1. ## Difficult integral!?!

I have no idea where to go on this one:

using substitution t = sinx evaluate the following integral within the limits pi/2 to pi/6:

4cosx /(3 + (cosx)^2 dx

i thought about using t^2 = s(inx)^2 but get in a right mess!

Can you help?

Thanks D

2. Originally Posted by dojo
I have no idea where to go on this one:

using substitution t = sinx evaluate the following integral within the limits pi/2 to pi/6:

4cosx /(3 + (cosx)^2 dx

i thought about using t^2 = s(inx)^2 but get in a right mess!

Can you help?

Thanks D

Send your question again and this time be careful with the parentheses!

Tonio

3. Hello, dojo!

$\displaystyle \int\frac{4\cos x}{3 + \cos^2\!x}\,dx$

We have: .$\displaystyle 4\int\frac{\cos x\,dx}{3 + (1-\sin^2\!x)} \;=\;4\int\frac{\cos x\,dx}{4-\sin^2\!x}$

Let: .$\displaystyle u \:=\:\sin x \quad\Rightarrow\quad du \:=\:\cos x\,dx$

Substitute: .$\displaystyle 4\int\frac{du}{4-u^2} \;=\;4\int\frac{du}{(2+u)(2-u)}$

Apply Partial Fractions and get: .$\displaystyle 4\int\bigg[\frac{\frac{1}{4}}{2+u} - \frac{\frac{1}{4}}{2-u}\bigg] du$

. . . . . . $\displaystyle =\;\ln|2+u| - \ln|2-u| + C \;=\; \ln\left|\frac{2+u}{2-u}\right| + C$

Back-substitute: .$\displaystyle \ln\left|\frac{2+\sin x}{2 - \sin x}\right| + C$