To find:

Formula Used:

1. sec² x = 1 + tan² x

2.

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,

… (1)

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,