Mathematics

# Integrate with respect to $x$:$\sec^{3}x$

##### SOLUTION
$I= \displaystyle \int\sec^3x\,\,dx$
$=\displaystyle \int\sec x. \, \, sec ^2x\,\,dx$
using formula of integration by parts:
$I=\sec x. \displaystyle \int\sec ^2dx-\displaystyle \int(\sec \,x)^1(\displaystyle \int\sec ^2x\,\,dx)$
$=\sec \, x.tan \,\, x-\displaystyle \int(\sec \, x \tan \, x)(\tan \, x)dx$
$\sec\, x \tan \, x- \displaystyle \int \sec \,x.\tan ^2xdx$
$\sec \,x \tan \,x - \displaystyle \int\sec \,x (\sec^2x-1)dx$
$\sec \,\, x. \tan \, x - \displaystyle \int\sec ^3xdx+ \displaystyle \int \sec \, x\, dx$
$I=\sec \, x .\tan \, x -I+ |n|\sec \,x + \tan \, x |+C$
$\Rightarrow 2 I = \sec \, x . \tan \, x + |n|\sec \, x+ \tan \, x|+C$
$\Rightarrow \boxed {I=\dfrac{1}{2}\sec \, x. \tan + \dfrac{1}{2}|n|\sec \, x+ \tan \, x|+C}$
Identities used:
$\displaystyle \int f(x) g(x) dx=f(x)\displaystyle \int g(x)-\displaystyle \int f(x)\left[\displaystyle \int g(x)dx\right] dx$
$\displaystyle \int\sec ^2xdx- \tan \,x +C$
$\displaystyle \int\sec x dx - In|\sec \, x+ \tan \, x|+C$

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Subjective Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

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