#### Explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.8 question 52 maths

$\frac{1}{4}[cos\, ecx-log\left | sec\, x+tan\, x \right |]+C$

Hint:

$Put \; \; 1=sin^{2}x+cos^{2}x$

Given:

$\int \! \frac{1}{cos3x-cos\, x}dx$

Explanation:

$\int \! \frac{sin^{2}x+cos^{2}x}{cos3x-cos\, x}dx$

$=\int \! \frac{sin^{2}x+cos^{2}x}{4cos^{3}x-3cos\, x-cos\, x}dx$            $cos3x=4cos^{3}x-3cos\, x$

$=\int \! \frac{sin^{2}x+cos^{2}x}{4cos^{3}x-4cos\, x}dx$

$=\int \! \frac{sin^{2}x+cos^{2}x}{4cos\, x(cos^{2}x-1)}dx$

$=-\int \! \frac{sin^{2}x+cos^{2}x}{4cos\, x(sin^{2}x)}dx$                $sin^{2}x+cos^{2}x=1$

$=-\int \! \frac{sin^{2}x}{4cos\, x\, sin^{2}x}dx+(-\int \! \frac{cos^{2}x}{4cos\, x\, sin^{2}x}dx)$

$=-\frac{1}{4}\int \! \frac{1}{cos\, x}dx-\frac{1}{4}\int \! \frac{cos\, x}{ sin^{2}x}dx$

$=-\frac{1}{4}\int \! sec\, xdx-\frac{1}{4}\int \! cot\, x\, cos\, e\, cxdx$

$=-\frac{1}{4}log\left | sec\, x+tan\, x \right |-\frac{1}{4}(-cos\, e\, cx)+C$

$=\frac{1}{4}[cos\, ecx-log\left | sec\, x+tan\, x \right |]+C$