bobhans-I-sin-2x-a-cos-2-x-b-sin-2-x-c-

Question Number 108444 by bobhans last updated on 17/Aug/20

\frac{b o b h a n s}{β ⊝ β}

I = \int \frac{\sin 2 x}{a co s^{2} x + b \sin^{2} x + c}

Answered by john santu last updated on 17/Aug/20

\frac{⊸ J S ⊸}{★}

r e c a l l \to {\begin{cases} \cos^{2} x = \frac{1}{2} + \frac{1}{2} \cos 2 x \\ \sin^{2} x = \frac{1}{2} - \frac{1}{2} \cos 2 x \end{cases}

n e x t t h e i n t e g r a l b e c o m e s

I = \int \frac{\sin 2 x d x}{\frac{1}{2} (a + b) + \frac{1}{2} (a - b) \cos 2 x + c}

I = \int \frac{2 \sin 2 x d x}{(a + b + 2 c) + (a - b) \cos 2 x}

s e t \cos 2 x = j \to - 2 \sin 2 x d x = d j

I = \int \frac{- d j}{(a + b + 2 c) + (a - b) j}

I = \frac{1}{b} \int \frac{d (a + b + 2 c + (a - b) j)}{(a + b + 2 c) + (a - b) j}

I = \frac{1}{b} \ln ∣ a + b + 2 c + (a - b) j ∣ + K

I = \frac{1}{b} \ln ∣ a + b + 2 c + (a - b) \cos 2 x ∣ + K