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Question

Column -I and Column -II contains four entries each. Entry of column -I are to be matched one or more than one entries of column -II.


Solution

$$A=\int { \cfrac { cosx-sinx }{ cosx+sinx } dx } $$
Substituting $$t=sinx+cosx\Rightarrow dt=\left( cosx-sinx \right) dx$$
$$A=\int { \cfrac { 1 }{ t } dt } =\log { t } +c=\log { \left( cosx+sinx \right)  } +c\\ cosx+sinx=\sqrt { 2 } \left( \cfrac { 1 }{ \sqrt { 2 }  } cosx+\cfrac { 1 }{ \sqrt { 2 }  } sinx \right) =\sqrt { 2 } \left( cos\left( x-\cfrac { \pi  }{ 4 }  \right)  \right) $$
$$A=\log { \left( \sqrt { 2 } \left( cos\left( x-\cfrac { \pi  }{ 4 }  \right)  \right)  \right)  } +c=\log { \left( cos\left( x-\cfrac { \pi  }{ 4 }  \right)  \right)  } +\log { \sqrt { 2 }  } +c\\ =\log { \left( cos\left( x-\cfrac { \pi  }{ 4 }  \right)  \right)  } +c$$
$$B=\int { \cfrac { cos2x }{ sin2x+1 } dx } $$
Substituting $$t=2x\Rightarrow dt=2dx$$, we get
$$B=\cfrac { 1 }{ 2 } \int { \cfrac { cost }{ sint+1 } dt } $$
Again substituting $$sint+1=u\Rightarrow costdt=du$$
$$B=\cfrac { 1 }{ 2 } \int { \cfrac { 1 }{ u } du } =\cfrac { \log { u }  }{ 2 } +c=\cfrac { 1 }{ 2 } \log { \left( sint+1 \right)  } +c\\ =\log { \left( sin2x+1 \right)  } +c=\log { \left( sinx+cosx \right)  } +c\\ =\log { \left( cos\left( x-\cfrac { \pi  }{ 4 }  \right)  \right)  } +c$$
$$C=\int { \cfrac { 1-sin2x }{ cos2x } dx } =\int { \left( 1-sin2x \right) sec2xdx } $$
Substituting $$t=2x\Rightarrow dt=2dx$$
$$C=\cfrac { 1 }{ 2 } \int { \left( 1-sint \right) sectdt } =\int { \left( sect+tant \right) dt } \\ =-\cfrac { 1 }{ 2 } \int { tantdt } +\cfrac { 1 }{ 2 } \int { sectdt } =\cfrac { 1 }{ 2 } \int { \cfrac { sint }{ cost } dt } +\cfrac { 1 }{ 2 } \int { sectdt } $$
Again substituting $$cost=u\Rightarrow -sintdt=du$$
$$C=-\cfrac { 1 }{ 2 } \int { -\cfrac { 1 }{ u } du } +\cfrac { 1 }{ 2 } \int { sectdt } =\cfrac { 1 }{ 2 } \int { \cfrac { 1 }{ u } du } +\cfrac { 1 }{ 2 } \int { sectdt } \\ =\cfrac { \log { u }  }{ 2 } +\cfrac { 1 }{ 2 } \log { \left( tant+sect \right)  } +c=\cfrac { 1 }{ 2 } \log { \left( cos2x \right)  } +\cfrac { 1 }{ 2 } \log { \left( tant+sect \right)  } +c\\ =\cfrac { 1 }{ 2 } \log { \left( sin2x+1 \right)  } +c=\log { \left( cosx+sinx \right)  } +c=\log { \left( cos\left( x-\cfrac { \pi  }{ 4 }  \right)  \right)  } +c$$
$$D=\int { \left( \cfrac { sin^{ -1 }\sqrt { x }  }{ \sqrt { x-x^{ 2 } }  }  \right) dx } $$
Substituting $$t=sin^{ -1 }\sqrt { x } \Rightarrow dt=\cfrac { 1 }{ 2\sqrt { -x\left( x-1 \right)  }  } dx$$
$$D=2\int { tdt } =t^{ 2 }+c=\left( sin^{ -1 }\sqrt { x }  \right) ^{ 2 }+c=\left( \pi -cos^{ -1 }\sqrt { x }  \right) ^{ 2 }+c\\ =\pi ^{ 2 }+\left( cos^{ -1 }\sqrt { x }  \right) ^{ 2 }-2\pi cos^{ -1 }\sqrt { x } +c=\left( cos^{ -1 }\sqrt { x }  \right) ^{ 2 }-2\pi cos^{ -1 }\sqrt { x } +c$$

Mathematics

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