The question is to find
$\displaystyle \int \frac {\sin (2x)}{1+\cos^2x}.$
Can anyone help me? I need all the steps, because I need to understand what to do. Thank you.
The question is to find
$\displaystyle \int \frac {\sin (2x)}{1+\cos^2x}.$
Can anyone help me? I need all the steps, because I need to understand what to do. Thank you.
You can use the identity: $ \sin(2x)=2\sin(x)\cos(x). $ Then use a $u$-substitution with $u=1+\cos^2(x)$.
By the double angle formula, $\sin(2x) = 2 \sin(x)\cos(x)$
$\int \frac{\sin(2x)}{1 + \cos^2(x)}dx = \int \frac{2\sin(x)\cos(x)}{1+\cos^2(x)}dx$
Let $u = 1 + \cos^2(x)$ $du = -2\sin(x)\cos(x) dx$
so...substituting, we get: $\int \frac{2\sin(x)\cos(x)}{1+\cos^2(x)}dx = \int -\frac{1}{u} du$
Can you take it from here?
Integrate with respect to $u$, then "back" substitute $u = 1 + \cos^2(x)$ into the result.
Note that $\sin(2x)=2\sin(x)\cos(x)$
Therefore the problem reduces to finding the integral: $\int \frac {2\sin(x)\cos(x)}{1+\cos^2(x)}dx=-\log(1+\cos^2(x))+C$