Question

Integrate (3sinx-2)cosx/13-cos^2x-7sinx

Answer 1

Please find the attachment

Answer 2

Solution:

→$\int \frac{(3sin x -2)cos x. dx}{13 - cos^2x - 7 sin x}=\int \frac{(3sin x -2)cos x. dx}{13 -1+sin^2x -7 sinx}\\\\=\int \frac{(3sin x -2)cos x. dx}{12 +sin^2x -7 sinx}\\\\,{\text{ put sinx = t,and Differentiating both sides}$

→$\text{ gives cos x dx = dt}\\\text{The given integral becomes}\\\\\int\frac{(3 t -2).dt}{t^2 -7 t +12}\\\\t^2-7t + 12 = (t-4)(t-3)\\\\\frac{(3 t -2)}{(t-4)(t-3)}= \frac{A}{t-4}+\frac{B}{t-3}\\\\3 t -2 = A(t-3) + B (t-4)\\\\ \text{put t=3, and t=4 to obtain B= -7,A=10}$

→$\int\frac{10.dt}{t-4}-\int\frac{7.dt}{t-3}= 10 \log\left | t-4 \right |- 7\log\left | t-3 \right | + k$ where k is any constant.

=$10 \log\left | sinx-4 \right |- 7\log\left | sin x-3 \right | + k$