AP EAPCET 08-Jul-2022 Shift 1 Paper

Section: Mathematics

Question : 71 of 160

Marks: +1, -0

0 < x < 1, \int \frac{dx}{\sqrt{x^{2} - x^{5}}} = \frac{1}{3} \log |f(x)| + C

f\left(\frac{1}{2}\right) =

Solution: 👈: Video Solution

Let

I = \int \frac{dx}{\sqrt{x^{2} - x^{5}}}

I = \int \frac{1}{x \sqrt{1 - x^{3}}} dx

Let

x^{3} = \sin^{2} t

3 x^{2} dx = 2 \sin t \cos t dt

\Rightarrow I = \frac{1}{3} \int \frac{3 x^{2} dx}{x^{3} \sqrt{1 - x^{3}}}

= \frac{2}{3} \int \frac{1}{\sin^{2} t \cos t} \sin t \cos t dt

= \frac{2}{3} \int \csc t dt

= \frac{2}{3} \log |\csc t - \cot t| + C

= \frac{2}{3} \log \left| \frac{1 - \cos t}{\sin t} \right| + C

= \frac{2}{3} \log \left| \frac{1 - \sqrt{1 - \sin^{2} t}}{\sqrt{\sin^{2} t}} \right| + C

= \frac{1}{3} \log \left( \left| \frac{1 - \sqrt{1 - x^{3}}}{\sqrt{x^{3}}} \right|^{2} \right) + C

Here,

f(x) = \left| \frac{1 - \sqrt{1 - x^{3}}}{\sqrt{x^{3}}} \right|^{2}

f\left(\frac{1}{2}\right) = \left| \frac{1 - \sqrt{1 - \left(\frac{1}{2}\right)^{3}}}{\left(\frac{1}{2}\right)^{3/2}} \right|^{2}

= \left( \frac{1 - \frac{\sqrt{7}}{\sqrt{8}}}{\frac{1}{\sqrt{8}}} \right)^{2} = (\sqrt{8} - \sqrt{7})^{2}

From option (a)

\colon \left( \frac{\sqrt{8} - \sqrt{7}}{\sqrt{8} + \sqrt{7}} \right) \times \left( \frac{\sqrt{8} - \sqrt{7}}{\sqrt{8} - \sqrt{7}} \right)

= (\sqrt{8} - \sqrt{7})^{2}

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