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  • Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise Revision Exercise Question 6

Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise Revision Exercise Question 6

Answers (1)

Answer: \frac{\pi}{2}-\log 2

Given: \int_{0}^{1} \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) d x

Hint: Use the substitution method and trigonometric identities

Solution:

\int_{0}^{1} \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) d x

Put \begin{aligned} &x=\tan \theta \\ & \end{aligned}

\Rightarrow d x=\sec ^{2} \theta d \theta

\begin{aligned} &=\int_{0}^{1} \cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right) \times \sec ^{2} \theta d \theta \\ & \end{aligned}

=\int_{0}^{1} \cos ^{-1}(\cos 2 \theta) \times \sec ^{2} \theta d \theta              \quad\left(\because \cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right)

\begin{aligned} &=2 \int_{0}^{1} \theta \sec ^{2} \theta d \theta \\ & \end{aligned}

=2\left[\theta \int_{0}^{1} \sec ^{2} \theta d \theta-\int_{0}^{1} \frac{d}{d \theta} \theta\left[\sec ^{2} \theta d \theta\right]_{0}^{1}\right] \\

=2\left[\theta \tan \theta-\int \tan \theta d \theta\right]_{0}^{1}

\begin{aligned} &=2(\theta \tan \theta)_{0}^{1}-2 \int_{0}^{1} \frac{\sin \theta}{\cos \theta} d \theta \\ & \end{aligned}

=2\left[\tan ^{-1} x \times \tan \left(\tan ^{-1} x\right)\right]_{0}^{1}-2 \int_{0}^{1} \frac{\sin \theta}{\cos \theta} d \theta

Let \begin{aligned} &\cos \theta=t \\ & \end{aligned}

\Rightarrow-\sin \theta d \theta=\mathrm{dt}                          (differentiate w.r.t \theta)

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

= 2 \times \frac{\pi}{4}+2[\log |t|]_{0}^{1} \\

=\frac{\pi}{2}+2[\log |\cos \theta|]_{0}^{1}

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

=\frac{\pi}{2}+2\left(\log \frac{1}{\sqrt{2}}-\log 1\right) \\

=\frac{\pi}{2}-\log 2

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