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Question Number 52947 by gunawan last updated on 15/Jan/19

∫_( 0) ^π ((x tan x)/(sec x+cos x)) dx =

$$\underset{\:\mathrm{0}} {\overset{\pi} {\int}}\:\:\frac{{x}\:\mathrm{tan}\:{x}}{\mathrm{sec}\:{x}+\mathrm{cos}\:{x}}\:{dx}\:= \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

∫_0 ^π ((x×((sinx)/(cosx)))/((1/(cosx))+cosx))dx ∫_0 ^π ((xsinx)/(1+cos^2 x))dx=∫_0 ^π (((π−x)sin(π−x))/(1+cos^2 (π−x)))dx 2I=∫_0 ^π ((xsinx+(π−x)sinx)/(1+cos^2 x))dx 2I=π∫_0 ^π ((sinx)/(1+cos^2 x))dx 2I=(−π)∫_0 ^π ((d(cosx))/(1+cos^2 x)) −2I=π×∣tan^(−1) (cosx)∣_0 ^π I=(((−π)/2) )×{tan^(−1) (−1)−tan^(−1) (1)} =(((−π))/2)×{−2tan^(−1) (1)} =(π/2)×2×(π/4)=(π^2 /4) pls check...

$$\int_{\mathrm{0}} ^{\pi} \frac{{x}×\frac{{sinx}}{{cosx}}}{\frac{\mathrm{1}}{{cosx}}+{cosx}}{dx} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{{xsinx}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}{dx}=\int_{\mathrm{0}} ^{\pi} \frac{\left(\pi−{x}\right){sin}\left(\pi−{x}\right)}{\mathrm{1}+{cos}^{\mathrm{2}} \left(\pi−{x}\right)}{dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\pi} \frac{{xsinx}+\left(\pi−{x}\right){sinx}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$\mathrm{2}{I}=\pi\int_{\mathrm{0}} ^{\pi} \frac{{sinx}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$\mathrm{2}{I}=\left(−\pi\right)\int_{\mathrm{0}} ^{\pi} \frac{{d}\left({cosx}\right)}{\mathrm{1}+{cos}^{\mathrm{2}} {x}} \\ $$$$−\mathrm{2}{I}=\pi×\mid{tan}^{−\mathrm{1}} \left({cosx}\right)\mid_{\mathrm{0}} ^{\pi} \\ $$$${I}=\left(\frac{−\pi}{\mathrm{2}}\:\right)×\left\{{tan}^{−\mathrm{1}} \left(−\mathrm{1}\right)−{tan}^{−\mathrm{1}} \left(\mathrm{1}\right)\right\} \\ $$$$=\frac{\left(−\pi\right)}{\mathrm{2}}×\left\{−\mathrm{2}{tan}^{−\mathrm{1}} \left(\mathrm{1}\right)\right\} \\ $$$$=\frac{\pi}{\mathrm{2}}×\mathrm{2}×\frac{\pi}{\mathrm{4}}=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:\:{pls}\:{check}... \\ $$

Commented by gunawan last updated on 16/Jan/19

Good bless you Sir

$$\mathrm{Good}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{Sir} \\ $$

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