The integral _ ^0(x+3) x1+3 ^2xdx is equal to

Question

The integral ${\int }_{\pi }^{0}\frac{(x+3)\sin x}{1+3{\cos }^{2}x}dx$ is equal to

Accepted Answer

$\begin{gathered} I={\int }_{\pi }^{0}\frac{(x+3)\sin x}{1+3{\cos }^{2}x}dx\, \ ={\int }_{\pi }^{0}\frac{(\pi -x+3)\sin (\pi -x)}{1+3{\cos }^{2}(\pi -x)}dx \ ={\int }_{\pi }^{0}\frac{(\pi +3)\sin x-x\sin x}{1+3{\cos }^{2}x}dx\, \end{gathered}$

Add (i) \& (ii)

$2I={\int }_{\pi }^{0}\frac{(\pi +6)\sin x}{1+3{\cos }^{2}x}dx$

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

$(\pi +6){\int }_{-1}^{1}\frac{-dt}{(1+3{t}^{2})}=(\frac{\pi +6}{3}){\int }_{1}^{-1}\frac{dt}{{t}^{2}+{(\frac{1}{\sqrt{3}})}^{2}}$

$\Rightarrow \, 2I={(\frac{\pi +6}{3})\cdot \frac{1}{(\frac{1}{\sqrt{3}})}\cdot {	an }^{-1}\frac{t}{\frac{1}{\sqrt{3}}}|}_{1}^{-1}$

$\begin{gathered} \Rightarrow \, 2I=(\frac{\pi +6}{\sqrt{3}})[{	an }^{-1}(\sqrt{3})-{	an }^{-1}(-\sqrt{3})] \ =(\frac{\pi +6}{\sqrt{3}})\cdot (\frac{\pi }{3}-(-\frac{\pi }{3})) \ =(\frac{\pi +6}{\sqrt{3}})\cdot \frac{2\pi }{3} \ \Rightarrow \, I=\frac{\pi (\pi +6)}{3\sqrt{3}} \end{gathered}$

Solutions · 1