1. ## Integrals!

$\int \frac{sin\;3x}{1+cos^2\;x}\;dx$

2. Originally Posted by polymerase
$\int \frac{sin\;3x}{1+cos^2\;x}\;dx$
You can easily show that $\sin 3x = 3\sin x - 4\sin ^3 x.$

Now try to find an arctangent:

$\int {\frac{{\sin 3x}}
{{1 + \cos ^2 x}}\,dx} = \int {\frac{{3\sin x - 4\sin ^3 x}}
{{1 + \cos ^2 x}}\,dx} = \int {\frac{{\left( {8\sin x - 4\sin ^3 x} \right) - 5\sin x}}
{{1 + \cos ^2 x}}\,dx} .$

Note that $8\sin x - 4\sin ^3 x = 4\sin x\left( {2 - \sin ^2 x} \right) = 4\sin x\left( {1 + \cos ^2 x} \right).$

The rest follows.

3. Hello, polymerase!

I found an approach . . . but there must be a better way!

$\int \frac{\sin3x}{1+\cos^2\!x}\,dx$
Multiple-angle identity: . $\sin3\theta \:=\:3\sin\theta - 4\sin^3\!\theta$

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

Long division: . $\int\left(4\sin x + \frac{5\sin x}{\sin^2\!x-2}\right)dx \;=\;\int\left(4\sin x + \frac{5\sin x}{(1-\cos^2\!x) - 2}\right)dx$

. . and we have: . $4\int\sin x\,dx - 5\int\frac{\sin x}{1 + \cos^2\!x}\,dx \;=\;-4\cos x - 5\int\frac{\sin x}{1+\cos^2\!x}\,dx$

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

Substitute: . $-4\cos x + 5\int\frac{du}{1+u^2} \;=\;-4\cos x + 5\arctan u + C$

Back-substitute: . $\boxed{-4\cos x + 5\arctan(\cos x) + C}$