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Show that: int0^(pi//2)f(sin2x)sinxdx...

Show that: `int_0^(pi//2)f(sin2x)sinxdx=sqrt(2)int_0^(pi//4)f(cos2x)cosxdxdot`

Text Solution

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Let `I=int_(0)^(pi//2)f(cos2x)cosxdx` . . . (i)
`rArr I=int_(0)^(pi//2)f(cos2((pi)/(2)-x))*cos((pi/(2)-x)dx`
[ using `int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx]`
`rArr I=int_(0)^(pi//2)f(cos2x)sinx dx` . . . (ii)
On adding Eqs. (i) and (ii) , we get
`2I=int_(0)^(pi//2)f(cos2x)(sinx+cosx)dx`
`=sqrt(2)int_(0)^(pi//2)f(cos2x)[cos(x-pi//4)] dx`
Put `-x+(pi)/(4)=trArr-dx =dt`
`:. 2I=-sqrt(2)int_(pi//4)^(pi//4)[cos((pi)/(2)-2t)] cos t dt`
`rArr 2I=sqrt(2)int_(-pi//4)^(pi//4)f(sin2 t )cos t dt`
`:. I=sqrt(2)int_(0)^(pi//4)f (sin 2 t)cos t dt`
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