Definite Integration Question 16
Question 16 - 31 January - Shift 1
The value of $\int _{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2+3 \sin x)}{\sin x(1+\cos x)} d x$ is equal to
(1) $\frac{7}{2}-\sqrt{3}-\log _{e} \sqrt{3}$
(2) $-2+3 \sqrt{3}+\log _{e} \sqrt{3}$
(3) $\frac{10}{3}-\sqrt{3}+\log _{e} \sqrt{3}$
(4) $\frac{10}{3}-\sqrt{3}-\log _{e} \sqrt{3}$
Show Answer
Answer: (3)
Solution:
Formula: Pythagorean Identities, Integration by substitution, Properties of definite integral
$I = \int _{\pi / 3}^{\pi / 2}(\frac{2+3 \sin x}{\sin x(1+\cos x)}) d x=2 \int _{\pi / 3}^{\pi / 2} \frac{d x}{\sin x+\sin x \cos x}+ 3 \int _{\pi / 3}^{\pi / 2} \frac{dx}{1+\cos x}$
$I = 2I_1 + 3I_2 $, where
$I_1 = \int _{\pi / 3}^{\pi / 2} \frac{d x}{1+\cos x}=\int _{\pi / 3}^{\pi / 2} \frac{1-\cos x}{\sin ^{2} x} d x$
$I_1=\int _{\pi / 3}^{\pi / 2}(cosec^{2} x-\cot x cosec x) d x$
$I_1=(cosec x-\cot x) \int _{\pi / 3}^{\pi / 2}=(1)-(\frac{2}{\sqrt{3}}-\frac{1}{\sqrt{3}})$
$I_1=1-\frac{1}{\sqrt{3}}$
and
$I_2 = \int _{\pi / 3}^{\pi / 2} \frac{d x}{\sin x(1+\cos x)}$
$I_2 = \int _{\pi / 3}^{\pi / 2} \frac{d x}{\frac{2tan\frac{x}{2}}{1+tan^2(\frac{x}{2})}(1+\frac{ 1-tan^2(\frac{x}{2})}{1+tan^2(\frac{x}{2})})}$
$I_2=\int \frac{(1+\tan ^{2} x / 2) \sec ^{2} x / 2 d x}{4 \tan x / 2}$
Substitute $\tan x / 2=t \Rightarrow \sec ^{2} \frac{x}{2} \times \frac{1}{2} dx=dt$
$I_2=\frac{1}{2} \int(\frac{1+t^{2}}{t}) dt=\frac{1}{2}[\ell nt+\frac{t^{2}}{2}] _{\frac{1}{\sqrt{3}}}^{1}$
$I_2=\frac{1}{2}[(0+\frac{1}{2})-(\ln \frac{1}{\sqrt{3}}+\frac{1}{6})]=(\frac{1}{3}+\ln \sqrt{3}) \frac{1}{2}$
$I_2=(\frac{1}{6}+\frac{1}{2} \ln \sqrt{3})$
Now, $I = 2I_1 + 3I_2 $
$I = 2(\frac{1}{6}+\frac{1}{2} \ln \sqrt{3})+3(1-\frac{1}{\sqrt{3}})$
$I =\frac{1}{3}+\ln \sqrt{3}+3-\sqrt{3}$
$I=\frac{10}{3}+\ln \sqrt{3}-\sqrt{3}$
