Mathematics

$$\int sec^{3}x dx$$


SOLUTION
$$I=\displaystyle\int{{\sec}^{3}{x}dx}$$
Integrating by parts, we have
$$u=\sec{x}$$,$$dv={\sec}^{2}{x}dx$$
$$\Rightarrow du=\sec{x}\tan{x}dx$$ ,$$\Rightarrow v=\tan{x}$$
$$=\sec{x}\tan{x}-\displaystyle\int{\sec{x}{\tan}^{2}{x}dx}$$
$$=\sec{x}\tan{x}-\displaystyle\int{\sec{x}{\left({\sec}^{2}{x}-1\right)}dx}$$
$$=\sec{x}\tan{x}-\displaystyle\int{{{\sec}^{3}{x}-\sec{x}}dx}$$
Since $$\displaystyle \int{{\sec}^{3}{x}dx}=I$$
$$=\sec{x}\tan{x}-I+\displaystyle\int{\sec{x}dx}$$
$$\Rightarrow 2I=\sec{x}\tan{x}+\ln{\left|\sec{x}+\tan{x}\right|}+{c}_{1}$$
$$\Rightarrow I=\dfrac{1}{2}\sec{x}\tan{x}+\dfrac{1}{2}\ln{\left|\sec{x}+\tan{x}\right|}+\dfrac{{c}_{1}}{2}$$
Hence $$\displaystyle \int{{\sec}^{3}{x}dx}=\dfrac{1}{2}\sec{x}\tan{x}+\dfrac{1}{2}\ln{\left|\sec{x}+\tan{x}\right|}+c$$

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Subjective Hard Published on 17th 09, 2020
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