Mathematics

# $\displaystyle\int \tan^32x\sec 2xdx$ is equal to

##### SOLUTION
$I = \int tan^3 2x sec 2x dx$

$I = \int tan^2 2x . sec 2x , tan 2x dx$

$I = \int (sec^2 2x -1 ) sec 2x . tan 2x dx$

$(as 1 + tan^2 \theta = sec^2 \theta$)

let $\sec 2x = t$

differentiate above wrt $x$, we get ,

$sec 2x . tan 2x (2) = \dfrac {dt}{dx}$

$2 sec 2x . tan 2x dx = dt$

$I =\displaystyle \int (t^2 -1 ) \frac {dt}{2}$

$I =\displaystyle \int \frac {t^2 }{2} dt - \int \frac {dt}{2}$

$I = \displaystyle\frac {t^3}{6} - \dfrac {t}{2} +C$$I = \dfrac {sec^3 2x }{6} - \dfrac {sec 2x}{2} +C$

where C is any arbitrary constant

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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