I have to evaluate:∫0π2sin⁡xsin⁡x+cos⁡xdx

Question

Jocelyn Harwood · Accepted Answer

Let I denote the integral and consider the substitution u=π2−x. ThenI=∫0π2cos⁡ucos⁡u+sin⁡udu and 2I=∫0π2cos⁡u+sin⁡ucos⁡u+sin⁡udu=π2. Hence I=π4In general, ∫0af(x)dx=∫0af(a−x)dx whenever f is integrable, and ∫0π2cosaxcosax+sinaxdx=∫0π2sinaxcosax+sinax=π4 for a&amp;gt;0

greikkar5bu · Accepted Answer

Note that sin⁡(π2−x)=cos⁡x and cos⁡(π2−x)=sin⁡x. The answer will exploit the symmetry.Break up the original integral into two parts,  from 0 to π4 and from π4 to π2.So our first integral is∫x=0π4sin⁡xsin⁡x+cos⁡xdx (1)For the second integral, make the change of variable u=π2−x. Using the fact that sin⁡x=sin⁡(π2−u)=cos⁡u and cos⁡x=cos⁡(π2−u)=sin⁡u, and the fact that dx=−du, we get after not much work∫u=π40−cos⁡ucos⁡u+sin⁡uduChange the dummy variable of integration variable to the name x. Also, do the integration in the &quot;right&quot; order, 0 to π4. That changes the sign, so our second integral is equal to∫x=0π4cos⁡xcos⁡x+sin⁡xdx (2)Our original integral is the sum of the integrals (1) and (2). Add, and note the beautiful cancellation sin⁡xsin⁡x+cos⁡x+cos⁡xcos⁡x+sin⁡x=1. Thus our original integral is equal to∫0π41dxThis is trivial to compute: the answer is π4

I have to evaluate: \int_0^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx

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