Mathematics

$$\displaystyle \int_{\pi /5}^{3\pi /10}\frac{\cos x}{\cos x+\sin x}dx $$is equal to


ANSWER

none of these


SOLUTION
Let $$\displaystyle I=\int _{ \frac { \pi  }{ 5 }  }^{ \frac { 3\pi  }{ 10 }  }{ \frac { \cos { x }  }{ \cos { x } +\sin { x }  }  } dx$$

Multiply numerator and denominator by $$sec^{ 3 }{ \left( x \right)  }$$
$$\displaystyle I=\int _{ \frac { \pi  }{ 5 }  }^{ \frac { 3\pi  }{ 10 }  }{ \frac { sec^{ 2 }{ x } }{ sec^{ 2 }{ x }+sec^{ 2 }{ x }\tan { x }  }  } dx=\int _{ \frac { \pi  }{ 5 }  }^{ \frac { 3\pi  }{ 10 }  }{ \frac { sec^{ 2 }{ x } }{ 1+\tan { x } +\tan ^{ 2 }{ x } +\tan ^{ 3 }{ x }  }  } dx$$

Put $$t=\tan { x } \Rightarrow dt=sec^{ 2 }{ x }dx$$
$$\displaystyle \therefore I=\int _{ \tan { \frac { \pi  }{ 5 }  }  }^{ \tan { \frac { 3\pi  }{ 10 }  }  }{ \frac { 1 }{ { u }^{ 2 }+{ u }^{ 3 }+u+1 }  } du$$

$$\displaystyle=\int _{ \tan { \frac { \pi  }{ 5 }  }  }^{ \tan \frac { 3\pi  }{ 10 }  }{ \left( \frac { 1-u }{ 2\left( { u }^{ 2 }+1 \right)  } +\frac { 1 }{ 2\left( u+1 \right)  }  \right) du } $$

$$\displaystyle =\frac { 1 }{ 2 } \int _{ \tan { \frac { \pi  }{ 5 }  }  }^{ \tan { \frac { 3\pi  }{ 10 }  }  }{ \left( \frac { 1 }{ { u }^{ 2 }+1 } -\frac { u }{ { u }^{ 2 }+1 }  \right)  } du+\frac { 1 }{ 2 } \int _{ \tan { \frac { \pi  }{ 5 }  }  }^{ \tan { \frac { 3\pi  }{ 10 }  }  }{\dfrac{1} {u+1} } du$$

$$\displaystyle =\left[ -\frac { 1 }{ 4 } \log { \left( { u }^{ 2 }+1 \right) +\frac { 1 }{ 2 }  } \log { \left( u+1 \right) +\frac { 1 }{ 2 } \tan ^{ -1 }{ u }  }  \right] _{ \log { \frac { \pi  }{ 5 }  }  }^{ \log { \frac { 3\pi  }{ 10 }  }  }=\frac { \pi  }{ 20 } $$

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Single Correct Medium Published on 17th 09, 2020
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