Question

The value of integral $i{\int }_0^{\pi /2}{e}^t\sin tdt$ will be

• A. $\frac{i}{2}(e^{-\pi /4}+1)$
• B. $\frac{i}{2}(e^{\pi /2}+1)$
• C. $\frac{i}{2}(e^{\pi /4}+1)$
• D. $\frac{i}{2}(e^{-\pi /2}+1)$

Answer 1

The correct option is B $\frac{i}{2}(e^{\pi /2}+1)$

We have find the integral of $i{\int }_0^{\frac{\pi }{2}}{e}^{t}sintdt$

The integral of $e^{t}sint$ is $\int e^{t}sintdt=sint\left(e^t\right)-\int e^{t}costdt=sint\left(e^t\right)-(e^{t}cost+\int e^{t}sintdt)$

$\Rightarrow \int e^{t}sintdt=\frac{e^t}{2}(sint-cost)$

If we apply integral limits from $0$ to $\frac{\pi }{2}$ , we get $\frac{e^{\frac{\pi }{2}}}{2}+\frac{1}{2}=\frac{1}{2}(e^{\frac{\pi }{2}}+1)$

Therefore $i\int e^{t}sintdt=\frac{i}{2}(e^{\frac{\pi }{2}}+1)$

Therefore the correct option is $B$