₀¹^{-1}≤ft({2x}{1+x²})dx - NCERT Class XII Mathematics Chapter

NCERT Class XII Mathematics Chapter - - Solutions

Exercise - 7.10
Evaluate the definite integrals in Exercises 198 to 205 using substitution.

Question : 200 of 271

Marks: +1, -0

\int\limits_{0}^{1}\sin^{-1}\left(\frac{2x}{1+x^{2}}\right)dx

Solution:

Let

I= \int\limits_{0}^{1}\sin^{-1}\left(\frac{2x}{1+x^{2}}\right)dx

Put

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

When

x = 0 \Rightarrow \theta = 0

and when

x = 1 \Rightarrow \theta = \frac{\pi}{4}

\therefore I = \int\limits_{0}^{\frac{\pi}{4}} \sin^{-1}\left(\frac{2\tan\theta}{1+\tan^{2}\theta}\right) \sec^{2}\theta d\theta

= \int\limits_{0}^{\frac{\pi}{4}} \sin^{-1}(\sin 2\theta) \cdot \sec^{2}\theta d\theta

= \int\limits_{0}^{\frac{\pi}{4}} 2\theta \sec^{2}\theta d\theta

= \left[ 2\theta \tan\theta \right]_{0}^{\frac{\pi}{4}} - 2 \int\limits_{0}^{\frac{\pi}{4}} \left( \frac{d}{d\theta}(\theta) \cdot \tan\theta \right) d\theta

= \left[ 2\theta \tan\theta - 2 \log \sec\theta \right]_{0}^{\frac{\pi}{4}}

= \left( 2 \cdot \frac{\pi}{4} \cdot \tan\frac{\pi}{4} - 2 \log \sec\frac{\pi}{4} \right) - \left( 0 - 2 \log 1 \right)

= \frac{\pi}{2} \times 1 - 2 \log \sqrt{2} = \frac{\pi}{2} \log 2

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