int_0^{2pi} e^{x/2} sin left( frac{x}{2} + fr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $I = \int_0^{2\pi} e^{x/2} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) \, dx$. Substitute $t = \frac{x}{2}$,then $dt = \frac{1}{2} dx$,whi

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) Let $I = \int_0^{2\pi} e^{x/2} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) \, dx$. Substitute $t = \frac{x}{2}$,then $dt = \frac{1}{2} dx$,which implies $dx = 2 dt$. When $x = 0$,$t = 0$. When $x = 2\pi$,$t = \pi$. So,$I = \int_0^{\pi} e^t \sin \left( t + \frac{\pi}{4} \right) (2 dt) = 2 \int_0^{\pi} e^t \sin \left( t + \frac{\pi}{4} \right) dt$. Using the standard integral formula $\int e^{at} \sin(bt + c) dt = \frac{e^{at}}{a^2 + b^2} [a \sin(bt + c) - b \cos(bt + c)]$,where $a=1, b=1, c=\frac{\pi}{4}$: $I = 2 \left[ \frac{e^t}{1^2 + 1^2} (\sin(t + \frac{\pi}{4}) - \cos(t + \frac{\pi}{4})) \right]_0^{\pi}$. $I = 2 \left[ \frac{e^t}{2} (\sin(t + \frac{\pi}{4}) - \cos(t + \frac{\pi}{4})) \right]_0^{\pi} = [e^t (\sin(t + \frac{\pi}{4}) - \cos(t + \frac{\pi}{4}))]_0^{\pi}$. At $t = \pi$: $e^{\pi} (\sin(\pi + \frac{\pi}{4}) - \cos(\pi + \frac{\pi}{4})) = e^{\pi} (-\sin \frac{\pi}{4} - (-\cos \frac{\pi}{4})) = e^{\pi} (-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = 0$. At $t = 0$: $e^0 (\sin \frac{\pi}{4} - \cos \frac{\pi}{4}) = 1 (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) = 0$. Thus,$I = 0 - 0 = 0$.

$\int_0^{2\pi} e^{x/2} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) \, dx = $

Similar Questions

The value of the integral $\int \limits_{-\log _{e} 2}^{\log _e 2} e^x \ln \left(e^x+\sqrt{1+e^{2 x}}\right) d x$ is equal to

$\int_{-1}^4 \sqrt{\frac{4-x}{x+1}} \, dx =$

$\int_1^2 \left( \tan ^{-1}\left(\frac{x}{x^2+1}\right)+\tan ^{-1}\left(\frac{x^2+1}{x}\right) \right) d x =$

$\int_{-2}^{2} |[x]| \, dx = $

The value of $\int_{-1}^{2} \frac{|x|}{x} d x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module