# ∫ ( 3 Sin X − 2 ) Cos X 5 − Cos 2 X − 4 Sin X D X - Mathematics

Sum
$\int\frac{\left( 3 \sin x - 2 \right) \cos x}{5 - \cos^2 x - 4 \sin x} dx$

#### Solution

 ∫   {( 3 sin x -2 ) cos x   dx}/{5 cos^2 x -4 sin x}
 ∫   {( 3 sin x -2 ) cos x   dx}/{5 -(1 - sin^2)-4 sin x}
 ∫   {( 3 sin x -2 ) cos x   dx}/{sin^2 x -4  sin x + 4}

\text{ Let sin x }= t
$\Rightarrow \text{ cos x dx} = dt$
$\int\frac{\left( 3t - 2 \right) dt}{t^2 - 4t + 4}$
$3t - 2 = A\frac{d}{dx}\left( t^2 - 4t + 4 \right) + B$
$3t - 2 = A \left( 2t - 4 \right) + B$
$3t - 2 = \left( 2 A \right) t + B - 4 A$

Comparing the Coefficients of like powers of t

$\text{ 2 A }= 3$
$A = \frac{3}{2}$
$\text{ B - 4 A }= - 2$
$B - 4 \times \frac{3}{2} = - 2$
$B = - 2 + 6$
$B = 4$

$3t - 2 = \frac{3}{2} \left( 2t - 4 \right) + 4$
$\therefore \int\frac{\left( 3t - 2 \right) dt}{t^2 - 4t + 4}$
$= \int\left( \frac{\frac{3}{2}\left( 2t - 4 \right) + 4}{t^2 - 4t + 4} \right)dt$
$= \frac{3}{2}\int\left( \frac{2t - 4}{t^2 - 4t + 4} \right)dt + 4\int\frac{dt}{t^2 - 4t + 4}$
  =  3/2    I_1 + 4      I_2  . . . ( 1 )
Where
$I_1 = \int\frac{\left( 2t - 4 \right) dt}{t^2 - 4t + 4}, I_2 = \int\frac{dt}{t^2 - 4t + 4}$
$I_1 = \int\frac{\left( 2t - 4 \right) dt}{t^2 - 4t + 4}$
$\text{ Let t }^2 - 4t + 4 = p$
$\Rightarrow \left( 2t - 4 \right) dt = dp$
$I_1 = \int\frac{\left( 2t - 4 \right) dt}{t^2 - 4t + 4}$
$= \int\frac{dp}{p}$
$= \text{ log }\left| p \right| + C_1$
$= \text{ log }\left| t^2 - 4t + 4 \right| + C_1 . . . \left( 2 \right)$
$I_2 = \int\frac{dt}{t^2 - 4t + 4}$
$I_2 = \int\frac{dt}{\left( t - 2 \right)^2}$
$I_2 = \int \left( t - 2 \right)^{- 2} dt$
$I_2 = \frac{\left( t - 2 \right)^{- 2 + 1}}{- 2 + 1} + C_2$
$I_2 = \frac{- 1}{t - 2} + C_2 . . . \left( 3 \right)$
$\text{ from }\left( 1 \right), \left( 2 \right) \text{ and }\left( 3 \right)$
 ∫   {( 3 sin x -2 ) cos x   dx}/{5 cos^2 x -4 sin x}
$= \frac{3}{2} \text{ log } \left| t^2 - 4t + 4 \right| + 4 \times \frac{- 1}{t - 2} + C_1 + C_2$
$= \frac{3}{2} \text{ log }\left| \sin^2 x - 4 \sin x + 4 \right| + \frac{4}{2 - t} + C \left( \text{ Where C }= C_1 + C_2 \right)$
$= \frac{3}{2}\text{ log }\left| \left( \sin x - 2 \right)^2 \right| + \frac{4}{2 - \sin x} + C$
$= \frac{3}{2} \times 2 \text{ log }\left| \sin x - 2 \right| + \frac{4}{2 - \sin x} + C$
$= 3 \text{ log }\left| 2 - \sin x \right| + \frac{4}{2 - \sin x} + C$

Concept: Indefinite Integral Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Exercise 19.19 | Q 10 | Page 104