Evaluate: ∫ (2 sin2Φ-cosΦ)/(6-cos^2 Φ-4 sin Φ)dΦ

Question

Evaluate:\(\int\cfrac{(2sin\,2Φ-cosΦ)}{6-cos^2\,Φ-4\,sin\,Φ}dΦ\)

∫ (2 sin2Φ-cosΦ)/(6-cos² Φ-4 sin Φ)dΦ

Answer 1

To find:\(\int\cfrac{(2sin\,2Φ-cosΦ)}{6-cos^2\,Φ-4\,sin\,Φ}dΦ\)

Formula Used:

1. sec² x = 1 + tan² x

2.\(\int\cfrac{1}{1+x^2}dx=tan^{-1}x+c\)

3. sin 2x = 2 sin x cos x

Rewriting the given equation,

Let y = sin ϕ 

dy = cos ϕ dϕ

Substituting in the original equation,

\(\Rightarrow\int\cfrac{4y-1}{y^2-4y+5}dy\)

Using partial fraction,

4y – 1 = A (2y - 4) + B 

Equating the coefficients of y, 

4 = 2A 

A = 2 

Also, -1 = -4A + B 

B = 7 

Substituting in (1),

⇒ 2 log |y² – 4y + 5| + 7 tan^-1(y – 2) + C

But y = sin ϕ

⇒ 2 log |sin²ϕ – 4 sin ϕ + 5| + 7 tan ^-1(sin ϕ – 2) + C

Therefore,