Mathematics

# $\displaystyle\int_{}^{} {x{{\sec }^2}xdx}$

##### SOLUTION
Now,
$\displaystyle\int_{}^{} {x{{\sec }^2}xdx}$
$=x.\displaystyle\int_{}^{} {{{\sec }^2}xdx}$$-\displaystyle\int 1. \left(\displaystyle\int_{}^{} {{{\sec }^2}xdx}\right) dx$ [ Using method of by parts]
$=x\tan x-\log|\sec x|+c$ [ Where $c$ is integrating constant]

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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1 Verified Answer | Published on 17th 09, 2020

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