Question #ccfbd

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Answer 1 · 2015-12-07T15:22:11

I suppose this is #cos(x^2)# instead of #cos^2(x)#

let's

#t = 1-cos(x^2)#

#dt = 2xsin(x^2)dx#

#1/2int2xsin(x^2)(1-cos(x^2))^5dx#

become

#1/2intt^5dt#

#1/12[t^6]+C#

Substitute back

#1/12(1-cos(x^2))^6+C#