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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

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