# How do I take the integral of this?

##
∫(x^2)/((16-x^2)^(3/2))dx

I keep on getting the wrong answer. Please help?

∫(x^2)/((16-x^2)^(3/2))dx

I keep on getting the wrong answer. Please help?

##### 1 Answer

#### Explanation:

#intx^2/(16-x^2)^(3/2)dx#

Let's use the trigonometric substitution

#=int(16sin^2theta)/(16-16sin^2theta)^(3/2)(4costhetad theta)#

#=64int(sin^2thetacostheta)/(16^(3/2)(1-sin^2theta)^(3/2))d theta#

Recall that

#=64/64int(sin^2thetacostheta)/cos^3thetad theta#

#=intsin^2theta/cos^2thetad theta#

#=inttan^2thetad theta#

Use the identity

#=int(sec^2theta-1)d theta#

Both of these terms can be integrated easily:

#=tantheta-theta+C#

From our substitution

Also, since

#=x/sqrt(16-x^2)-sin^-1(x/4)+C#