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Showthat: `int_0^(pi//2)f(sin2x)sinxdx=sqrt(2)int_0^(pi//4)f(cos2x)cosxdxdot` |
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Answer» Let `I=int_(0)^(pi//2)f(sin2x)sinx dx` `=int_(0)^(pi//2)f{"sin"2(1/2pi-x)}sin(1/2 pi-x)dx` `:. I=int_(0)^(pi//2)f(sin2x)cosx dx` Then adding 1 and 2 we have `2I=int_(0)^(pi//2)f(sin2x)(sinx+cosx)dx` `=sqrt(2)int_(0)^(pi//2)f(sin2x)sin(x+(pi)/4)dx` To get `f(cos 2x)` on RHS, we have to substitute `(pi)/2-2theta =2x`. `:. dx=d theta` Also, when `x=0, theta pi//4`. and when `x=pi//2, theta=-pi//4`. `:. 2I=sqrt(2)int_(-pi//4)^(pi//4)f(cos 2theta) cos theta d theta ` `=2sqrtint_(0)^(pi//4) f(cos 2 theta) cos theta d theta` [as `g(theta)=f(cos 2theta) cos theta` is an even function) `:.I=sqrt(2)int_(0)^(pi//4)f(cos2x)cosx dx` |
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