_ ^ ^3 2x 2x ;dx = - Vidyadip

MCQ

$\int_{}^{} {{{\tan }^3}} 2x\sec 2x\;dx = $

✓
$\frac{1}{6}{\sec ^3}2x - \frac{1}{2}\sec 2x + c$
B
$\frac{1}{6}{\sec ^3}2x + \frac{1}{2}\sec 2x + c$
C
$\frac{1}{9}{\sec ^2}2x - \frac{1}{3}\sec 2x + c$
D
None of these

✓

Answer

Correct option: A.

$\frac{1}{6}{\sec ^3}2x - \frac{1}{2}\sec 2x + c$

a
(a)$\int_{}^{} {{{\tan }^3}2x\sec 2x\,dx} = \int_{}^{} {({{\sec }^2}2x - 1)\sec 2x\tan 2x\,dx} $
$\int_{}^{} {({{\sec }^3}2x\tan 2x - \sec 2x\tan 2x)dx} $
$ = \int_{}^{} {{{\sec }^3}2x\tan 2x\,dx} - \int_{}^{} {\sec 2x\tan 2x\,dx} $ ..$(i)$
Now, we take $\int_{}^{} {{{\sec }^3}2x\tan 2x\,dx} $
Put $\sec 2x = t \Rightarrow \sec 2x\tan 2x = \frac{{dt}}{2},$ then it reduces to
$\frac{1}{2}\int_{}^{} {{t^2}dt} = \frac{{{t^3}}}{6} = \frac{{{{\sec }^3}2x}}{6}$
From $(i),$ $\int_{}^{} {{{\sec }^3}2x\tan 2x\,dx} - \int_{}^{} {\sec 2x\tan 2x\,dx} $
$ = \frac{{{{\sec }^3}2x}}{6} - \frac{{\sec 2x}}{2} + c.$
Trick : Let $\sec 2x = t,$ then $\sec 2x\tan 2x\,dx = \frac{1}{2}dt$
$\frac{1}{2}\int_{}^{} {({t^2} - 1)\,dt} = \frac{1}{6}{t^3} - \frac{1}{2}t + c = \frac{1}{6}{\sec ^3}2x - \frac{1}{2}\sec 2x + c$.

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