SATHEE: Definite Integration Question 53

Definite Integration Question 53

Question 53

The value of the integral $\int_{0}^{1 / 2} \frac{1 + \sqrt{3}}{{((x + 1)^{2} (1 - x)^{6})}^{1 / 4}} d x$ is

(2018 Adv.)

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Solution:

(2) Let $I = \int_{0}^{1 / 2} \frac{1 + \sqrt{3}}{{[(x + 1)^{2} (1 - x)^{6}]}^{1 / 4}} d x$

$\Rightarrow I = \int_{0}^{1 / 2} \frac{1 + \sqrt{3}}{(1 - x)^{2} \frac{1 - x}{1 + x}} d x$

Put $\frac{1 - x}{1 + x} = t \Rightarrow \frac{- 2 d x}{(1 + x)^{2}} = d t$

when $x = 0, t = 1, x = \frac{1}{2}, t = \frac{1}{3}$

$∴$

$I = \int_{1}^{1 / 3} \frac{(1 + \sqrt{3}) d t}{- 2 (t)^{6 / 4}}$

$\Rightarrow I = \frac{- (1 + \sqrt{3})}{2} {\frac{- 2}{\sqrt{t}}}_{1}^{1 / 3}$

$\Rightarrow I = (1 + \sqrt{3}) (\sqrt{3} - 1) \Rightarrow I = 3 - 1 = 2$

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Definite Integration Question 53

Question 53

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