# ∫ Sin X Cos 2 X Dx - Mathematics

Sum
$\int\frac{\sin x}{\cos 2x} \text{ dx }$

#### Solution

$\text{ Let I} = \int\frac{\sin x}{\text{ cos 2 x}} dx$
$= \int\left( \frac{\sin x}{2 \cos^2 x - 1} \right) dx ...................\left[ \because \cos 2x = 2 \cos^2 x - 1 \right]$
$\text{ Putting cos x = t}$
$\Rightarrow - \text{ sin x dx = dt}$
$\Rightarrow \text{ sin x dx = - dt}$
$\therefore I = \int\frac{- dt}{2 t^2 - 1}$
$= \frac{1}{2}\int\frac{- dt}{t^2 - \frac{1}{2}}$
$= \frac{- 1}{2}\int\frac{dt}{t^2 - \left( \frac{1}{\sqrt{2}} \right)^2}$
 =   -1 / 2  ×  1/ 2 × 1/1\sqrt2    In    |{t -1/\sqrt2}/{t+1/\sqrt2 }| + C      ..... [ ∵ ∫ {1}/{x^2 - a^2} = {1}/{2a}\text{ ln } | {x - a}/{x + a} | + C ]
$= - \frac{1}{2\sqrt{2}} \text{ ln} \left| \frac{\sqrt{2}t - 1}{\sqrt{2}t + 1} \right| + C$
$= - \frac{1}{2\sqrt{2}} \text{ ln }\left| \frac{\sqrt{2} \cos x - 1}{\sqrt{2} \cos x + 1} \right| + C ............\left[ \because t = \cos x \right]$
$= \frac{1}{2\sqrt{2}} \text{ ln} \left| \frac{\sqrt{2} \cos x + 1}{\sqrt{2} \cos x - 1} \right| + C$

Concept: Indefinite Integral Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Revision Excercise | Q 27 | Page 203