Question

Evaluate ${\int }_0^{\pi }{e}^{|cosx|}\left(2\sin \right(\frac{1}{2}\cos x)+3\cos (\frac{1}{2}\cos x\left)\right)\sin xdx$

Answer 1

Let $I={\int }_0^{\pi }{e}^{\left|\cos x\right|}\left(2\sin \left(\frac{1}{2}\cos x\right)+3\cos \left(\frac{1}{2}\cos x\right)\right)\sin xdx$

$\Rightarrow I={\int }_0^{\pi }{e}^{\left|\cos x\right|}.\sin x.2\sin \left(\frac{1}{2}\cos x\right)dx+{\int }_0^{\pi }{e}^{\left|\cos x\right|}.3\cos \left(\frac{1}{2}\cos x\right).\sin xdx$

$\Rightarrow I=I_1+I_2$ ...(i)

Using ${\int }_0^{2a}f\left(x\right)dx=\{\begin{array}{c}0,f(2a-x)=-f\left(x\right)\\ 2{\int }_0^{a}f\left(x\right)dx,f(2a-x)=+f\left(x\right)\end{array}$

Here, $I_1=0$ $[\because f(\pi -x)=-f\left(x\right)]$ ...(ii)

$I_2=6{\int }_0^{\frac{\pi }{2}}{e}^{\cos x}.\sin x.\cos \left(\frac{1}{2}\cos x\right)dx$ $[\because f(\pi -x)=f\left(x\right)]$

Put $\cos x=t\Rightarrow -\sin xdx=dt$

When $x=0\Rightarrow t=1$

When $x=\frac{\pi }{2}\Rightarrow t=0$

$I_2=6{\int }_0^1{e}^t.\cos \left(\frac{t}{2}\right)dt$

Applying integration by parts
$=6{\left[e^t\cos \left(\frac{t}{2}\right)+\frac{1}{2}\int e^t\sin \left(\frac{t}{2}\right)dt\right]}_0^1$

Again applying integration by parts, we get
$=6{\left[e^t\cos \left(\frac{t}{2}\right)+\frac{1}{2}\left(e^t\sin \frac{t}{2}-\int \frac{e^t}{2}\cos \frac{t}{2}dt\right)\right]}_0^1$
$I_2=6{\left[e^t\cos \frac{t}{2}+\frac{1}{2}{e}^t\sin \frac{t}{2}\right]}_0^1-\frac{I_2}{4}$

$I_2+\frac{I_2}{4}=6{\left[e^t\cos \frac{t}{2}+\frac{1}{2}{e}^t\sin \frac{t}{2}\right]}_0^1$

$\Rightarrow I_2=\frac{24}{5}\left(e\cos \left(\frac{1}{2}\right)+\frac{e}{2}\sin \left(\frac{1}{2}\right)-1\right)$ ...(iii)

From eq.(i), we get
$I=\frac{24}{5}\left(e\cos \left(\frac{1}{2}\right)+\frac{e}{2}\sin \left(\frac{1}{2}\right)-1\right)$