First slide

Methods of integration

Question

$\int \frac{d x}{\sqrt{\sin^{3} x \cos^{5} x}} =$

Moderate

A

$\frac{2}{3} [\frac{\tan^{2} x - 3}{\sqrt{\tan x}}] + C$
B

$\frac{3}{2} [\frac{\tan^{2} x - 3}{\sqrt{\tan x}}] + C$
C

$\frac{- 1}{3} [\frac{\tan x + 3}{\tan^{2} x}] + C$
D

$\frac{- 2}{5} [\frac{\tan^{2} x - 3}{\sqrt{\tan x}}] + C$

Solution

$I = \int \frac{d x}{\sqrt{\sin^{3} x \cos^{5} x}}$
$\begin{array}{l} Dividing N r \in D r by \cos^{4} x \\ = \int \frac{\sec^{4} x}{\sqrt{\frac{\sin^{3} x \cos^{5} x}{\cos^{8} x}} d x} \\ = \int \frac{\sec^{4} x}{\sqrt{\tan^{3} x}} d x \\ = \int \frac{\sec^{2} x \sec^{2} x}{\sqrt{\tan^{2} x \cdot \tan x}} d x \\ = \int \frac{(1 + \tan^{2} x) \sec^{2} x}{\tan x \cdot \sqrt{\tan x}} d x \\ Put \tan x = t \\ \sec^{2} x d x = d t \\ = \int \frac{(1 + t^{2}) d t}{t \sqrt{t}} \end{array}$
$\begin{array}{l} = \int (t^{\frac{- 3}{2}} + t^{\frac{1}{2}}) d t \\ = \frac{t^{\frac{- 3}{2} + 1}}{\frac{- 3}{2} + 1} + \frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + c \end{array}$
$\begin{array}{l} = \frac{t^{\frac{- 1}{2}}}{\frac{- 1}{2}} + \frac{t^{\frac{3}{2}}}{\frac{3}{2}} + c \\ = + 2 [\frac{- 1}{\sqrt{t}} \cdot + \frac{t \sqrt{t}}{3}] + c \\ = 2 [\frac{- 3 + t^{2}}{3 \sqrt{t}}] + c \\ = \frac{2}{3} [\frac{t^{2} - 3}{\sqrt{t}}] + c \\ = \frac{2}{3} [\frac{\tan^{2} x - 3}{\sqrt{\tan x}}] + c \end{array}$

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