How can I evaluate the following integral?
I=∫π/20cosx2−sin2xdx
I tried it with Wolfram Alpha, it gave me a numerical solution: 0.785398.
Although I immediately know that it is equal to π/4, I fail to obtain the answer with pen and paper.
I tried to use substitution u=tanx, but I failed because the upper limit of the integral is π/2 and tanπ/2 is undefined.
So how are we going to evaluate this integral? Thanks.
Answer
Hint:
Knowing that sin2x=2sinxcosx and sin2x+cos2x=1. The integral can be expressed as
I=∫π/20cosx1+(sinx−cosx)2 dx
then use substitution x↦π2−x, we have
I=∫π/20sinx1+(sinx−cosx)2 dx
Add the two I's and let u=sinx−cosx.
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