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
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