what is the best way to solve $\int\frac{\cos^2x\,\mathrm dx}{\sin^2x+4\sin x\cos x}$?

Question

I have the integral:

$$\int\frac{\cos^2x\,\mathrm dx}{\sin^2x+4\sin x\cos x}$$

of course there is $u = \tan\frac{x}{2}$

But I want to avoid it if possible, also neither $\cos x = u$ nor $\sin x = u$ made the life easier :(

Answer 1

Hint:

Divide numerator & denominator by $\cos^2x$ and set $\tan x=u$

to find $$\int\dfrac{du}{(u^2+1)u(u+4)}$$

Now use Partial fraction decomposition

$$\dfrac1{(u^2+1)u(u+4)}=\dfrac Au+\dfrac B{u+4}+\dfrac{Cu+D}{u^2+1}$$

Answer 2

I must say Lab bhattachargee shew a method that is good, here is my method which may also be used. First we split the numerator using $\sin^2 x + \cos ^2 x = 1$.

$$\begin{align} I &=\int\frac{ dx}{\sin^2x+4\sin x\cos x}-\int\frac{\sin^2x dx}{\sin^2x+4\sin x\cos x}\\ &= \int \frac{\csc^2 x}{1+4 \cot x}dx - \int \frac{\sin x}{\sin x + 4 \cos x}dx \\ &= \frac{-1}{4} \ln|1+4\cot x|- \frac{1}{17}\left(\int \frac{\sin x + 4 \cos x}{\sin x + 4 \cos x} dx -4\int\frac{\cos x - 4 \sin x}{\sin x + 4 \cos x} dx \right) \end{align}$$

You can see why we split the second integral into two parts, in one numerator contains the denominator $(\sin x + 4 \cos x)$ and in other numerator is its derivative $(\cos x - 4 \sin x)$. So we had to solve the equation :

$$\sin x = a(\sin x + 4 \cos x)+b(\cos x + 4 \sin x)$$

to find the coefficients $a,b$ by comparing coefficients of $\sin x$ and $\cos x$.

The final antiderivative is simple to write:

$$\frac{-1}{4} \ln|1+4\cot x|- \frac{1}{17}\left(x -4 \ln|\sin x + 4 \cos x| \right)$$