InterviewSolution
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InterviewSolution
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`int_0^pi(xtanx)/(tanx+secx).dx=[pi(pi-2)]/2` |
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Answer» `I=int_0^pi (xtanx)/(tanx+secx)dx` `=int_0^pi (xsinx/cosx)/(sinx/cosx+1/cosx)dx` `I=int_0^pi (xsinx)/(1+sinx)-(1)` `I=int_0^pi((pi-x)sin(pi-x))/(1+sin(pi-x)` `I=int_0^pi((pi-x)sinx)/(1+sinx)-(2)` `I=int_0^pi[(xsinx)/(1+sinx)+((pi-x)sinx)/(1+sinx)]dx` `2I=int_0^pi[(xsinx+pisinx-xsinx)/(1+sinx)]dx` `2I=pi int_0^pi sinx/(1+sinx)dx` `I=pi/2 int_0^pi sinx/(1+sinx)dx` `[secpi-sec0]-int_0^pi sec^2x dx+int_0^pi 1dx` `-1-1-[tanx]_0^pi+(x)_0^pi` `-2+pi` `pi-2` `I=pi/2(pi-2)`. |
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