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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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