Mathematics

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


ANSWER

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


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
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