Mathematics

$$\displaystyle\int x\cos^3x\sin x dx$$.


SOLUTION
We are given, $$ I= \int { x\quad cos^{ 3 }.sinxdx. }  $$

as we know, $$ \int { uvdx\quad =\quad u\int { vdx\quad -\quad \int { \left( \frac { dy }{ dx } -\int { vdx }  \right)  }  }  } dx $$ 

let $$  u=x$$ and $$v =cos^{ 3 }x.sinx $$

so, $$I=x\int { cos^{ 3 }x.sindx-\int { \left( 1.\int { cos^{ ^{ 3 } }sinx.dx }  \right)  } dx }  $$

let $$cosx=t\Rightarrow -sin dx = dt $$
 
so, differentiating wrt $$x$$, we get, 

$$ I=x\int { t^{ 3 } } (-dt)-\int { \left[ \int { t^{ 3 } } (-dt) \right]  } dx $$

$$ =x\left[ \frac { -t^{ 4 } }{ 4 }  \right] -\int { \frac { t^{ 4 } }{ 4 }  } dx $$ 

$$ \therefore I=-\frac { x }{ 4 } cos^{ 4 }x-\frac { 1 }{ 4 } \int { cos^{ 4 } } xdx $$

$$ \therefore I=-\frac { x }{ 4 } cos^{ 4 }x-\frac { 1 }{ 4 } \int { \left( \frac { cos2x+1 }{ 2 }  \right) ^{ 2 }dx }  $$

$$\displaystyle =-\frac { x }{ 4 } cos^{ 4 }x-\frac { 1 }{ 4\times 4 } \left[ \int { \left( cos^{ 2 }2x+2cos2x+1 \right)  } dx \right]  $$

$$\displaystyle =-\frac { x }{ 4 } cos^{ 4 }x-\frac { 1 }{ 16 } \left[ \int { \left( \frac { cos4x+1 }{ 2 } +2cos2x+1 \right)  } dx \right]  $$

$$\displaystyle =-\frac { x }{ 4 } cos^{ 4 }x-\frac { 1 }{ 32 } \int { cos4xdx\quad -\frac { 1 }{ 8 } \int { cos2xdx }  }   -\int { \frac { 3 }{ 32 } dx } $$

$$\displaystyle I=\frac { -x }{ 4 } .cos^{ 4 }x-\frac { 1 }{ 32 } .\frac { sin^{ 4 }x }{ 4 } -\frac { 1 }{ 8 } .\frac { sin^{ 2 }x }{ 2 } -\frac { 3 }{ 32 } x+c $$

$$\displaystyle I=\quad \frac { -sin^4x }{ 128 } -\frac { sin^2x }{ 16 } -\frac { x.cos^{ 2 }x }{ 4 } -\frac { 3x }{ 32 } +c $$

$$\displaystyle \int { xcos^{ 3 }x \, sinx \, dx=\frac { -sin^{ 4 }x }{ 128 }  } -\frac { sin^{ 2 }x }{ 16 } -\frac { x.cos^{ 4 }x }{ 4 } -\frac { 3x }{ 32 } +c $$ 

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Subjective Medium Published on 17th 09, 2020
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