Mathematics

The value of intergral $$\displaystyle \int_{\dfrac{\pi}{4}}^{\dfrac{3\pi}{4}}\dfrac{x}{1+\sin x}dx$$ is :


ANSWER

$$\pi(\sqrt{2}-1)$$


SOLUTION
Given the integral,
$$\int _{ \dfrac { \pi  }{ 4 }  }^{ \dfrac { 3\pi  }{ 4 }  }{ \dfrac { x }{ \sin { x } +1 }  } dx\\ =\int _{ \dfrac { \pi  }{ 4 }  }^{ \dfrac { 3\pi  }{ 4 }  }{ \dfrac { x\dfrac { 1 }{ \cos { x }  }  }{ \dfrac { \sin { x }  }{ \cos { x }  } +\dfrac { 1 }{ \cos { x }  }  }  } dx\\ =\int _{ \dfrac { \pi  }{ 4 }  }^{ \dfrac { 3\pi  }{ 4 }  }{ \dfrac { x\sec { x }  }{ \tan { x } +\sec { x }  }  } dx$$
Using integration by parts we get,
$$=-\dfrac { x }{ \tan { x } +\sec { x }  } -\int _{ \dfrac { \pi  }{ 4 }  }^{ \dfrac { 3\pi  }{ 4 }  }{ \dfrac { 1 }{ -\tan { x } -\sec { x }  }  } dx$$
For,
$$\int { \dfrac { 1 }{ -\tan { x } -\sec { x }  }  } dx\\ =-\int { \dfrac { 1 }{ \tan { x } +\sec { x }  }  } dx\\ =-\int { \cos { x }  } \dfrac { 1 }{ \sin { x } +1 } dx$$
Let,
$$u=\sin { x } \\ \Rightarrow \dfrac { du }{ dx } =\cos { x } \\ \Rightarrow du=\cos { x } dx$$
substituting these values we get,
$$-\int { \cos { x }  } \dfrac { 1 }{ \sin { x } +1 } dx\\ =-\int { \dfrac { 1 }{ u+1 }  } du$$
Again let,
$$v=u+1\\ \Rightarrow \dfrac { dv }{ du } =1\\ \Rightarrow dv=du$$
Substituting these values we get,
$$\int { \dfrac { 1 }{ u+1 }  } du\\ =\int { \dfrac { 1 }{ v }  } dv\\ =\ln { (v) } \\ =\ln { (u+1) } \\ =\ln { (\sin { x } +1) } \\ \therefore \int { \dfrac { 1 }{ -\tan { x } -\sec { x }  }  } dx\\ =-\ln { (\sin { x } +1) } $$
So,
$$-\dfrac { x }{ \tan { x } +\sec { x }  } -\int { \dfrac { 1 }{ -\tan { x } -\sec { x }  }  } dx\\ =\ln { (\sin { x } +1) } -\dfrac { x }{ \tan { x } +\sec { x }  } \\ \therefore \int _{ \dfrac { \pi  }{ 4 }  }^{ \dfrac { 3\pi  }{ 4 }  }{ \dfrac { x }{ \sin { x } +1 }  } dx\\ ={ \left[ \ln { (\sin { x } +1) } -\dfrac { x }{ \tan { x } +\sec { x }  }  \right]  }_{ \dfrac { \pi  }{ 4 }  }^{ \dfrac { 3\pi  }{ 4 }  }\\ =\ln { (\sin { \dfrac { 3\pi  }{ 4 }  } +1) } -\ln { (\sin { \dfrac { \pi  }{ 4 }  } +1) } -\left[ \dfrac { \dfrac { 3\pi  }{ 4 }  }{ \tan { \dfrac { 3\pi  }{ 4 }  } +\sec { \dfrac { 3\pi  }{ 4 }  }  } -\frac { \dfrac { \pi  }{ 4 }  }{ \tan { \dfrac { \pi  }{ 4 }  } +\sec { \dfrac { \pi  }{ 4 }  }  }  \right] \\ =\pi (\sqrt { 2 } -1).$$

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Single Correct Medium Published on 17th 09, 2020
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