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what will be its answer .....
its a bit confusing one....
∫(Sin X2)2.(CosX2)2 dx
Dear Pawan deep,
Ans:- f(x)=∫(sinX²)²(cosX²)²dx
=1/4∫(sin2X²)²dx
=1/16∫(1 - cos4X²)²dx
=1/16∫(1 - 2cos4X²+ cos²4X²)dx
=1/16∫(3/2 - 2cos4X² +1/2cos8X²)
=3/32 X -1/8∫cos4X²dx +1/16 ∫cos8X²
Now the value of the 2nd and the third integral can not be found unless it is a definte integral.
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.
All the best Pawan deep!!!
Regards,
Askiitians Experts
SOUMYAJIT IIT_KHARAGPUR
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