First slide
Evaluation of definite integrals
Question

The value of ππcos2x1+axdx,a>0 is

Easy
Solution

Let I=ππcos2x1+axdx

Put x=θ to obtain

I=ππcos2θ1+aθ(1)dθ=ππaxcos2x1+axdx

Adding (1) and (2), we get 

2I=ππcos2xdx=20πcos2xdx=πI=π/2

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