# ∫ ( 3 Sin X − 2 ) Cos X 13 − Cos 2 X − 7 Sin X D X - Mathematics

Sum
$\int\frac{\left( 3\sin x - 2 \right)\cos x}{13 - \cos^2 x - 7\sin x}dx$

#### Solution

I= $\int\frac{\left( 3\sin x - 2 \right)\cos x}{13 - \cos^2 x - 7\sin x}dx$

=  $\int\frac{\left( 3\sin x - 2 \right)\cos x}{13 -(1 - \ sin ^2 x) - 7\sin x}dx$    (∵  cos^2x =1 - sin^2 x)
$= \int\frac{\left( 3\sin x - 2 \right) \cos x}{\text{ sin^}2 x - 7\sin x + 12}dx$
$= \int\frac{\left( 3\sin x - 2 \right) \cos x}{\text{ sin^}2 x - 4\sin x - 3\text{ sin } x + 12}dx$
$= \int\frac{\left( 3\sin x - 2 \right) \cos x}{\sin x\left( \sin x - 4 \right) - 3\left( \sin x - 4 \right)}dx$
$= \int\frac{\left( 3\sin x - 2 \right)\cos x}{\left( \sin x - 3 \right)\left( \sin x - 4 \right)}dx$

$\text{ Let sin x }= t$
$\Rightarrow \text{ cos x dx }= dt$
$\therefore I = \int\frac{\left( 3t - 2 \right)}{\left( t - 3 \right)\left( t - 4 \right)}dt$

Using partial fraction, we get

$\frac{\left( 3t - 2 \right)}{\left( t - 3 \right)\left( t - 4 \right)} = \frac{A}{\left( t - 3 \right)} + \frac{B}{\left( t - 4 \right)} = \frac{A\left( t - 4 \right) + B\left( t - 3 \right)}{\left( t - 3 \right)\left( t - 4 \right)}$
$\Rightarrow 3t - 2 = (A + B)t - 4A - 3B$

Comparing coefficients, we get

A = - 7 and = 10

So,

$I = - 7\int\frac{1}{\left( t - 3 \right)}dt + 10\int\frac{1}{\left( t - 4 \right)}dt$

$\Rightarrow I = - 7\text{ ln }\left| t - 3 \right| + 10\text{ ln}\left| t - 4 \right| + c$
$\therefore I = - 7\text{ ln }\left| \sin x - 3 \right| + 10 \text{ ln }\left| \sin x - 4 \right| + c$

Concept: Indefinite Integral Problems
Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Exercise 19.19 | Q 14 | Page 104