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Evaluate: `int_0^(pi/2) sin^2x/(sinx+cosx)dx` |
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Answer» Let `I=int(sin^(2)xdx)/(sinx+cosx)`………………….1 `:.I=int_(0)^(pi//2)(sin^(2)(1/2pi-x)dx)/("sin"(1/2pi-x)+"cos"(1/2pi-x))` `=int_(0)^(pi//2)(cos^(2)xdx)/(cosx+sinx)` ………..2 Now adding 1 and 2 we get `2I=int_(0)^(pi//2)((sin^(2)x+cos^(2)x)dx)/(sinx+cosx)` or `I=1/2int_(0)^(pi//2)(dx)/(sinx+cosx)` `=1/(2sqrt(2))int_(0)^(pi//2)(dx)/(sin(x+1/4pi))` `=(1/(2sqrt(2)))int_(0)^(pi//2)cosec(x+1/4)dx` `=(1/(2sqrt(2)))[log{cosec(x+1/4pi)-cot(x+1/4pi)}]_(0)^(pi//2)` `=(1//2sqrt(2))[log{cosec(1/2pi+1/4pi)-cot(1/2pi+1/4pi)}` `-log{cosec(1/4pi)-cot(1/4pi)}]` `=(1//2sqrt(2)[log{sec(1/4pi)+tan(1/4pi)}-log(sqrt(2)-1)]` `=(1//2sqrt(2))[log(sqrt(2)+1)-log(sqrt(2)-1)]` `=(1/(2sqrt(2)))log((sqrt(2)+1)/(sqrt(2)-1))` |
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