Mathematics

# ${\tan ^3}2x\sec 2x$

##### SOLUTION
$\displaystyle \int{\tan^3 2x\sec 2x dx}$
$2x=t$
$2.dx=dt$
$\Rightarrow \dfrac{1}{2}\displaystyle \int{\tan^3t\sec t. dt}$
$\sec t=y$
$\sec t \tan t.dt=dy$
$\tan^2t=1-\sec^2t$
$=1-y^2$
$\Rightarrow \dfrac{1}{2}\displaystyle \int{(1-y^2).dy}$
$=\dfrac{1}{2}\left(y-\dfrac{y^3}{3}\right)=\dfrac{1}{2}\left(\sec 2x-\dfrac{\sec^3 2x}{3}\right)$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

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