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,