Mathematics

Solve:
 $$\displaystyle \int tan^{3}2x\sec 2xdx=$$


ANSWER

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


SOLUTION
$$\int tan^{3}2x sec2x dx \cdot$$

$$\int (tan 2x sec2x)\cdot(tan^{2}2x)dx \cdot$$

Let $$sec 2x=t$$

$$2sec 2x tan 2x \cdot dx=dt$$

$$sec2x \cdot tan2x \cdot dx =1/2dt\cdot$$

$$\int (t^{2}-1)^1/_2 dt \cdot$$

$$\because tan^{2}2x=1+sec^{2}2x$$

$$^1/_2 \int (1+t^{2}) dt$$

$\Rightarrow ^1/_2 [-t+^t\ t^{3}/_3 ] +c$$

$$\Rightarrow ^1/_2 [t\ ^{3}/_3 -t]_{+c}=\ ^1/_2 \left [ \dfrac{sec^{3}\ 2x}{3}-sec2x \right ]{+c}$$

$$ =\left [ \dfrac{sec^{3}2x}{6}-\dfrac{sec2x}{2}+c \right ]$$
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Single Correct Medium Published on 17th 09, 2020
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