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  • Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 42 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 42 Maths Textbook Solution.

Answers (1)

Answer: 2\: x\tan ^{-1}x-2log\left ( \sqrt{1+x^{2}} \right )+c

Given: \int \cos ^{-1}\left ( \frac{1-x^{2}}{1+x^{2}} \right )dx

Hint: Put x=\tan \Theta

Solution:

            I=\int \cos ^{-1}\left ( \frac{1-x^{2}}{1+x^{2}} \right )dx

    Put x=\tan \theta\Rightarrow dx=\sec ^{2}\theta d\theta

                      \begin{aligned} &I=\int \cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right) \sec ^{2} \theta d \theta \\ &{\left[\cos 2 \theta=\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right)\right]} \\ &I=\int \cos ^{-1}(\cos 2 \theta) \sec ^{2} \theta d \theta \end{aligned}

 \begin{aligned} &=\int 2 \theta \sec ^{2} \theta d \theta \\ &=2 \int \theta \sec ^{2} \theta d \theta \\ &=2\left[\theta \int \sec ^{2} \theta d \theta-\int \frac{d}{d x} \theta\left(\int \sec ^{2} \theta d \theta\right) d \theta\right] \\ &=2 \theta \times \tan \theta-2 \int \tan \theta d \theta \\ &=2 \theta \tan \theta-2 \theta+c \\ &\theta=\tan ^{-1} x \\ &=2 x \tan ^{-1} x-2 \log \left(\sqrt{1+x^{2}}\right)+c \\ &=2 x \tan ^{-1} x-2 \log \left(\sqrt{1+x^{2}}\right)+c \end{aligned}

 

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