- #1
nightingale123
- 25
- 2
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