Question Number 13623 by Tinkutara last updated on 21/May/17
$$\mathrm{Evaluate}:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} {e}^{{x}} \:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}\:+\:\frac{{x}}{\mathrm{2}}\right){dx} \\ $$
Answered by ajfour last updated on 21/May/17
![I=∫_0 ^( 2π) e^x cos ((π/4)+(x/2))dx =[e^(2π) (−(1/(√2)))−(1/(√2))]+(1/2)∫_0 ^( 2π) e^x sin ((π/4)+(x/2))dx =−(1/(√2))(e^(2π) +1)+((1/2))[−(1/(√2))(e^(2π) +1)] −(1/4)∫_0 ^( 2π) e^x cos ((π/4)+(x/2))dx ⇒ I=−(3/(2(√2)))(e^(2π) +1)−(I/4) ⇒ I=−((3(√2))/5)(e^(2π) +1) .](Q13631.png)
$${I}=\int_{\mathrm{0}} ^{\:\:\mathrm{2}\pi} {e}^{{x}} \mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}+\frac{{x}}{\mathrm{2}}\right){dx} \\ $$$$\:\:=\left[{e}^{\mathrm{2}\pi} \left(−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right]+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\:\mathrm{2}\pi} {e}^{{x}} \mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}+\frac{{x}}{\mathrm{2}}\right){dx} \\ $$$$\:=−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\left({e}^{\mathrm{2}\pi} +\mathrm{1}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\left[−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\left({e}^{\mathrm{2}\pi} +\mathrm{1}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\:\mathrm{2}\pi} {e}^{{x}} \mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}+\frac{{x}}{\mathrm{2}}\right){dx} \\ $$$$\Rightarrow\:{I}=−\frac{\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{2}}}\left({e}^{\mathrm{2}\pi} +\mathrm{1}\right)−\frac{{I}}{\mathrm{4}} \\ $$$$\Rightarrow\:\boldsymbol{{I}}=−\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{5}}\left({e}^{\mathrm{2}\pi} +\mathrm{1}\right)\:. \\ $$
Commented by ajfour last updated on 22/May/17
$${correct}\:{answer}\:{please}\:? \\ $$
Commented by Tinkutara last updated on 22/May/17
$$\mathrm{Thanks}\:\mathrm{Sir}!\:\mathrm{Your}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{right}. \\ $$