SATHEE: Indefinite Integration 1 Question 70

Indefinite Integration 1 Question 70

71. Prove that $\int_{0}^{1} \tan^{- 1} \frac{1}{1 - x + x^{2}} d x = 2 \int_{0}^{1} \tan^{- 1} x d x$ .

Hence or otherwise, evaluate the integral

$\int_{0}^{1} \tan^{- 1} (1 - x + x^{2}) d x .$

$(1998, 8$ M)

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Solution:

$\int_{0}^{1} \tan^{- 1} \frac{1}{1 - x + x^{2}} d x = \int_{0}^{1} \tan^{- 1} \frac{1 - x + x}{1 - x (1 - x)} d x$

$\begin{aligned} = \int_{0}^{1} [\tan^{- 1} (1 - x) - \tan^{- 1} x] d x \\ = \int_{0}^{1} \tan^{- 1} [1 - (1 - x)] d x + \int_{0}^{1} \tan^{- 1} x d x \\ = 2 \int_{0}^{1} \tan^{- 1} x d x ∵ \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x \end{aligned}$

Now, $\int_{0}^{1} \tan^{- 1} \frac{1}{1 - x + x^{2}} d x$

$\begin{aligned} = \int_{0}^{1} \frac{π}{2} - \cot^{- 1} \frac{1}{1 - x + x^{2}} d x \\ = \frac{π}{2} - \int_{0}^{1} \tan^{- 1} (1 - x + x^{2}) d x \\ ∴ \int_{0}^{1} \tan^{- 1} (1 - x + x^{2}) d x = \frac{π}{2} - \int_{0}^{1} \tan^{- 1} \frac{1}{(1 - x + x^{2})} d x \\ = \frac{π}{2} - 2 I_{1} \end{aligned}$

where, $I_{1} = \int_{0}^{1} \tan^{- 1} x d x = {[x \tan^{- 1} x]}_{0}^{1} - \int_{0}^{1} \frac{x d x}{1 + x^{2}}$

$\begin{matrix} = \frac{π}{4} - \frac{1}{2} {[\log (1 + x^{2})]}_{0}^{1} = \frac{π}{4} - \frac{1}{2} \log 2 \\ ∴ \int_{0}^{1} \tan^{- 1} (1 - x + x^{2}) d x = \frac{π}{2} - 2 \frac{π}{4} - \frac{1}{2} \log 2 = \log 2 \end{matrix}$

Indefinite Integration 1 Question 70

71. Prove that $\int_{0}^{1} \tan^{- 1} \frac{1}{1 - x + x^{2}} d x = 2 \int_{0}^{1} \tan^{- 1} x d x$ .

Solution:

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Indefinite Integration 1 Question 70

71. Prove that ∫01tan−1⁡11−x+x2dx=2∫01tan−1⁡xdx.

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71. Prove that $\int_{0}^{1} \tan^{- 1} \frac{1}{1 - x + x^{2}} d x = 2 \int_{0}^{1} \tan^{- 1} x d x$ .