InterviewSolution
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InterviewSolution
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Evaluate `int_0^(pi/4)(sqrt(tanx)+sqrt(cotx))dx` |
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Answer» `I = int (sqrt(sin x/cos x) + sqrt(cos x/sinx) ) dx` `= int( sinx + cosx)/(sqrtcosx sqrtsinx) dx` `= int(sin x + cos x)/(sqrt(cos x sin x))dx` now , `sin x + cosx ` integral is `- cosx + sinx` now, doing `(sin x - cos x)^2` `= sin^2x + cos^2 x - 2 sinx cos x` `(sin x -cos x)^2 = 1- 2 sinxcosx` `2sinxcos x = 1-(sinx - cosx)^2` `sqrt(sinx cosx) = sqrt((1- (sinx-cosx)^2)/2)` putting it in the equation, we get `= sqrt2 int_0 ^(pi/4) (sinx + cos x)/sqrt(1 - (sinx - cosx)^2)dx` let `sinx - cosx = t` `(cos x + sinx)dx = dt` `x=0 ; t= sin0-cos 0 = -1` `x = pi/4 ; t= 0` now, `sqrt2 int_-1^0 dt/sqrt(1-t^2)` `= sqrt2 [ sin^-1t]_-1^0` `= sqrt2 [ sin^-1 0 - sin^-1(-1)] ` `= sqrt2 [ 0 - (-pi/2)]` `= sqrt2 [xx pi/2 = pi xx sqrt2 /2` `= pi/sqrt2` Answer |
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