Mathematics

If $$\displaystyle I=\int \frac{\sin x\left ( \cos x \right )^{-5/2}dx}{\sqrt{\sin x+3\cos x}+\sqrt{\sin x+4\cos x}}$$$$\displaystyle= \frac { A }{ 1550 } \left( \left( \tan  x+4 \right) ^{ 5/2 }-\left( \tan  x+3 \right) ^{ 3/2 } \right) -\frac { 2 }{ 3 } \left[ 4\left( \tan  x+4 \right) ^{ 3/2 }-3{ \left( \tan { x } +3 \right)  }^{ 3/2 } \right] +C$$,
then the value of A is equal to


ANSWER

620


SOLUTION
$$I=\int { \cfrac { \sin  x\left( \cos  x \right) ^{ -5/2 }dx }{ \sqrt { \sin  x+3\cos  x } +\sqrt { \sin  x+4\cos  x }  }  } =\int { \cfrac { \sin { x } \left( \cos { x }  \right) ^{ -3 } }{ \sqrt { \tan { x } +3 } +\sqrt { \tan { x } +4 }  }  } $$
Let $$\tan { x } =t$$
$$I=\int { \cfrac { t }{ \sqrt { t+3 } +\sqrt { t+4 }  } dt } ={ I }_{ 4 }-{ I }_{ 3 }$$
Where $${ I }_{ a }=\int { t\sqrt { t+a }  } dt$$
Susbtituting $$t+a={ u }^{ 2 }$$
$${ I }_{ a }=\int { \left( { u }^{ 2 }-a \right) u\left( 2u \right) du } =\cfrac { 2{ u }^{ 5 } }{ 5 } -\cfrac { 2a{ u }^{ 3 } }{ 3 } =\cfrac { 2{ \left( t+a \right)  }^{ 5/2 } }{ 5 } -\cfrac { 2a{ \left( t+a \right)  }^{ 3/2 } }{ 3 } $$
Therefore
$$I=\cfrac { 2 }{ 5 } \left( { \left( \tan { x } +4 \right)  }^{ 5/2 }-{ \left( \tan { x } +3 \right)  }^{ 5/2 } \right) -\cfrac { 2 }{ 3 } \left( 4{ \left( \tan { x } +4 \right)  }^{ 3/2 }-3{ \left( \tan { x } +3 \right)  }^{ 3/2 } \right) +c$$
Therefore, $$A=620$$
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