How do you find the antiderivative of ${cos}^{4} (2 x) \cdot {sin}^{3} (2 x)$ $cos^4 (2x) * sin^3 (2x)$ ?

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How do you find the antiderivative of ${cos}^{4} (2 x) \cdot {sin}^{3} (2 x)$ $cos^4 (2x) * sin^3 (2x)$ ?

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Answer 1 · 2017-10-17T17:20:58

$\int {cos}^{4} (2 x) \cdot {sin}^{3} (2 x) d x = \frac{{cos}^{7} (2 x)}{14} - \frac{{cos}^{5} (2 x)}{10} + C$ $intcos^4(2x)*sin^3(2x)dx=cos^7(2x)/14-cos^5(2x)/10+C$

Explanation:

We have the integral $I_{1} = \int {cos}^{4} (2 x) \cdot {sin}^{3} (2 x) d x$ $I_(color(red)1 )=intcos^4(2x)*sin^3(2x)dx$

Here, let

$(2 x) = t$ $(2x)=t$

$\frac{d}{d x} (2 x) = \frac{d}{d x} t$ $d/dx(2x)=d/dxt$

$2 = \frac{d t}{d x}$ $2=dt/dx$

Therefore, $d x = \frac{d t}{2}$ $dx=dt/2$

Put this in $I_{1}$ $I_(color(red)1$

$I_{2} = \frac{1}{2} \int {cos}^{4} (t) \cdot {sin}^{3} (t) d t$ $I_(color(red)2 )=1/2intcos^4(t)*sin^3(t)dt$

$I_{2} = \frac{1}{2} \int {cos}^{4} (t) \cdot {sin}^{2} (t) \cdot sin (t) d t$ $I_(color(red)2 )=1/2intcos^4(t)*sin^2(t)*sin(t)dt$

We know $({sin}^{2} t = 1 - {cos}^{2} t)$ $(sin^2t=1-cos^2t)$

Put this in $I_{2}$ $I_(color(red)2$

$I_{2} = \frac{1}{2} \int {cos}^{4} (t) \cdot (1 - {cos}^{2} t) \cdot sin (t) d t$ $I_(color(red)2 )=1/2intcos^4(t)*(1-cos^2t)*sin(t)dt$

Here, let

$cos t = v$ $cost=v$

$\frac{d}{d t} cos t = \frac{d}{d t} v$ $d/dtcost=d/dtv$

$- sin t = \frac{d v}{d t}$ $-sint=(dv)/dt$

Therefore,

$sin (t) d t = - d v$ $sin(t)dt=-dv$

Put this in $I_{2}$ $I_(color(red)2$

$I_{3} = \frac{- 1}{2} \int v^{4} \cdot (1 - v^{2}) \cdot d v$ $I_(color(red)3 )=(-1)/2intv^4*(1-v^2)*dv$

$I_{3} = \frac{- 1}{2} \int (v^{4} - v^{6}) \cdot d v$ $I_(color(red)3 )=(-1)/2int(v^4-v^6)*dv$

$I_{3} = (\frac{- 1}{2}) \frac{v^{5}}{5} - (\frac{- 1}{2}) \frac{v^{7}}{7} + C$ $I_(color(red)3 )=((-1)/2)(v^5)/5-((-1)/2)v^7/7+C$

Now put in the value of $v$ $v$ .

$I_{3} = (\frac{- 1}{2}) \frac{{cos}^{5} (t)}{5} - (\frac{- 1}{2}) \frac{{cos}^{7} (t)}{7} + C$ $I_(color(red)3 )=((-1)/2)(cos^5(t))/5-((-1)/2)cos^7(t)/7+C$

Now put in the value of $t$ $t$ .

$I_{3} = \frac{- {cos}^{5} (2 x)}{10} + \frac{{cos}^{7} (2 x)}{14} + C$ $I_(color(red)3 )=(-cos^5(2x))/10+cos^7(2x)/14+C$

Therefore,

$\int {cos}^{4} (2 x) \cdot {sin}^{3} (2 x) d x = \frac{{cos}^{7} (2 x)}{14} - \frac{{cos}^{5} (2 x)}{10} + C$ $intcos^4(2x)*sin^3(2x)dx=cos^7(2x)/14-cos^5(2x)/10+C$

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How do you find the antiderivative of ${cos}^{4} (2 x) \cdot {sin}^{3} (2 x)$ $cos^4 (2x) * sin^3 (2x)$ ?

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