Tuesday, March 6, 2012

`int_0^(pi/2)(sinx)^4/(sinx+cosx)dx`

The indefinite integral is complex. I don't want to work to the end, but there is an "industrial" way to convert this integral into an integral of a rational function: the tangent half-angle substitution.


Substitute `t=tan(x/2),` then `sin(x)=(2t)/(1+t^2),` `cos(x)=(1-t^2)/(1+t^2)` and `dx=(2dt)/(1+t^2).` The limits `[0,pi/2]` for `x` become `[0,1]` for `t,` so the integral becomes


`int_0^1 ((2t)/(1+t^2))^4*1/((1+2t-t^2)/(1+t^2))*2/(1+t^2) dt = 32int_0^1 (t^4 dt)/((1+t^2)^4(1+2t-t^2)),`


which may be with the general method for rational functions. The answer is about `0.48.`

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