The integral _0^ ((x+3) x)/(1+3 ^2 x) d x is equal to

Question

The integral $\int_0^\pi \frac{(x+3) \sin x}{1+3 \cos ^2 x} d x$ is equal to

Accepted Answer

$$\begin{aligned}
& I=\int_0^\pi \frac{(x+3) \sin x}{1+3 \cos ^2 x} d x \quad\text{..... (i)}\
& =\int_0^\pi \frac{(\pi-x+3) \sin (\pi-x)}{1+3 \cos ^2(\pi-x)} d x \
& =\int_0^\pi \frac{(\pi+3) \sin x-x \sin x}{1+3 \cos ^2 x} d x \quad\text{..... (ii)}
\end{aligned}$$

Add (i) \& (ii)

$$2 I=\int_0^\pi \frac{(\pi+6) \sin x}{1+3 \cos ^2 x} d x$$

Let $\cos x=t \Rightarrow-\sin x d x=d t$

$(\pi+6) \int_1^{-1} \frac{-d t}{\left(1+3 t^2\right)}=\left(\frac{\pi+6}{3}\right) \int_{-1}^1 \frac{d t}{t^2+\left(\frac{1}{\sqrt{3}}\right)^2}$

$\Rightarrow \quad 2 I=\left.\left(\frac{\pi+6}{3}\right) \cdot \frac{1}{\left(\frac{1}{\sqrt{3}}\right)} \cdot \tan ^{-1} \frac{t}{\frac{1}{\sqrt{3}}}\right|_{-1} ^1$

$$\begin{aligned}
& \Rightarrow \quad 2 I=\left(\frac{\pi+6}{\sqrt{3}}\right)\left[\tan ^{-1}(\sqrt{3})-\tan ^{-1}(-\sqrt{3})\right] \
&=\left(\frac{\pi+6}{\sqrt{3}}\right) \cdot\left(\frac{\pi}{3}-\left(-\frac{\pi}{3}\right)\right) \
&=\left(\frac{\pi+6}{\sqrt{3}}\right) \cdot \frac{2 \pi}{3} \
& \Rightarrow \quad I=\frac{\pi(\pi+6)}{3 \sqrt{3}}
\end{aligned}$$

Solutions · 1