Question

1/sin x(3+2cosx) ,Integrate the given function defined over a proper domain w.r.t. x.

Answer 1

HELLO DEAR,


I = $\int$[sinx/sin²x(3 + 2cosx)] dx

I = $\int$[sinx/(1 - cos²x)(3 + 2cosx)]dx

put cosx = t so that -sinx.dx = dt.

therefore, I = $\int$[-dt/(1 - t^2)(3 + 2t)]

solving by partial fraction
-1/(1 - t²)(3 + 2t) = A/(1 - t) + B/(1 + t) + C/(3 + 2t)

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

-1 = A(2t² + 5t + 3) + B(-2t² - t + 3) + C(1 - t²)

-1 = (2A - 2B - C)t² + (5A - B) + (3A + 3B + C)

(2A - 2B - C) = 0 , 5A - B = 0 , 3A + 3B + C = -1

2A - 10B - C = 0 , B = 5/A , 3A + 15A - 8A = -1

-8A = C , B = 5/A , 10A - 1

C = 4/5 , B = -1/2 , A = -1/10


therefore, I = $\int$[-dt/10(1 - t) - dt/2(1 + t) + 4dt/5(3 + 2t)]

I = 1/10 log|t - 1| - 1/2 log|1 + t| + 4/(2*5) log|3 + 2t| + C

I = 1/10 log|cosx - 1| - 1/2 log|cosx + 1| + 2/5 log|3 + 2cosx| + C



I HOPE ITS HELP YOU DEAR,
THANKS