Question

What is $\int\limits_{0}^{\pi }{{{e}^{x}}\sin xdx}$ equal to?
A. $\dfrac{{{e}^{\pi }}+1}{2}$
B. $\dfrac{{{e}^{\pi }}-1}{2}$
C. ${{e}^{\pi }}+1$
D. $\dfrac{{{e}^{\pi }}+1}{4}$

Answer 1

Use integration by parts to solve the given integral. By using the ILATE rule, take first function as $\sin x$ and second function as${{e}^{x}}$. Simplify the expression and then apply the required formula. After the simplification apply the limits$(O,\pi )$.

Complete step-by-step answer:
We have been given to solve the integral$\int\limits_{0}^{\pi }{{{e}^{x}}\sin xdx}$.
$\therefore I=\int\limits_{0}^{\pi }{{{e}^{x}}\sin xdx}$.
Let us do partial integration or integration by parts. It is a process to find the integral of a product of functions in terms of the integral of the product of their derivative and antiderivative.
$\therefore \int{uv.dx=u\int{v.dx-\int{u'\left( \int{v.dx} \right)}}}dx$
Here${{I}_{n}}=\int{{{e}^{x}}\sin x}dx$.
We can use the ILATE rule here. In integration by parts we take the product of two functions. The integral of two functions are taken by considering the left term as the first function and the second term as the second function.
ILATE stands for
I - Inverse function,
L - Logarithmic function,
A - Algebraic function,
T - Trigonometric function,
E - Exponential function.
Now, if we have $\int{u(x)v(x)dx}$ , then priority of selection of function u ( x ) decreases from Inverse function to Exponential and value of integration is $\int{u(x)v(x)dx}=u(x)\int{v(x)dx-\int{\left( \dfrac{d}{dx}u(x)\int{v(x)dx} \right)}}dx$
We have a trigonometric and an exponential function. So, let us take the first function as $u=\sin x$ and the second function as $v={{e}^{x}}$ .
$u'=\dfrac{du}{dx}=\dfrac{d}{dx}\sin x=\cos x$ (We know that $\int{{{e}^{x}}dx={{e}^{x}})}$

& I=\sin x\int{{{e}^{x}}}dx-\int{\dfrac{d}{dx}\sin x\left( \int{{{e}^{x}}dx} \right)}dx \\\ & I={{e}^{x}}\sin x-\int{\cos x.{{e}^{x}}.dx} \\\ \end{aligned}$$ $$\begin{aligned} & I={{e}^{x}}\sin x-\int{{{e}^{x}}\cos x.dx} \\\ & I={{e}^{x}}\sin x-\left[ \cos x\int{{{e}^{x}}dx-\int{\dfrac{d}{dx}\cos x\left( \int{{{e}^{x}}dx} \right)}}dx \right] \\\ \end{aligned}$$ $$\begin{aligned} & I={{e}^{x}}\sin x-\left[ {{e}^{x}}\cos x-\int{-\sin x.{{e}^{x}}dx} \right] \\\ & \Rightarrow I={{e}^{x}}\sin x-\left[ {{e}^{x}}\cos x+\int{\sin x.{{e}^{x}}dx} \right] \\\ \end{aligned}$$ $$\int{{{e}^{x}}\sin x.dx={{e}^{x}}\sin x-{{e}^{x}}\cos x-\int{{{e}^{x}}\sin x.dx}}$$ $$\begin{aligned} & \Rightarrow \int{{{e}^{x}}\operatorname{sinxdx}}+\int{{{e}^{x}}\operatorname{sinx}}dx={{e}^{x}}(\sin x-\cos x) \\\ & 2\int{{{e}^{x}}\sin x.dx={{e}^{x}}(\sin x-\cos x)} \\\ & \therefore \int{{{e}^{x}}\sin xdx}=\dfrac{{{e}^{x}}}{2}(\sin x-\cos x)={{I}_{n}} \\\ & \therefore \int\limits_{0}^{\pi }{{{e}^{x}}\sin x.dx}=\left[ \dfrac{{{e}^{x}}}{2}(\sin x-\cos x) \right]_{0}^{\pi } \\\ \end{aligned}$$ Now let us apply the limits$$(0,\pi )$$. $$\begin{aligned} & =\left[ \dfrac{{{e}^{\pi }}}{2}(\sin \pi -\cos \pi ) \right]-\dfrac{{{e}^{0}}}{2}(\sin 0-\cos 0) \\\ & =\dfrac{{{e}^{\pi }}}{2}\left[ 0-(-1) \right]-\dfrac{1}{2}(1-0) \\\ & =\dfrac{{{e}^{\pi }}}{2}(1)-\dfrac{1}{2}(1)=\dfrac{{{e}^{\pi }}-1}{2} \\\ \end{aligned}$$ We have used the values as $$\left[ \begin{aligned} & \sin \pi =0 \\\ & \cos \pi =-1 \\\ & \sin 0=1 \\\ & \cos 0=0 \\\ & {{e}^{0}}=1 \\\ \end{aligned} \right]$$ Hence we got $$\int\limits_{0}^{\pi }{{{e}^{x}}\sin xdx}=\dfrac{{{e}^{\pi }}-1}{2}.$$ Hence, option B is the correct answer. **So, the correct answer is “Option B”.** **Note:** We used partial integration in a question like this. This concept is very important in integration. In this question, take $$u=\sin x$$ and$$v={{e}^{x}}$$. By this you will get the answer easily, but if you take $$u={{e}^{x}}$$ and$$v=\sin x$$, it makes the integration complex. If any trigonometric function comes along with$${{e}^{x}}$$, then always take the trigonometric function first. For integration, the ILATE method must be remembered. Try not to make any calculation mistakes.