First slide

Methods of integration

Question

$\int \frac{d x}{(1 + \sqrt{x}) \sqrt{(x - x^{2})}} is equal to$

Moderate

A

$\frac{1 + \sqrt{x}}{(1 - x)^{2}} + c$
B

$\frac{1 + \sqrt{x}}{(1 + x)^{2}} + c$
C

$\frac{1 - \sqrt{x}}{(1 - x)^{2}} + c$
D

$\frac{2 (\sqrt{x} - 1)}{\sqrt{(1 - x)}} + c$

Solution

$Let I = \int \frac{d x}{(1 + \sqrt{x}) \sqrt{(x - x^{2})}}$
$\begin{array}{l} If \sqrt{x} = \sin p, then \frac{1}{2 \sqrt{x}} dx = \cos pdp \\ I = \int \frac{2 \sin pcos pdp}{(1 + \sin p) \sin p \cdot \cos p} \end{array}$
$\begin{array}{l} = 2 \int \frac{dp}{(1 + \sin p)} = 2 \int \frac{(1 - \sin p) dp}{\cos^{2} p} \\ = 2 \{\int \sec^{2} pdp - \int (\tan psec p) dp\} \\ = 2 (\tan p - \sec p) = 2 (\sqrt{\frac{x}{(1 - x)}} - \frac{1}{\sqrt{(1 - x)})} + C \\ = \frac{2 (\sqrt{x} - 1)}{\sqrt{(1 - x)}} + C \end{array}$

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