$\int_{0}^{\frac{\pi }{2}} e^{\sin x} \cos x \; dx$ = ...

 

For finding the value of the integral $\int_{0}^{\frac{\pi }{2}} e^{\sin x} \cos x \; dx$  we are going to use substitution method.

i.e., Put  $\sin x=u$ 

Now by differentiating with respect to x we will get:

\[\begin{align*} \frac{\mathrm{d} (\sin x)}{\mathrm{\! d} x} &= \frac{\mathrm{d} u}{\mathrm{d} x} \Rightarrow \cos x = \frac{\mathrm{d} u}{\mathrm{d} x}\\ & \; \;\; \; \; \; \; \; \; \Rightarrow \cos x \; dx = du \end{align*}\]

Now by substituting  u instead of $\sin x$ we will get

\[\int e^{\sin x} \cos x \; dx = \int e^{u} \; du.\]

Next we have to apply limit for finding the value of integral.We have two methods:

Method1: Applying limit after integrating the function

\[\int e^{u} \; du. = e^{u} = e^{\sin x} (\because definite \; not \; using\; constant C) \]

\[\int_{0}^{\frac{\pi }{2}} e^{\sin x} \cos x\; dx. = \left [e^{\sin x} \right ]_{0}^{\frac{\pi }{2}}\]

\[\begin{align*} \int_{0}^{\frac{\pi }{2}} e^{\sin x} \cos x\; dx. &=\left [e^{\sin x} \right ]_{0}^{\frac{\pi }{2}} \\ &= e^{\sin \frac{\pi }{2}} - e^{0}\\ &= e^{1} - 1 \end{align*}\]

$\displaystyle \therefore \int_{0}^{\frac{\pi }{2}} e^{\sin x} \cos x\; dx. = e - 1$ 

Method2: Changing the limit before evaluating integral.

At first we have to find limits of integration before evaluating it.

i.e,We have to find value of u when $\displaystyle x=0 \; and \; x=\pi /2$ .

$x=0 \; \Rightarrow u= \sin 0=0 $ 

$\\ x=\pi /2 \; \Rightarrow u=\sin \frac{\pi }{2}=1$ 

Hence 

\[\int_{0}^{\frac{\pi }{2}} e^{\sin x} \cos x \; dx =\int_{0}^{1} e^{u} du\]

\[\int_{0}^{1} e^{u} \; du =\left [e^{u} \right ]_{0}^{1}=e^{1} - e^{0}=e-1\]

$\displaystyle \therefore \int_{0}^{\frac{\pi }{2}} e^{\sin x} \cos x\; dx. = e - 1$ 

😀😀😀