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Need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.8 question 11

Answers (1)

Answer:

$log\left | cos(\frac{\pi }{4}-x) \right |+C$

Hint:

$sin^{2}x+cos^{2}x=1\; and\; sin\: 2x=2sin\: x\, cos\: x$

Given:

$\int \! \sqrt{\frac{1-sin2x}{1+sin2x}}dx$

Explanation:

$\int \! \sqrt{\frac{sin^{2}x+cos^{2}x-2sin\: x\: cos\: x}{sin^{2}x+cos^{2}x+2sin\: x\: cos\: x}}dx$

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

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

$=\int \! {\frac{sin\: x-cos\: x}{sin\: x+cos\: x}}dx$

$=\int \!\frac{\frac{sin\: x}{cos\: x}-\frac{cos\: x}{cos\: x}}{\frac{sin\: x}{cos\: x}+\frac{cos\: x}{cos\: x}}dx$

$=\int \!\frac{tan\: x-1}{tan\: x+1}dx$

$=\int \!\frac{tan\: x-tan\frac{\pi }{4}}{1+tan\: x\: tan\frac{\pi }{4}}dx$ $[tan\frac{\pi }{4}=1]$

$=\int \! tan(x-\frac{\pi }{4})dx$

$=-\int \! tan(\frac{\pi }{4}-x)dx$

$=-[-log\left | cos(\frac{\pi }{4}-x) \right |+C]$

$=log\left | cos(\frac{\pi }{4}-x) \right |+C$

Posted by

Gurleen Kaur

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