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  • Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.3 Question 26

Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.3 Question 26

Answers (1)

Answer: -\infty

Given:  \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{-\frac{\pi}{2}}{\sqrt{\cos x \sin ^{2} x}} d x

Solution:

Let

\begin{aligned} &I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{-\frac{\pi}{2}}{\sqrt{\cos x \sin ^{2} x}} d x \\ & \end{aligned}

=-\frac{\pi}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{\sqrt{\cos x \sin ^{2} x}} d x

\begin{aligned} &=-\frac{\pi}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{\sqrt{\cos x}|\sin x|} d x\\ \end{aligned}

=-\frac{\pi}{2} \times 2 \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{\cos x}|\sin x|} d x\\

\begin{aligned} &|\sin x|=\sin x, 0 \leq x \leq \frac{\pi}{2} \\ & \end{aligned}

=-\pi \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{\cos x} \sin x} d x \\

=-\pi \int_{0}^{\frac{\pi}{3}} \frac{1}{\sqrt{\cos x}\left(1-\cos ^{2} x\right)} d x

Put \cos x=t^{2}  and differentiate on both sides

\begin{aligned} &-\sin x d x=2 t d t \\ & \end{aligned}

\text { when } x \rightarrow 0, t \rightarrow 1 \\

\text { when } x \rightarrow \frac{\pi}{2}, t \rightarrow 0

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

=2 \pi \int_{1}^{0} \frac{d t}{\left(1-t^{4}\right)} \\

=2 \pi \int_{1}^{0} \frac{d t}{(1-t)(1+t)\left(1+t^{2}\right)}

Now by partial fraction

\begin{aligned} &\frac{1}{(1-t)(1+t)\left(1+t^{2}\right)}=\frac{A}{(1-t)}+\frac{B}{(1+t)}+\frac{C}{\left(1+t^{2}\right)} \\ & \end{aligned}

1=A(1+t)\left(1+t^{2}\right)+B(1-t)\left(1+t^{2}\right)+C(1-t)(1+t)

\begin{aligned} &\text { Put } t=1, \text { we get } \\ & \end{aligned}

A=\frac{1}{4} \\

\text { Putting } t=-1, \text { we get }

\begin{aligned} &1=A+B+D \\ & \end{aligned}

D=1-\frac{1}{4}-\frac{1}{4} \\

=\frac{1}{2}

Equating coefficients of t^{3} both sides

\begin{aligned} &A-B+C=0 \\ & \end{aligned}

\frac{1}{4}-\frac{1}{4}+C=0 \\

C=0

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

=2 \pi \int_{1}^{0} \frac{\frac{1}{4}}{1-t^{2}} d t+2 \pi \int_{1}^{0} \frac{\frac{1}{4}}{1+t} d t+2 \pi \int_{1}^{0} \frac{\frac{1}{2}}{1+t^{2}} d t

\begin{aligned} &=\frac{2 \pi}{4} \times\left[\frac{\log (1-t)}{-1}\right]_{1}^{0}+\frac{2 \pi}{4} \times[\log (1+t)]_{1}^{0}+\frac{2 \pi}{2} \times\left[\tan ^{-1} t\right]_{1}^{0} \\ & \end{aligned}

=\frac{-\pi}{2}(\log 1-\log 0)+\frac{\pi}{2}(\log 1-\log 2)+\pi\left(\tan ^{-1} 0-\tan ^{-1} 1\right)

\begin{aligned} &=\frac{-\pi}{2}[0-(-\infty)]+\frac{\pi}{2}[0-\log 2]+\frac{\pi}{2}\left(0-\frac{\pi}{4}\right) \\ & \end{aligned}

=-\infty-\frac{\pi}{2} \log 2-\frac{\pi^{2}}{4}

= -\infty [any validated up with \infty becomes \infty]

 


Posted by

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