The value of ₀¹ ^{-1}(1-x+x²) \, dx is

Correct Answer is : π2−log⁡2π/2 - 22π−log2

Correct Answer is : π2−log⁡2π/2 - 22π−log2

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BITSAT 2018 Slot 2 Solved Paper

Section: Mathematics

Question : 144 of 150

Marks: +1, -0

The value of

\int\limits_{0}^{1} \cot^{-1}(1-x+x^{2}) \, dx

Solution:

\cot^{-1}(1-x+x^{2}) dx

=\tan^{-1}\left(\frac{1}{1-x+x^{2}}\right)

=\tan^{-1}\left(\frac{1}{1-x(1-x)}\right)

=\tan^{-1}\left(\frac{x+(1-x)}{1-x(1-x)}\right)

\Rightarrow \cot^{-1}(1-x+x^{2}) = \tan^{-1} x

-\tan^{-1}(1-x)

\therefore \int\limits_{0}^{1} \cot^{-1}(1-x+x^{2}) \, dx = \int\limits_{0}^{1} \tan^{-1} x \, dx - \int\limits_{0}^{1} \tan^{-1}(1-x) \, dx

= \int\limits_{0}^{1} \tan^{-1} x \, dx + \int\limits_{0}^{1} \tan^{-1} x \, dx

\left[\because \int\limits_{0}^{a} f(x) \, dx = -\int\limits_{0}^{a} f(a-x) \, dx\right]

= 2 \int\limits_{0}^{1} \tan^{-1} x \, dx

On evaluating by integration by parts, we have

=2\left\{ \left[ \tan^{-1} x \cdot x \right]_{0}^{1} - \int\limits_{0}^{1} \frac{x}{1+x^{2}} \, dx \right\}

=2\left\{ \frac{\pi}{4} - \left[ \frac{1}{2} \ln(1+x^{2}) \right]_{0}^{1} \right\}

=2\left[ \frac{\pi}{4} - \frac{1}{2} \log 2 \right] = \frac{\pi}{2} - \log 2

Hence,

\int\limits_{0}^{1} \cot^{-1}(1-x+x^{2}) \, dx = \frac{\pi}{2} - \log 2

So, option (b) is correct.

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