First slide

Methods of integration

Question

$If \int \frac{1}{\cos^{3} x \sqrt{2 \sin 2 x}} d x = (\tan x)^{A} + C (\tan x)^{B} + k, where k is constant of integration then A + B + C is$

Moderate

A

$\frac{16}{5}$
B

$\frac{27}{10}$
C

$\frac{7}{10}$
D

$\frac{25}{5}$

Solution

$I = \int \frac{d x}{\cos^{3} x \sqrt{2 \sin 2 x}} = \int \frac{d x}{\cos^{3} x \sqrt{2 \cdot 2 s in x \cos x}} = \frac{1}{2} \int \frac{1}{{(\cos x)}^{\frac{7}{2}} . {(\sin x)}^{\frac{1}{2}}} d x$
$\begin{array}{l} = \frac{1}{2} \int \frac{d x}{{(\frac{\sin x}{\cos x})}^{\frac{1}{2}} \cdot {(\cos x)}^{\frac{7}{2} + \frac{1}{2}}} = \frac{1}{2} \int \frac{1}{(\tan x)^{\frac{1}{2}} \cdot \cos^{4} x} d x = \frac{1}{2} \int \frac{1 + \tan^{2} x}{(\tan x)^{\frac{1}{2}}} \sec^{2} x d x \\ = \frac{1}{2} \int \frac{1 + t^{2}}{t^{\frac{1}{2}}} d t, w h e r e t = \tan x \\ d t = s e c^{2} x d x \\ = \frac{1}{2} [2 \sqrt{t} + \frac{2}{5} t^{\frac{5}{2}}] + k \\ = (\tan x)^{\frac{1}{2}} + \frac{(\tan x)^{\frac{5}{2}}}{5} + k \\ = (\tan x)^{A} + C (\tan x)^{B} + k (Given) \\ ∴ A = \frac{1}{2} B = \frac{5}{2} C = \frac{1}{5} \\ ∴ A + B + C = \frac{1}{2} + \frac{5}{2} + \frac{1}{5} = \frac{16}{5} \end{array}$

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