#### Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 8 maths

$\log \left(\frac{4}{3}\right)$

Hint:

You must know about log properties and using partial fraction.

Given:

$\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{(2+\sin x)+(1+\sin x)} d x$

Explanation:

Let

\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{(2+\sin x)(1+\sin x)} d x \\\\ &I=\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{(1+1+\sin x)(1+\sin x)} d x \end{aligned}

Put

\begin{aligned} &1+\sin x=t \\\\ &\cos x \; d x=d t \end{aligned}

When $x=0$  then $t=1$

When $x=\frac{\pi }{2}$  then $t=2$

\begin{aligned} &=\int_{0}^{2} \frac{d t}{(t+1) t} \\\\ &\frac{1}{(t+1) t}=\frac{A}{t}+\frac{B}{t+1} \end{aligned}

Multiplying by $t(t+1)$

$1=A(t+1)+B(t)$

Putting $t=-1$

$1=A(0)+B(-1)$

$B=-1$

Putting  $t=-1$

\begin{aligned} &1=A(1)+B(0) \\\\ &A=1 \end{aligned}

\begin{aligned} &\frac{1}{(t+1) t}=\frac{1}{t}-\frac{1}{t+1} \\\\ &\int_{0}^{2} \frac{d t}{(t+1) t}=\int_{0}^{2} \frac{1}{t}-\frac{1}{t+1} d t \end{aligned}

\begin{aligned} &=\int_{0}^{2} \frac{1}{t} d t-\int_{0}^{2} \frac{1}{t+1} d t \\\\ &=[\log |t|]_{1}^{2}-[\log |t+1|]_{1}^{2} \\\\ &=[\log 2-\log 1]-[\log 3-\log 2] \end{aligned}

\begin{aligned} &=\log 2-0-\log 3+\log 2 \\\\ &=2 \log 2-\log 3 \\\\ &=\log 2^{2}-\log 3 \\\\ &=\log 4-\log 3 \\\\ &=\log \left(\frac{4}{3}\right) \end{aligned}