Integration by Parts Problem

[Jason76]

Would this use integration by parts?

\(\displaystyle \int \cos^{3}(x)\sin(x) dx\)

 

[MarkFL]

I would just use a \(\displaystyle u\)-substitution:

\(\displaystyle u=\cos(x)\,\therefore\,du=-\sin(x)\,dx\)

and we now have:

\(\displaystyle \displaystyle -\int u^3\,du\)

 

[Jason76]

I would just use a \(\displaystyle u\)-substitution:

\(\displaystyle u=\cos(x)\,\therefore\,du=-\sin(x)\,dx\)

and we now have:

\(\displaystyle \displaystyle -\int u^3\,du\)

Let's try it with integration by parts (which should be one way to do it):

\(\displaystyle \int u(dv) = uv - \int v(du)\)

\(\displaystyle \int \cos^{3}(x)\sin(x) dx\)

\(\displaystyle u = \cos^{3}(x)\)

\(\displaystyle du = -3\sin^{2}(x) \cos(x)\)

\(\displaystyle v = \sin(x)\)

\(\displaystyle dv = \cos(x)\)

\(\displaystyle \int \cos^{3}(x) \cos(x) = \cos^{3}(x)\sin(x) - \int \sin(x)(-3\sin^{2}(x) \cos(x))\)

\(\displaystyle \int \cos^{3}(x) \cos(x) = \cos^{3}(x)\sin(x) - \int -3\sin(x) \sin^{2}(x) \cos(x)\)

How to get to the answer of \(\displaystyle -\dfrac{cos^{3}(x) }{3} + C\)

 

[Deleted member 4993]

How to get to the answer of \(\displaystyle -\dfrac{cos^{3}(x) }{3} + C\)

Answer is NOT \(\displaystyle -\dfrac{cos^{3}(x) }{3} + C\)

\(\displaystyle \displaystyle \int cos^3(x)*sin(x) dx \ \ne \ -\dfrac{cos^{3}(x) }{3} + C\)

 

[MarkFL]

Let's try it with integration by parts (which should be one way to do it):

\(\displaystyle \int u(dv) = uv - \int v(du)\)

\(\displaystyle \int \cos^{3}(x)\sin(x) dx\)

\(\displaystyle u = \cos^{3}(x)\)

\(\displaystyle du = -3\sin^{2}(x) \cos(x)\)

\(\displaystyle v = \sin(x)\)

\(\displaystyle dv = \cos(x)\)

\(\displaystyle \int \cos^{3}(x) \cos(x) = \cos^{3}(x)\sin(x) - \int \sin(x)(-3\sin^{2}(x) \cos(x))\)

\(\displaystyle \int \cos^{3}(x) \cos(x) = \cos^{3}(x)\sin(x) - \int -3\sin(x) \sin^{2}(x) \cos(x)\)

How to get to the answer of \(\displaystyle -\dfrac{cos^{3}(x) }{3} + C\)

You are not properly applying the technique. We are given:

\(\displaystyle I=\displaystyle \int \cos^3(x)\sin(x)\,dx\)

Using IBP, we could choose to let:

\(\displaystyle u=\cos^3(x)\,\therefore\,du=-3\cos^2(x)\sin(x)\,dx\)

\(\displaystyle dv=\sin(x)\,dx\,\therefore\,v=-\cos(x)\)

And so we have:

\(\displaystyle \displaystyle I=-\cos^4(x)-3\int \cos^3(x)\sin(x)\,dx=-\cos^4(x)-3I+C\)

Now we may solve for \(\displaystyle I\):

\(\displaystyle 4I=-\cos^4(x)+C\)

\(\displaystyle I=-\dfrac{1}{4}\cos^4(x)+C\)

 

[Jason76]

You are not properly applying the technique. We are given:

\(\displaystyle I=\displaystyle \int \cos^3(x)\sin(x)\,dx\)

Using IBP, we could choose to let:

\(\displaystyle u=\cos^3(x)\,\therefore\,du=-3\cos^2(x)\sin(x)\,dx\)

\(\displaystyle dv=\sin(x)\,dx\,\therefore\,v=-\cos(x)\)

And so we have:

\(\displaystyle \displaystyle I=-\cos^4(x)-3\int \cos^3(x)\sin(x)\,dx=-\cos^4(x)-3I+C\)

Now we may solve for \(\displaystyle I\):

\(\displaystyle 4I=-\cos^4(x)+C\)

\(\displaystyle I=-\dfrac{1}{4}\cos^4(x)+C\)

\(\displaystyle \displaystyle I=-\cos^4(x)-3\int \cos^3(x)\sin(x)\,dx=-\cos^4(x)-3I+C\)

\(\displaystyle uv\) should be \(\displaystyle -\cos^3(x)\cos(x)\)

 

[MarkFL]

Why write it like that when they can be combined as I did? You do realize that:

\(\displaystyle a^b\cdot a^c=a^{b+c}\)

Right?

 

[Jason76]

Why write it like that when they can be combined as I did? You do realize that:

\(\displaystyle a^b\cdot a^c=a^{b+c}\)

Right?

yes

 

[MarkFL]

yes

Okay, good!

Thus:

\(\displaystyle -\cos^3(x)\cos(x)=-\cos^{3+1}(x)=-\cos^4(x)\)

et voila! :mrgreen: