Mathematics

# $\displaystyle \int x^{2}\sec x^{3}dx.$

$\displaystyle \frac{1}{3}\log \left ( \sec x^{3} +\tan x^{3}\right ).$

##### SOLUTION
Let $\displaystyle I=\int { { x }^{ 2 }\sec ^{ 2 }{ { x }^{ 3 } } dx }$
Put $t={ x }^{ 3 }\Rightarrow dt=3{ x }^{ 2 }dx$

$\displaystyle I=\frac { 1 }{ 3 } \int { \sec{ t }dt } =\frac { 1 }{ 3 } \int { \frac { \sec ^{ 2 }{ t } +\sec{ t }\tan { t } }{ \tan { t } +\sec{ t } } dt }$

Now put $\tan { t } +\sec{ t }=u\Rightarrow \left( \sec ^{ 2 }{ t } +\sec{ t }\tan { t } \right) dt=du$

$\displaystyle I=\frac { 1 }{ 3 } \int { \frac { du }{ u } } =\frac { 1 }{ 3 } \log { u } =\frac { 1 }{ 3 } \log { \left( \tan { t } +\sec{ t } \right) }$

$\displaystyle =\frac { 1 }{ 3 } \log { \left( \tan { { x }^{ 3 } } +\sec{ { x }^{ 3 } } \right) }$
Hence, option 'B' is correct.

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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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