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?

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