Mathematics

Evaluate$$\displaystyle \int_{0}^{\sin ^{2}t}\sin ^{-1}\sqrt{x}dx+\int_{0}^{\cos ^{2}t}\cos ^{-1}\sqrt{x}dx = k$$, then $$tan(k) =?$$


ANSWER

1


SOLUTION
Let $$\displaystyle I=\int _{ 0 }^{ \sin ^{ 2 } t } \sin ^{ -1 } \sqrt { x } dx+\int _{ 0 }^{ \cos ^{ 2 } t } \cos ^{ -1 } \sqrt { x } dx$$
$$I=I_{ 1 }+I_{ 2 }$$
For $$I_{ 1 }$$ put $$\sqrt { x } =\sin  \theta \Rightarrow x=\sin ^{ 2 } \theta \Rightarrow dx=\sin  2\theta d\theta $$
$$\therefore I_{ 1 }=\int _{ 0 }^{ t } \theta \sin  2\theta d\theta $$
And for $$I_{ 2 }$$ put $$\sqrt { x } =\cos  \phi $$
$$\therefore I_{ 2 }=\int _{ \pi /2 }^{ t } -\phi \sin  2\phi d\phi =\int _{ t }^{ \pi /2 } \theta \sin  2\theta d\theta $$
Hence
$$\displaystyle I=\int _{ 0 }^{ t } \theta \sin  2\theta d\theta +\int _{ t }^{ \pi /2 } \theta \sin  2\theta d\theta =\int _{ 0 }^{ \pi /2 } \theta \sin  2\theta d\theta $$
$$\displaystyle \\ =\left[ \theta \left( -\dfrac { \cos  2\theta  }{ 2 }  \right) -\left( 1 \right) \left( -\dfrac { \sin  2\theta  }{ 4 }  \right)  \right] _{ 0 }^{ \pi /2 }=\dfrac { \pi  }{ 2 } .\dfrac { 1 }{ 2 } +.0=\dfrac { \pi  }{ 4 } $$
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