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  • Please solve RD Sharma class 12 Chapter Definite integrals exercise 19.2 question 40 maths textbook solution.

Please solve RD Sharma class 12 Chapter Definite integrals exercise 19.2 question 40 maths textbook solution.

Answers (1)

Answer   : \frac{\pi}{6}

Hint         : use indefinite integral formula and the limits to solve this integral

Given      : \int_{0}^{\frac{\pi}{2}}\frac{\cos ^2 x}{1 +3 \sin ^2 x}

Solution : \int_{0}^{\frac{\pi}{2}}\frac{\cos ^2 x}{1 +3 \sin ^2 x}

on multiplying and dividing by sec4x , we get

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}}\left(\frac{\cos ^{2} x \cdot \sec ^{4} x}{\sec ^{4} x\left(1+3 \sin ^{2} x\right)}\right) d x=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x \cdot \frac{1}{\cos ^{2} x} \cdot \sec ^{2} x}{\sec ^{2} x \cdot \frac{1}{\cos ^{2} x}\left(1+3 \sin ^{2} x\right)} d x \\ &=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{\sec ^{2} x\left(\frac{1}{\cos ^{2} x}+3 \frac{\sin ^{2} x}{\cos ^{2} x}\right)} d x \end{aligned}
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{\sec ^{2} x\left(\sec ^{2} x+3 \tan ^{2} x\right)} d x \\ &=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{\left(1+\tan ^{2} x\right)\left(1+\tan ^{2} x+3 \tan ^{2} x\right)} d x \\ &=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{\left(1+\tan ^{2} x\right)\left(1+4 \tan ^{2} x\right)} d x \end{aligned}
put \tan x=t \Rightarrow \sec^2x dx=dt
when x=0 then t=0 , when x =\frac{\pi}{2} then t=\infty
therefore ,

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{\left(1+\tan ^{2} x\right)\left(1+4 \tan ^{2} x\right)} d x\\ &=\int_{0}^{\infty} \frac{1}{\left(1+t^{2}\right)\left(1+4 t^{2}\right)} d t \end{aligned}

To solve this integral, first we need to find its partial fraction then integrate  it using indefinite integral formula then put the limits to get required answer.
therefore ,

\begin{aligned} &\frac{1}{\left(1+t^{2}\right)\left(1+4 t^{2}\right)}=\frac{A t+B}{\left(1+t^{2}\right)}+\frac{c t+D}{1+4 t^{2}} \\ &\Rightarrow 1=\frac{(A t+B)\left(1+t^{2}\right)\left(1+4 t^{2}\right)}{\left(1+t^{2}\right)}+\frac{(C t+D)\left(1+t^{2}\right)\left(1+4 t^{2}\right)}{\left(1+4 t^{2}\right)} \\ &\Rightarrow 1=(A t+B)\left(1+4 t^{2}\right)+(C t+D)\left(1+t^{2}\right) \\ &\Rightarrow 1=A t+4 A t^{3}+B+4 B t^{2}+C t+C t^{3}+D+D t^{2} \\ &\Rightarrow 1=(4 A+C) t^{3}+(4 B+D) t^{2}+(A+C) t+(B+D) \end{aligned}

Equating the coefficient of t3,t2,t and constant term respectively then 

0=4A+C             … a
0=4B+D            … b
0=A+C               … c
1=B+D             …(d)
-4A=C
-4B=D

Since A= -C=0
 Substracting (d) by (b), we get
4B+D=0
B+D   =13B= -1
B=-13

 1= -13+D ? 1+13=D?3+13=D


Therefore A=C=0, B=-\frac{1}{3} , D=\frac{4}{3}
Therefore  \frac{1}{\left.(1+t^{2}\right)\left(1+4 t^{2}\right)}=\frac{0 . t-\frac{1}{3}}{1+t^{2}}+\frac{0 . t+\frac{4}{3}}{1+4 t^{2}}=\frac{-1}{3\left(1+t^{2}\right)}+\frac{4}{3\left(1+4 t^{2}\right)}
Now (i)?

\begin{aligned} &\int_{0}^{\infty} \frac{1}{\left(1+t^{2}\right)\left(1+4 t^{2}\right)} d t \\ &=\int_{0}^{\infty}\left\{-\frac{1}{3} \frac{1}{\left(1+t^{2}\right)}+\frac{4}{3\left(1+4 t^{2}\right)}\right\} d t \\ &=\frac{-1}{3} \int_{0}^{\infty} \frac{1}{1+t^{2}} d t+\frac{4}{3} \int_{0}^{\infty} \frac{1}{1+4 t^{2}} d t \\ &=\frac{-1}{3} \int_{0}^{\infty} \frac{1}{1+t^{2}} d t+\frac{4}{3} \int_{0}^{\infty} \frac{1}{1+(2 t)^{2}} d t \end{aligned}

put 2t=u ⇒2dt=du ⇒ dt=du2 in second integral, then

\begin{aligned} &\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{1+3 \sin ^{2} x} d x=\frac{-1}{3}\left[\tan ^{-1} t\right]_{0}^{\infty}+\frac{4}{3} \int_{0}^{\infty} \frac{1}{1+u^{2}} \frac{d u}{2} \\ &=\frac{-1}{3}\left[\tan ^{-1} \infty-\tan ^{-1} 0\right]+\frac{2}{3}\left[\tan ^{-1} u\right]_{0}^{\infty} \\ &=\frac{-1}{3}\left[\frac{\pi}{2}-0\right]+\frac{2}{3}\left[\tan ^{-1}(2 t)\right]_{0}^{\infty} \\ &=\frac{-1}{3} \frac{7}{2}+\frac{2}{3}\left[\tan ^{-1} \infty-\tan ^{-1} 2 \times 0\right] \\ &=\frac{-\pi}{6}+\frac{2}{3}\left[\tan ^{-1} \infty-\tan ^{-1} 0\right] \end{aligned}

\begin{aligned} &=\frac{-\pi}{6}+\frac{2}{3}\left[\frac{\pi}{2}-0\right] \\ &=\frac{-\pi}{6}+\frac{2}{3} \frac{\pi}{2} \\ &=\frac{-\pi}{6}+\frac{\pi}{3}=\frac{-\pi+2 \Pi}{6} \\ &=\frac{\pi}{6} \end{aligned}

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