Mathematics

$$\int ^{\pi /2}_{-\pi /2} cost . \sin(2t-\displaystyle \frac{\pi}{4})dt=$$


ANSWER

$$-\displaystyle \frac{\sqrt{2}}{3}$$


SOLUTION
$$I=\int { \left( cost \right) \left( sin\left( 2t-\cfrac { \pi  }{ 4 }  \right)  \right)  } dt=-\int { sin\left( \cfrac { \pi  }{ 4 } -2t \right) \left( cost \right) dt } \\ =-\cfrac { 1 }{ 2 } \int { \left( sin\left( \cfrac { 1 }{ 4 } \left( \pi -12t \right)  \right) +sin\left( \cfrac { 1 }{ 4 } \left( \pi -4t \right)  \right)  \right) dt } \\ =-\cfrac { 1 }{ 2 } \int { sin\left( \cfrac { 1 }{ 4 } \left( \pi -12t \right)  \right) dt } -\cfrac { 1 }{ 2 } \int { sin\left( \cfrac { 1 }{ 4 } \left( \pi -4t \right)  \right) dt } $$
Let $${ I }={ I }_{ 1 }+{ I }_{ 2 }$$, such that
$${ I }_{ 1 }=-\cfrac { 1 }{ 2 } \int { sin\left( \cfrac { 1 }{ 4 } \left( \pi -12t \right)  \right) dt } $$
Substituting $$x=\pi -4t\Rightarrow dx=-4dt$$
$${ I }_{ 1 }=\cfrac { 1 }{ 8 } \int { sin\cfrac { x }{ 4 } dx } =-\cfrac { 1 }{ 2 } cos\cfrac { x }{ 4 } =-\cfrac { 1 }{ 2 } sin\left( 3t+\cfrac { \pi  }{ 4 }  \right) $$ ...(1)
And 
$${ I }_{ 2 }=-\cfrac { 1 }{ 2 } \int { sin\left( \cfrac { 1 }{ 4 } \left( \pi -4t \right)  \right) dt } $$
Substituting $$y=\pi -12t\Rightarrow dy=-12dt$$, we get
$${ I }_{ 2 }=\cfrac { 1 }{ 24 } \int { sin\cfrac { y }{ 4 } dy } =-\cfrac { 1 }{ 6 } cos\cfrac { y }{ 4 } \quad =-\cfrac { 1 }{ 6 } sin\left( t+\cfrac { \pi  }{ 4 }  \right) $$...(2)
Therefore from (1) and (2)
$$\int _{ -\cfrac { \pi  }{ 2 }  }^{ \cfrac { \pi  }{ 2 }  }{ \left( cott \right) sin\left( 2t-\cfrac { \pi  }{ 4 }  \right) dt } =\left[ -\cfrac { 1 }{ 2 } sin\left( 3t+\cfrac { \pi  }{ 4 }  \right) -\cfrac { 1 }{ 6 } sin\left( t+\cfrac { \pi  }{ 4 }  \right)  \right] _{ -\cfrac { \pi  }{ 2 }  }^{ \cfrac { \pi  }{ 2 }  }=-\cfrac { \sqrt { 2 }  }{ 3 } $$
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Single Correct Hard Published on 17th 09, 2020
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