How do you find the antiderivative of #cos^4 (2x) * sin^3 (2x)#?

Question

Christopher Agresta · Accepted Answer

#intcos^4(2x)*sin^3(2x)dx=cos^7(2x)/14-cos^5(2x)/10+C#

We have the integral #I_(color(red)1 )=intcos^4(2x)*sin^3(2x)dx#

Here, let

#(2x)=t#

#d/dx(2x)=d/dxt#

#2=dt/dx#

Therefore, #dx=dt/2#

Put this in #I_(color(red)1#

#I_(color(red)2 )=1/2intcos^4(t)*sin^3(t)dt#

#I_(color(red)2 )=1/2intcos^4(t)*sin^2(t)*sin(t)dt#

We know #(sin^2t=1-cos^2t)#

Put this in #I_(color(red)2#

#I_(color(red)2 )=1/2intcos^4(t)*(1-cos^2t)*sin(t)dt#

Here, let

#cost=v#

#d/dtcost=d/dtv#

#-sint=(dv)/dt#

Therefore,

#sin(t)dt=-dv#

Put this in #I_(color(red)2#

#I_(color(red)3 )=(-1)/2intv^4*(1-v^2)*dv#

#I_(color(red)3 )=(-1)/2int(v^4-v^6)*dv#

#I_(color(red)3 )=((-1)/2)(v^5)/5-((-1)/2)v^7/7+C#

Now put in the value of #v#.

#I_(color(red)3 )=((-1)/2)(cos^5(t))/5-((-1)/2)cos^7(t)/7+C#

Now put in the value of #t#.

#I_(color(red)3 )=(-cos^5(2x))/10+cos^7(2x)/14+C#

Therefore,

#intcos^4(2x)*sin^3(2x)dx=cos^7(2x)/14-cos^5(2x)/10+C#

Aurora Alvarado · Answer

undefined To find the antiderivative of $ \cos^4(2x) \cdot \sin^3(2x) $, you can use the method of substitution. Let $ u = \cos(2x) $. Then, $ du = -2\sin(2x) \, dx $.

Using this substitution, rewrite the integral in terms of $ u $ and $ du $:

$$
\int \cos^4(2x) \cdot \sin^3(2x) \, dx = \int u^4 \cdot \left(\frac{-du}{2}\right)^3
$$

Now, simplify and integrate the expression in terms of $ u $:

$$
= -\frac{1}{8} \int u^4 \, du
$$
$$
= -\frac{1}{8} \cdot \frac{u^5}{5} + C
$$

Finally, substitute back for $ u $:

$$
= -\frac{1}{8} \cdot \frac{\cos^5(2x)}{5} + C
$$

So, the antiderivative of $ \cos^4(2x) \cdot \sin^3(2x) $ is $ -\frac{1}{40} \cos^5(2x) + C $, where $ C $ is the constant of integration.