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1/sin^4+cos^4 dx.integrate

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1/sin^4+cos^4 dx.integrate

Posted by Ashutosh Jha 7 years ago

CBSE > Class 12 > Mathematics

4 answers

Jashaswi Talukdar 7 years ago

I think I skipped one step, sorry. The second step would be =1/(sin^2x+cos^2x)^2-2sin^2xcos^2x =1/1-2sin^2xcos^2x

S M 7 years ago

Sorry 1/ 1-2sin^2xcos^2x

S M 7 years ago

I didn't understand the first step i think it would be 1- 2sin^2xcos^2x

Jashaswi Talukdar 7 years ago

1/sin^4x+cos^4x =1/1-sin^2x =sec^2x/sec^2x-tan^2x (dividing both numerator and denominator by cos^2x) Since, sec^2x=tan^2x+1 Let tan2x be t So, dt/dx=sec^2x/2 2dt=sec^2xdx so, it will come out as 2dt after putting t in place of tan2x Then after integration of 2dt U will get 2tan2x+c Please comment, if I have done something wrong in this.

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