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  4. ∫ ( sin 2 x+ sin 4 x- sin 6 x/1+ cos 2 x+ cos 4 x+ cos 6 x) d x is equal to (where C is indefinite integration constant.)

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Q. $\int \frac{\sin 2 x+\sin 4 x-\sin 6 x}{1+\cos 2 x+\cos 4 x+\cos 6 x} d x$ is equal to (where $C$ is indefinite integration constant.)

Integrals

Solution:

$ I=\int \frac{\sin 2 x+\sin 4 x-\sin 6 x}{1+\cos 2 x+\cos 4 x+\cos 6 x} d x=\int \frac{2 \sin 3 x \cos x-2 \sin 3 x \cos 3 x}{2 \cos ^2 x+2 \cos x \cos 5 x} d x $
$=\int \frac{\sin 3 x(\cos x-\cos 3 x)}{\cos x(\cos x+\cos 5 x)} d x=\int \frac{2 \sin 3 x \sin 2 x \sin x}{2 \cos x \cos 2 x \cos 3 x} d x=\int \tan x \tan 2 x \tan 3 x d x$
$\Theta 3 x=2 x+x \Rightarrow \tan 3 x=\frac{\tan 2 x+\tan x}{1-\tan 2 x \tan x} \Rightarrow \tan 3 x-\tan 3 x \tan 2 x \tan x=\tan 2 x+\tan x $
$\Rightarrow \tan 3 x \tan 2 x \tan x=\tan 3 x-\tan 2 x-\tan x $
$\therefore I=\int(\tan 3 x-\tan 2 x-\tan x) d x=\frac{1}{3} \ln |\sec 3 x|-\frac{1}{2} \ln |\sec 2 x|-\ln |\sec x|+C $