# Thread: Integrals, substitution, and partial fractions!

1. ## Integrals, substitution, and partial fractions!

I am stuck on a couple questions on my homework due tomorrow...if anyone can help me that would be great! I've been working on them for way too long. Thanks!

#6: evaluate: integral sec(x)/4tan^2(x)
I have narrowed it down to: 1/4integral 1/tan^3(x) ...but I'm not sure if that is right.

#4: if we use the substitution x=2sin(theta), then which of the following integrals is equivalent to: integral (x^2)/sqrt(4-x^2) dx ?

a) integral 4sin^2(theta)
b) integral sin^2(theta)/cos^2(theta)
c) integral 4*[sin^2(theta)]*cos(theta)

2. Originally Posted by amaya
I am stuck on a couple questions on my homework due tomorrow...if anyone can help me that would be great! I've been working on them for way too long. Thanks!

#6: evaluate: integral sec(x)/4tan^2(x)
I have narrowed it down to: 1/4integral 1/tan^3(x) ...but I'm not sure if that is right.

#4: if we use the substitution x=2sin(theta), then which of the following integrals is equivalent to: integral (x^2)/sqrt(4-x^2) dx ?

a) integral 4sin^2(theta)
b) integral sin^2(theta)/cos^2(theta)
c) integral 4*[sin^2(theta)]*cos(theta)
#6 Using the usual definitions, note that the integrand is equivalent to $\displaystyle \frac{\cos x}{\sin^2 x}$. Now make the substitution $\displaystyle u = \sin x$.

#4 $\displaystyle \int \frac{4 \sin^2 \theta}{\sqrt{4 - 4 \sin^2 \theta}} (2 \cos \theta) \, d\theta = \int \frac{4 \sin^2 \theta (2 \cos \theta)}{2 \cos \theta} \, d\theta = .....$

Note: $\displaystyle x = 2 \sin \theta \Rightarrow \frac{dx}{d\theta} = 2 \cos \theta \Rightarrow dx = (2 \cos \theta) \, d\theta$.