Question

If $\int \frac{\cos 2x+\sin 2x}{{(2\cos x-\sin x)}^2}dx=\frac{-A}{1725}x-\frac{2}{5}\log |2\cos x-\sin x|+2\frac{1}{2-\tan x}+C$ then A is equal to

Answer 1

$I=\int \frac{{\cos }^{2}x+\sin 2x}{{(2\cos x-\sin x)}^2}dx$

Divide numerator and denominator by ${\cos }^{2}x$

$I=\int \frac{1+2\tan x}{{(2-\tan x)}^2}dx=\int \frac{{\sec }^{2}x-{\tan }^{2}x+2\tan x}{{(2-\tan x)}^2}dx$

$=\int \frac{{\sec }^{2}x-\tan x(\tan x-2)}{{(2-\tan x)}^2}dx=\int \frac{{\sec }^{2}x}{{(2-\tan x)}^2}dx+\int \frac{\tan x}{2-\tan x}dx=I_1+I_2$

Where

$I_1=\int \frac{{\sec }^{2}x}{{(2-\tan x)}^2}dx$

Put $t=\tan x$

$I_1=\int \frac{dt}{{(2-t)}^2}=\frac{1}{2-t}=\frac{1}{2-\tan x}$
And

$I_2=\int \frac{\tan x}{2-\tan x}dx=\int \frac{\sin x}{2\cos x-\sin x}dx$

Put $\sin x=a(2\cos x-\sin x)+b(-2\sin x-\cos x)$

Equating coefficients of $\sin x$ and $\cos x$, we get

$0=2a-b$ and $1=-a-2b$, on solving we get $a=-1/5$ and $b=-2/5$

$I_2=-\frac{1}{5}\int \frac{2\cos x-\sin x}{2\cos x-\sin x}dx-\frac{2}{5}\int \frac{-2\sin x-\cos x}{2\cos x-\sin x}dx$

$I_2=-\frac{1}{5}x-\frac{2}{5}\log |2\cos x-\sin x|+c$

Therefore

$I=\frac{1}{2-\tan x}-\frac{1}{5}x-\frac{2}{5}\log |2\cos x-\sin x|+c$

Hence $A=345$