Mathematics

$$\int e^{\tan \theta}(\sec \theta -\sin \theta)d\theta$$ equals


ANSWER

$$e^{\tan \theta}\cos \theta+c$$


SOLUTION
Let $$\displaystyle I=\int { { e }^{ \tan { \theta  }  } } \left( \sec { \theta  } -\sec { \theta  }  \right) d\theta $$
Substitute $$\tan { \theta  } =t\Rightarrow \sec^{ 2 }{ \theta  }d\theta =dt$$
$$\therefore I=\int { { e }^{ t } } \left( \cos { \theta  } -\tan { \theta  } \sec { \theta  }  \right) dt$$
$$\displaystyle =\int { { e }^{ t } } \left( \frac { 1 }{ \sqrt { 1+{ t }^{ 2 } }  } -t\sqrt { 1+{ t }^{ 2 } }  \right) dt$$
As $$\displaystyle \frac { d }{ dx } \left( \frac { 1 }{ \sqrt { 1+{ t }^{ 2 } }  }  \right) =t\sqrt { 1+{ t }^{ 2 } } $$
$$\displaystyle \therefore I={ e }^{ t }\left( \frac { 1 }{ \sqrt { 1+{ t }^{ 2 } }  }  \right) ={ e }^{ \tan { \theta  }  }\cos { \theta  } +c$$
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Single Correct Medium Published on 17th 09, 2020
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