# Integration using Trig Substitution

I was given the following problem: $$\int\sqrt{1-7w^2}\ dw$$ I used the sin substitution - getting $$w=\frac1{\sqrt{7}}\sin\theta$$. I then needed to change the $$dw$$ to a $$d\theta$$, so I got this: $$dw=\frac1{\sqrt{7}}\cos\theta\ d\theta$$. My new problem looks like this: $$\int\sqrt{1-7(\frac1{\sqrt{7}}\sin\theta)^2}(\frac1{\sqrt{7}}\cos\theta)\ d\theta$$

Continuing, I get: $$\int\sqrt{1-\sin^2\theta}(\frac1{\sqrt{7}}\cos\theta)\ d\theta$$ $$=\int\sqrt{\cos^2\theta}(\frac1{\sqrt{7}}\cos\theta)\ d\theta$$$$=\int\frac1{\sqrt7}|\sin^2\theta|\cos\theta\ d\theta$$ If I now set $$u=\sin\theta$$ I get: $$\frac1{\sqrt7}\int u^2 \ du$$$$=\frac1{3\sqrt7}u^3$$$$=\frac{\sin\theta}{3\sqrt7}+c$$
This is not the correct answer. Why not? Where did I go wrong? What is the proper way to do a problem like this one?

• What do you mean by it "looks wrong"? – Ninad Munshi Sep 17 '19 at 1:01
• To me it doesn't look like I'm doing the right thing – Burt Sep 17 '19 at 1:13
• How can you know that if you don't have a lot of experience with these types of integrals? Experience is everything. Keep going and keep simplifying, don't just give up. – Ninad Munshi Sep 17 '19 at 1:15

Let us consider the general case where $$a > 0$$, $$b > 0$$, $$a^{2}-b^{2}x^{2} \geq 0$$ and $$a/b \leq 1$$: \begin{align*} \int\sqrt{a^{2} - b^{2}x^{2}}\mathrm{d}x \end{align*}

According to the substitution $$\displaystyle x = \frac{a\sin(\theta)}{b}$$, we get $$\displaystyle\mathrm{d}x = \frac{a\cos(\theta)}{b}\mathrm{d}\theta$$. Thus we have \begin{align*} \int\sqrt{a^{2}-b^{2}x^{2}}\mathrm{d}x & = a\int\sqrt1 - \left(\frac{bx}{a}\right)^{2}} = \frac{a^{2}}{b}\int\sqrt{1 - \sin^{2}(\theta)}\cos(\theta)\mathrm{d}\theta\\\\ & = \frac{a^{2}}{b}\int\cos^{2}(\theta)\mathrm{d}\theta = \frac{a^{2}}{b}\int\frac{\cos(2\theta) + 1}{2}\mathrm{d}\theta\\\\ & = \frac{a^{2}}{b}\left[\frac{\sin(2\theta)}{4} + \frac{\theta}{2}\right] = \left[\frac{x\sqrt{a^{2}-b^{2}x^{2}}}{2} + \frac{a^{2}\arcsin\left(\displaystyle\frac{bx}{a}\right)}{2b}\right] \end{align*

In your case, $$a = 1$$ and $$b = \sqrt{7}$$.

It may be more desirable to integrate directly without substitution.

$$I= \int \sqrt{1-x^2}dx = x\sqrt{1-x^2} + \int \frac{x^2}{\sqrt{1-x^2}}dx$$

$$=x\sqrt{1-x^2} -I + \int \frac{dx}{\sqrt{1-x^2}}=x\sqrt{1-x^2} -I + \sin^{-1}x+C$$

Thus,

$$I = \frac 12 \left( x\sqrt{1-x^2}+ \sin^{-1}x \right)+C$$

With $$x=\sqrt{7}w$$, the original integral,

$$\int\sqrt{1-7w^2}\ dw=\frac{1}{\sqrt{7}}I =\frac{1}{2\sqrt{7}} \left( \sqrt{7}w\sqrt{1-7w^2}+ \sin^{-1}(\sqrt{7}w)\right) +C$$

• Integration by parts...well played, sir! – bjcolby15 Sep 17 '19 at 22:01

$$\int\sqrt{1-7w^2}\ dw$$

$$w=\frac{\sin(x)}{\sqrt{7}}$$

$$dw=\frac{\cos(x)}{\sqrt{7}} dx$$

$$\int\sqrt{1-7w^2}\ dw$$=$$\int\sqrt{1-7(\frac{\sin(x)}{\sqrt{7}})^2} \frac{\cos(x)}{\sqrt{7}}\ dx$$=$$\int\sqrt{1-\sin^2(x)} \frac{\cos(x)}{\sqrt{7}}\ dx$$=$$\int \frac{\cos^2(x)}{\sqrt{7}}\ dx$$=$$\frac{\cos(x)\sin(x)+x}{2\sqrt{7}}$$.

Then use $$\sqrt{1-\sin^2x}=\cos(x)$$ to put it in terms of $$w$$ if you have to

• The point is that you put it back in terms of the original variable? Up till that point was I correct? – Burt Sep 19 '19 at 3:44