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Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise Revision Exercise Question 8

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Answer:  \frac{\pi}{2 \sqrt{3}}-\frac{3}{2} \log \frac{4}{3}

Given:  \int_{0}^{\frac{1}{\sqrt{3}}} \tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right) d x

Hint: You must know about trigonometric identities and formula of  \int u v d x


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

Let  \tan x=\theta

d x=\sec ^{2} \theta d \theta                     (differentiate w.r.t x )

We know ,

\begin{aligned} &0<\tan \theta<\frac{1}{\sqrt{3}} \\ & \end{aligned}


\therefore \int_{0}^{\frac{\pi}{6}} \tan ^{-1}\left(\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\right) \times \sec ^{2} \theta d \theta

\begin{aligned} &=\int_{0}^{\frac{\pi}{6}} \tan ^{-1}(\tan 3 \theta) \times \sec ^{2} \theta d \theta \\ & \end{aligned}

=3 \int_{0}^{\frac{\pi}{6}} \theta \sec ^{2} \theta d \theta

We know that  \left[\because \int u v d x=u \int v d x-\int \frac{d}{d x} u \int v d x\right]

\begin{aligned} &=3(\theta \tan \theta)_{0}^{\frac{\pi}{6}}-3 \int_{0}^{\frac{\pi}{6}} \tan \theta d \theta \\ & \end{aligned}

=3\left(\frac{\pi}{6} \times \frac{1}{\sqrt{3}}-0\right)-3 \int_{0}^{\frac{\pi}{6}} \tan \theta d \theta

Let \cos \theta=t

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

\begin{aligned} &=\frac{\pi}{2 \sqrt{3}}+3 \int_{0}^{\frac{\pi}{6}} \frac{1}{t} d t \\ & \end{aligned}

=\frac{\pi}{2 \sqrt{3}}+3(\log |t|)_{0}^{\frac{\pi}{6}}=\frac{\pi}{2 \sqrt{3}}+3(\log |\cos \theta|)_{0}^{\frac{\pi}{6}}

\begin{aligned} &=\frac{\pi}{2 \sqrt{3}}+3\left(\log \frac{\sqrt{3}}{2}-\log 1\right)_{0}^{\frac{\pi}{6}} \\ & \end{aligned}

\frac{\pi}{2 \sqrt{3}}-\frac{3}{2} \log \frac{4}{3}



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