Question Number 75082 by Rio Michael last updated on 07/Dec/19
$${find} \\ $$$$\int_{\mathrm{0}} ^{\pi} {e}^{{cosx}} {sinx}\:{dx} \\ $$
Answered by mr W last updated on 07/Dec/19
![∫_0 ^π e^(cosx) sinx dx =−∫_0 ^π e^(cosx) d(cos x) =−[e^(cos x) ]_0 ^π =e−(1/e)](Q75086.png)
$$\int_{\mathrm{0}} ^{\pi} {e}^{{cosx}} {sinx}\:{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\pi} {e}^{{cosx}} {d}\left(\mathrm{cos}\:{x}\right) \\ $$$$=−\left[{e}^{\mathrm{cos}\:{x}} \right]_{\mathrm{0}} ^{\pi} \\ $$$$={e}−\frac{\mathrm{1}}{{e}} \\ $$
Commented by Rio Michael last updated on 07/Dec/19
$${i}\:{don}'{t}\:{understand}\:{step}\:\mathrm{2}\:{sir} \\ $$
Commented by MJS last updated on 07/Dec/19
![it′s short for ∫_0 ^π e^(cos x) sin x dx= [t=cos x → dx=−(dt/(sin x))] =−∫_1 ^(−1) e^t dt=[−e^t ]_1 ^(−1) =[e^t ]_(−1) ^1 =e−(1/e)](Q75094.png)
$$\mathrm{it}'\mathrm{s}\:\mathrm{short}\:\mathrm{for} \\ $$$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\mathrm{e}^{\mathrm{cos}\:{x}} \mathrm{sin}\:{x}\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{cos}\:{x}\:\rightarrow\:{dx}=−\frac{{dt}}{\mathrm{sin}\:{x}}\right] \\ $$$$=−\underset{\mathrm{1}} {\overset{−\mathrm{1}} {\int}}\mathrm{e}^{{t}} {dt}=\left[−\mathrm{e}^{{t}} \right]_{\mathrm{1}} ^{−\mathrm{1}} =\left[\mathrm{e}^{{t}} \right]_{−\mathrm{1}} ^{\mathrm{1}} =\mathrm{e}−\frac{\mathrm{1}}{\mathrm{e}} \\ $$
Commented by Rio Michael last updated on 07/Dec/19
$${thank}\:{you}\:{sir},{thank}\:{you}\:{so}\:{much}\:{sir} \\ $$