Finding an indefinite integral

[nightingale123]

Homework Statement

Calculate the indefinite integral of the function $\int\frac{3x^3}{\sqrt{1-x^2}}$
my book gives the answer $-(2+x^2)\sqrt{1-x^2}+C$

Homework Equations

The Attempt at a Solution

So I started trying to calculate this indefinite integral by using a substitution $x=\sin(\theta)$

so $dx=\cos{\theta}d\theta$

then I substituted this term into my integral and got

$3\int\frac{\sin^3\theta*\cos\theta*d\theta}{\cos\theta}$

Then I canceled out the $\cos\theta$ and wrote $\sin^2\theta=1-\cos^2\theta$

$3\int\sin\theta*(1-\cos^2\theta*)d\theta$

here I substituted another new variable $\cos\theta=t\Rightarrow dt=-\sin\theta d\theta$ and I substituted that into my integral and got

$3\int(1-t^2)dt$

which is equal to $-3t+t^3+C$
and when I inserted all my renamed variables back into the equation I got
$\int\frac{3x^3}{\sqrt{1-x^2}}=-3\cos{(\sin^{-1}x)}+(\cos({\sin^{-1}x}))^3$
Which is nowhere near the answer that is given in my textbook so I don't even know where I made my mistake
Thanks

 

[fresh_42]

What do you get, if you differentiate both solutions?
How do you know, that they aren't the same?

 

[John Park]

Try putting x=sinθ again and simplifying. Note that you seem to have made a sign error in your last integration.

 

[fresh_42]

Try putting x=sinθ again and simplifying. Note that you seem to have made a sign error in your last integration.

No sign error in the integration result, only in the line before. But that's not the point. You simply have two expressions of possibly the same result.

 

[nightingale123]

Thank you. I completely forgot that I could check my answer by just deriving what I got and checking if it is the same as the equation I started with