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Question Number 64335 by Chi Mes Try last updated on 16/Jul/19

∫_( 0) ^π x sin x cos^4 x dx =

$$\underset{\:\mathrm{0}} {\overset{\pi} {\int}}\:\:{x}\:\mathrm{sin}\:{x}\:\mathrm{cos}^{\mathrm{4}} {x}\:{dx}\:= \\ $$

Answered by Tanmay chaudhury last updated on 17/Jul/19

I=∫_0 ^π (π−x)sin^ (π−x){cos(π−x)}^4 dx =∫_0 ^π (π−x)sinxcos^4 xdx 2I=∫_0 ^π πsinxcos^4 xdx ((2I)/π)=(−1)∫_0 ^π cos^4 xd(cosx) ((−2I)/π)=∣((cos^5 x)/5)∣_0 ^π ((−2I)/π)=((−1−1)/5) I=(π/5)

$${I}=\int_{\mathrm{0}} ^{\pi} \left(\pi−{x}\right){sin}^{} \left(\pi−{x}\right)\left\{{cos}\left(\pi−{x}\right)\right\}^{\mathrm{4}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi} \left(\pi−{x}\right){sinxcos}^{\mathrm{4}} {xdx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\pi} \pi{sinxcos}^{\mathrm{4}} {xdx} \\ $$$$\frac{\mathrm{2}{I}}{\pi}=\left(−\mathrm{1}\right)\int_{\mathrm{0}} ^{\pi} {cos}^{\mathrm{4}} {xd}\left({cosx}\right) \\ $$$$\frac{−\mathrm{2}{I}}{\pi}=\mid\frac{{cos}^{\mathrm{5}} {x}}{\mathrm{5}}\mid_{\mathrm{0}} ^{\pi} \\ $$$$\frac{−\mathrm{2}{I}}{\pi}=\frac{−\mathrm{1}−\mathrm{1}}{\mathrm{5}} \\ $$$${I}=\frac{\pi}{\mathrm{5}} \\ $$

Commented by Chi Mes Try last updated on 17/Jul/19

wow fnx boss♯♯

$${wow}\:\:{fnx}\:\:{boss}\sharp\sharp \\ $$

Commented by Tanmay chaudhury last updated on 17/Jul/19

thank you sir...

$${thank}\:{you}\:{sir}... \\ $$

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