We want to use the substitution $$x=a\sin\theta$$. Therefore we need to change the interval of integration from $$-a\le x\le a$$ to $$-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$$.
• Via the exact same calculation, we get $=ab\int_{-\pi/2}^{\pi/2}1+\cos2\theta\,\mathrm d\theta=ab\pi+\frac{b}{2a}(0-0)=\pi ab.$

## $$\sqrt{a^2-x^2}$$.

• $$\int\frac{\sqrt{9-x^2}}{x^2}\,\mathrm dx$$.
• We use the substitution $$x=3\sin\theta$$ to get $=\int\frac{3\cos\theta}{9(\sin\theta)^2}3\cos\theta\,\mathrm d\theta=\int(\cot\theta)^2\,\mathrm d\theta.$
• This is a trigonometric integral of the form $$\int(\cot\theta)^m(\csc\theta)^n\,\mathrm d\theta$$ with even $$n$$, therefore using a trigonometric formula we get $=\int(\csc\theta)^2-1\,\mathrm d\theta=-\cot\theta-\theta+C.$
• Now we need to change the variable back to $$x$$. We have $\cot\theta=\frac{\cos\theta}{\sin\theta}=\frac{\sqrt{9-x^2}/3}{x/3}=\frac{\sqrt{9-x^2}}{x^2},$ therefore the final answer is $\int\frac{\sqrt{9-x^2}}{x^2}\,\mathrm dx=-\frac{\sqrt{9-x^2}}{x^2}-\sin^{-1}\frac{x}{3}+C.$

## $$\sqrt{a^2-x^2}$$.

• $$\int\frac{x}{\sqrt{3-2x-x^2}}\,\mathrm dx$$.
• Here, the additional step is that we need to complete the square under the root sign: $3-2x-x^2=3-(2x+x^2)=3-((x+1)^2-1)=4-(x+1)^2.$
• This shows that we should substitute $$u=x+1$$ first. Then we get $=\int\frac{u-1}{\sqrt{4-u^2}}\,\mathrm du.$
• From here we proceed as usual: subsituting $$u=2\sin\theta$$ gives $=\int\frac{2\sin\theta-1}{2\cos\theta}2\cos\theta\,\mathrm d\theta=\int2\sin\theta-1\,\mathrm d\theta=-2\cos\theta-\theta+C=-\sqrt{4-u^2}-\sin^{-1}\frac{u}{2}+C=-\sqrt{4-(x+1)^2}-\sin^{-1}\frac{x+1}{2}+C.$
• Exercises. 7.3: 4, 5, 8, 10, 25, 26.

## $$\sqrt{x^2+a^2}$$.

• For integrals with $$\sqrt{x^2+a^2}$$, we can make the substitution $$x=a\tan\theta$$ with $$-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$$ and $$a>0$$: $\sqrt{x^2+a^2}=\sqrt{a^2((\tan\theta)^2+1)}=\sqrt{a^2(\sec\theta)^2}=a\sec\theta.$
• $$\int\frac{1}{x^2\sqrt{x^2+4}}\,\mathrm dx$$.
• Making the substitution $$x=2\tan\theta$$, we get $=\int\frac{1}{4(\tan\theta)^22\sec\theta}2(\sec\theta)^2\,\mathrm d\theta=\frac14\frac{\cos\theta}{(\sin\theta)^2}\,\mathrm d\theta.$
• Making the substitution $$u=\sin\theta$$, we get $=\frac14\int u^{-2}\,\mathrm du=-\frac14u^{-1}+C=-\frac{1}{4\sin\theta}+C=-\frac{\csc\theta}{4}+C.$
• To convert this back to $$x$$, we use the following argument. The assignment $$\tan\theta=\frac{x}{2}$$ means that there is a right triangle AOB with $$\measuredangle AOB=\theta$$, $$\overline{AB}=x$$ and $$\overline{OB}=2$$. Therefore, $$\overline{AO}=\sqrt{x^2+4}$$, and thus $\csc\theta=\frac{\overline{AO}}{\overline{AB}}=\frac{\sqrt{x^2+4}}{x}.$
• This shows that the final answer is $\int\frac{1}{x^2\sqrt{x^2+4}}\,\mathrm dx=-\frac{\sqrt{x^2+4}}{4x}+C.$
• Exercises. 7.3: 7, 12, 19, 23.

## $$\sqrt{x^2-a^2}$$.

• For integrals with $$\sqrt{x^2-a^2}$$, we can make the substitution $$x=a\sec\theta$$ with 0<<\$ or $$\pi<\theta<\frac{3\pi}{2}$$ and $$a>0$$: $\sqrt{x^2-a^2}=\sqrt{a^2((\sec\theta)^2-1)}=\sqrt{a^2(\tan\theta)^2}=a\tan\theta.$
• $$\int\frac{\mathrm dx}{\sqrt{x^2-a^2}}$$, $$a>0$$.
• Using the substitution $$x=a\sec\theta$$, we get $=\int\frac{a\tan\theta\sec\theta}{a\tan\theta}\,\mathrm d\theta=\int\sec\theta\,\mathrm d\theta=\ln|\tan\theta+\sec\theta|+C=\ln\left|\frac{\sqrt{x^2-a^2}}{a}+\frac{x}{a}\right|+C.$
• Exercises. 7.3: 8, 9, 16, 27.
• More exercises. 7.3: 34, 35, 37, 40.