Question

1/sin x-sin 2x ,Integrate the given function defined over a proper domain w.r.t. x.

Answer 1

HELLO DEAR,




GIVEN:-
∫dx/(sinx - sin2x)

=> I = ∫dx/sinx(1 - 2cosx)

=> I = ∫sinx.dx/sin²x(1 - 2cosx)

put cosx = t
=> -sinx.dx = dt

therefore,
I = -∫dt/(1 - t²)(1 - 2t)


solving by partial fraction;


-1/(1 - t)(1 + t)(1 - 2t) = A/(1 - t) + B/(1 + t) + C/(1 - 2t)

-1 = A(1 + t)(1 - 2t) + B(1 - t)(1 - 2t) + C(1 - t²)

-1 = A(1 - 2t + t - 2t²) + B(1 - 2t - t + 2t²) + C(1 - t²)

-1 = t²(-2A + 2B - C) + t(-A - 3B) + (A + B + C)

on comparing both side like power of cofficient of x.

-2A + 2B - C = 0 , -A - 3B = 0 , A + B + C = -1

on solving we get,

A = -1/2 , B = -1/6 , C = -4/3


therefore, I = -1/2∫dt/(t - 1) - 1/6∫dt/(1 + t) - 4/3∫dt/(1 - 2t)

=> I = -1/2log|t - 1| - 1/6log|t + 1| + 2/3∫(-2dt)/(1 - 2t)

=> I = -1/2log|t - 1| - 1/6log|t + 1| + 2/3log|1 - 2t| + C.

=> I = -1/2log|cosx - 1| - 1/6log|cosx + 1| + 2/3log|1 - 2cosx| + C.



I HOPE ITS HELP YOU DEAR,
THANKS