Question

Integration of the following expression \[\int\limits_{0}^{{}^{\pi }/{}_{2}}{{{e}^{x}}}\sin xdx\] is equal to
 $ \begin{align}
  & A.\text{ }\dfrac{1}{2}({{e}^{{}^{\pi }/{}_{2}}}-1) \\
 & B.\text{ }\dfrac{1}{2}({{e}^{{}^{\pi }/{}_{2}}}+1) \\
 & C.\text{ }\dfrac{1}{2}(1-{{e}^{{}^{\pi }/{}_{2}}}) \\
 & D.\text{ }2({{e}^{{}^{\pi }/{}_{2}}}+1) \\
\end{align} $

Answer 1

Hint: In order to evaluate the integration of product of two functions we use integration by part
 $ \int{f(x)\text{g(x)dx = f(x )}\int{g(x)dx}}-\int{\left[ \dfrac{d\text{ f(x)}}{dx}\int{g(x)dx} \right]}\text{ }dx $
Here we select $ f(x) $ and $ g(x) $ in accordance with ILATE rule (INVERSE, LOGARITHM, ALGEBRAIC, TRIGONOMETRIC, EXPONENTIAL) taking first and second function as ILATE sequence
So here we take trigonometric function sinx as first function f(x) and exponential function \[{{e}^{x}}\] as second function g(x). Once we found the integration of given function we put the limit by using Newton-Leibniz formula.
\[\int\limits_{a}^{b}{f(x)dx=\left[ F(x) \right]_{a}^{b}=F(b)-F(a)}\] Where F(x) is one of the antiderivatives of function f(x), i.e $ {F}'(x)\equiv f(x)\text{ (a}\le \text{x}\le \text{b)} $

Complete step-by-step answer:
Here we take first function $ f(x)=\sin x $ and second function $ g(x)={{e}^{x}} $
Now we use here integration by parts in order to find the given integral using ILATE rule so we can write
\[\int{{{e}^{x}}}\sin xdx=\sin x\int{{{e}^{x}}}dx-\int{[\dfrac{d}{dx}}(\sin x)\int{{{e}^{x}}dx]}dx\]
As we know that
  $ \begin{align}
  & \int{{{e}^{x}}}dx={{e}^{x}}+c \\
 & \dfrac{d}{dx}(\sin x)=\cos x \\
\end{align} $
So we can write
\[\Rightarrow \int{{{e}^{x}}}\sin xdx=\sin x({{e}^{x}})-\int{\cos x.{{e}^{x}}}dx\]
Now we get again integration that is product of two function cosx and $ {{e}^{x}} $
So we have to integrate\[\int{\cos x.{{e}^{x}}}dx\]
Here we again use integration by parts taking trigonometric function $ \cos x $ as first function f(x) and exponential function $ {{e}^{x}} $ as second function g(x). So we can write
\[\Rightarrow \int{{{e}^{x}}}\sin xdx=\sin x({{e}^{x}})-\left[ \cos x\int{{{e}^{x}}}dx-\int{[\dfrac{d}{dx}}(\cos x)\int{{{e}^{x}}dx]}dx \right]\]
As we know that
 $ \begin{align}
  & \dfrac{d}{dx}\left( \cos x \right)=-\sin x \\
 & \int{{{e}^{x}}}={{e}^{x}}+c \\
\end{align} $
So we get the above integration as
\[\Rightarrow \int{{{e}^{x}}}\sin xdx=\sin x({{e}^{x}})-\cos x({{e}^{x}})-\int{(\sin x){{e}^{x}}}dx\]
Now suppose $ I=\int{{{e}^{x}}}\sin xdx $
So we can write
\[\Rightarrow I=\sin x({{e}^{x}})-\cos x({{e}^{x}})-I\]
Now here we can transpose right hand I to left hand side. Hence we can write
\[\begin{align}
  & \Rightarrow I+I=\sin x({{e}^{x}})-\cos x({{e}^{x}}) \\
 & \Rightarrow 2I=\sin x({{e}^{x}})-\cos x({{e}^{x}}) \\
\end{align}\]
Dividing both side by 2
\[\Rightarrow \dfrac{2I}{2}=\left( \dfrac{\sin x({{e}^{x}})-\cos x({{e}^{x}})}{2} \right)\]
Hence we can write
 \[\Rightarrow I=\left( \dfrac{\sin x({{e}^{x}})-\cos x({{e}^{x}})}{2} \right)\]
As we assume $ I=\int{{{e}^{x}}}\sin xdx $
So we can write
\[\int{{{e}^{x}}\sin x}dx=\left( \dfrac{\sin x({{e}^{x}})-\cos x({{e}^{x}})}{2} \right)\]
Now we have to put the limit of the integration, as we know that if
\[\begin{align}
  & \int{f(x)dx=F(x)} \\
 & \Rightarrow \int\limits_{a}^{b}{f(x)dx=F(b)-F(a)} \\
\end{align}\]
So we put the given limit here we can write
\[\begin{align}
  & \int\limits_{0}^{\dfrac{\pi }{2}}{{{e}^{x}}}\sin xdx=\left[ \dfrac{1}{2}(\sin x-\cos x){{e}^{x}} \right]_{0}^{\dfrac{\pi }{2}} \\
 & \Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{{{e}^{x}}}\sin xdx=\left[ \dfrac{1}{2}(\sin \dfrac{\pi }{2}-\cos \dfrac{\pi }{2}){{e}^{\dfrac{\pi }{2}}}-\dfrac{1}{2}(\sin 0-\cos 0){{e}^{0}} \right] \\
\end{align}\]
As we know that
  $ \cos \dfrac{\pi }{2}=0,\text{ }\cos 0=1,\sin \dfrac{\pi }{2}=1,\text{ }\sin 0=0,{{e}^{0}}=1 $
So putting all the above value we can write
\[\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{{{e}^{x}}}\sin xdx=\left[ \dfrac{1}{2}(1-0){{e}^{\dfrac{\pi }{2}}}-\dfrac{1}{2}(0-1)1 \right]\]
So we get the value
\[\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{{{e}^{x}}}\sin xdx=\left[ \dfrac{1}{2}{{e}^{\dfrac{\pi }{2}}}+\dfrac{1}{2} \right]\]
Taking $ \dfrac{1}{2} $ outside the bracket we get the final value of integration as
\[\int\limits_{0}^{\dfrac{\pi }{2}}{{{e}^{x}}}\sin xdx=\dfrac{1}{2}\left[ {{e}^{\dfrac{\pi }{2}}}+1 \right]\]
Hence option B is correct.

NOTE: It is important to select the first function and second function so that integration will be easy. For this we follow ILATE rule (INVERSE, LOGARITHM, ALGEBRAIC, TRIGONOMETRIC, And EXPONENTIAL). When we have to find out definite integration we do not put integration constant C, as we do in case of indefinite integration. When we put limits we use Newton-Leibniz formula.