Mathematics

$\displaystyle \int \sec^{3}x\: dx=\frac{1}{2}\left [ \sec x\: \tan x+\log\left ( ? +\tan x \right ) \right ]+c$Choose the appropriate option to replace the question mark in the above equation.

$\sec x$

SOLUTION
$\displaystyle I=\int \sec^{3}x\: dx d$
$\displaystyle =\int \sec x\sec^{2}x\:dx$
$\displaystyle =\int \sqrt{1+\tan^{2}x}\:\:.\sec^{2}\:xdx$
Put  $\displaystyle \: \tan x=z\:or\: \sec^{2}xdx=dz$
$\displaystyle \therefore I=\int \sqrt{1+z^{2}}\:dz$
$\displaystyle =\frac{z\sqrt{z^{2}+1}}{2}+\frac{1}{2}\log\left | z+\sqrt{z^{2}+1} \right |+c$
$\displaystyle =\frac{\tan x\: \sec x}{2}+\frac{1}{2}\log\left ( \tan x+\sec x \right )+c$
$\displaystyle =\frac{1}{2}\left [ \sec x\: \tan x+\log\left ( \sec x+\tan x \right ) \right ]+c$

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

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