The value of [_()/(3)^()/(2) ((2+3 x))/( x(1+ x)) d x] is equal to :

Question

The value of $$\int_\limits{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2+3 \sin x)}{\sin x(1+\cos x)} d x$$ is equal to :

Accepted Answer

Let I = $$\int_\limits{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2+3 \sin x)}{\sin x(1+\cos x)} d x$$

$$
=\int\limits_{\pi / 3}^{\pi / 2} \frac{2}{\sin x(1+\cos x)} d x+\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{1+\cos x} d x
$$

$$
\begin{aligned}
=\int\limits_{\pi / 3}^{\pi / 2} \frac{2(1-\cos x)}{\sin x\left(1-\cos ^2 x\right)} & d x +\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{1+\cos x} d x
\end{aligned}
$$

$$
=\int\limits_{\pi / 3}^{\pi / 2} \frac{2(1-\cos x)}{\sin ^3 x} d x+\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{2 \cos ^2 \frac{x}{2}} d x
$$

$$
\[\begin{gathered} =\int\limits_{\pi / 3}^{\pi / 2} \frac{2}{\sin ^3 x} d x-\int\limits_{\pi / 3}^{\pi / 2} 2 \cot x \operatorname{cosec}^2 x d x
+\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{2} \sec ^2 \frac{x}{2} d x \end{gathered}$$
\]

$$
=2 I_1-2 I_2+\frac{3}{2} I_3
$$

Now,

$$
 \begin{aligned}
I_1 & =\int\limits_{\pi / 3}^{\pi / 2} \operatorname{cosec}^3 x d x \\
& =\int\limits_{\pi / 3}^{\pi / 2} \sqrt{1+\cot ^2 x} \operatorname{cosec}^2 x d x
\end{aligned}
$$

Put $\cot x=t \Rightarrow-\operatorname{cosec}^2 x d x=d t$

When, $x=\frac{\pi}{3} \Rightarrow t=\frac{1}{\sqrt{3}} \text { and } x=\frac{\pi}{2}
 \Rightarrow t=0$

$$
\begin{aligned}
\therefore & I_1=-\int\limits_{\frac{1}{\sqrt{3}}}^0 \sqrt{1+t^2} d t \\
= & \int\limits_0^{\frac{1}{\sqrt{3}}} \sqrt{1+t^2} d t \\
= & {\left[\frac{t}{2} \sqrt{1+t^2}+\frac{1}{2} \log \left[t+\sqrt{1+t^2}\right]\right]_0^{\frac{1}{\sqrt{3}}} } \\
= & \frac{1}{3}+\frac{1}{2} \log \sqrt{3}
\end{aligned}
$$

$$
I_2=\int\limits_{\pi / 3}^{\pi / 2} \cot x \operatorname{cosec}^2 x d x
$$

Put $\cot x=t \Rightarrow \operatorname{cosec}^2 x d x=-d t$

When, $x=\frac{\pi}{3}$
$\Rightarrow t=\frac{1}{\sqrt{3}}$ and $x=\frac{\pi}{2} \Rightarrow t=0$

$$
\therefore \begin{aligned}
I_2 & =-\int\limits_{\frac{1}{\sqrt{3}}}^0 t d t \\
& =\int\limits_0^{\frac{1}{\sqrt{3}} t} t d t=\left[\frac{t^2}{2}\right]_0^{\frac{1}{\sqrt{3}}}=\frac{1}{6}
\end{aligned}
$$

$$
\begin{aligned}
& I_3=\int\limits_{\frac{\pi}{3}}^{\frac{\pi}{2}} \sec ^2 \frac{x}{2} d x \\
= & 2\left(\tan \frac{\pi}{4}-\tan \frac{\pi}{6}\right)=2\left(1-\frac{1}{\sqrt{3}}\right)
\end{aligned}
$$

$$
\begin{aligned}
& \therefore \quad I=2 I_1-2 I_2+\frac{3}{2} I_3 \\
& =2\left(\frac{1}{3}+\frac{1}{2} \log \sqrt{3}\right)-2\left(\frac{1}{6}\right) +\frac{3}{2}\left(2\left(1-\frac{1}{\sqrt{3}}\right)\right)
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{2}{3}+\log \sqrt{3}-\frac{1}{3}+3-\sqrt{3} \\
& =\frac{10}{3}-\sqrt{3}+\log _e \sqrt{3}
\end{aligned}
$$

Solutions · 1