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
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