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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Hard
Evaluate: $\int _{ 1/3 }^{ 1 }{ \cfrac { { \left( x-{ x }^{ 3 } \right) }^{ 1/3 } }{ { x }^{ 4 } } } dx=$
• A. $3$
• B. $0$
• C. $4$
• D. $6$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
Find  $\int \cfrac {dx} {{x}^{2}-{a}^{2}}$ are hence evaluate
• A. $\int \cfrac {dx} {{x}^{2}- 8x+5}$
• B. $\int \cfrac {dx} {{3x}^{2}+13x-10}$
• C. None of these
• D. $\int \cfrac {dx} {{x}^{2}-36}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Evaluate the following integral
$\int { \cfrac { \cot { x } }{ \log { \sin { x } } } } dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Solve:
$\int {\frac{{dx}}{{x + {x^{ - 3}}}}}$

$\displaystyle\int\dfrac{\cos x-\sin x}{\sqrt{\sin 2x}}dx$.