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?

  • $\begingroup$ What do you mean by it "looks wrong"? $\endgroup$ – Ninad Munshi Sep 17 '19 at 1:01
  • $\begingroup$ To me it doesn't look like I'm doing the right thing $\endgroup$ – Burt Sep 17 '19 at 1:13
  • 1
    $\begingroup$ 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. $\endgroup$ – 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\sqrt{\displaystyle 1 - \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.

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

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


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

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

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

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

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


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

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

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

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