Evaluate the given Integrals

[Sakurazaki]

I have done everything on this packet except for 3 that I cannot figure out.

Directions: Evaluate the given Integrals

 

[galactus]

\(\displaystyle \int cos^{-1}(7x)dx\)

Using parts:

\(\displaystyle u=cos^{-1}(7x), \;\ dv=dx, \;\ du=\frac{-7}{\sqrt{1-49x^{2}}}dx, \;\ v=x\)

\(\displaystyle xcos^{-1}(x)+7\int\frac{x}{\sqrt{1-49x^{2}}}}dx\)

For the right integral, let \(\displaystyle u=1-49x^{2}\) and it becomes very easy.

 

[Sakurazaki]

Hmm I got

for the integral of ArcCos[7x]...is this still an acceptable answer?

 

[galactus]

Yes, it is. It would appear you are studying parts. Try the others and lets see how you do.

 

[Sakurazaki]

Well I tried rapid repeating on this one >.>' probably screwed something up.

 

[galactus]

I see you're using tabular integration.

Here it is. Looks good. Looks like you have it.

\(\displaystyle \frac{x^{3}sin(5x)}{5}+\frac{3x^{2}cos(5x)}{25}-\frac{6xsin(5x)}{125}-\frac{6cos(5x)}{625}\)

 

[galactus]

For the first one, lets use a method known as 'integration by recognition'.


\(\displaystyle \int xe^{2x}dx\)

Since we have a power of x that is 1, we use a linear set up:

\(\displaystyle (Ax+B)e^{2x}=xe^{2x}\)

Differentiate:

\(\displaystyle 2Axe^{2x}+Ae^{2x}+2Be^{2x}=xe^{2x}\)

equate coefficients:

\(\displaystyle A+2B=1\)
\(\displaystyle 2A=1\)

\(\displaystyle A=\frac{1}{2}, \;\ B=\frac{-1}{4}\)

So, we have:

\(\displaystyle \boxed{\frac{1}{2}xe^{2x}-\frac{1}{4}e^{2x}}\)

You can throw the 3 back in there.

 

[Sakurazaki]

For the last one I did the same thing or tried to.

 

[BigGlenntheHeavy]

\(\displaystyle \int cos^{-1}(7x)dx = xcos^{-1}(7x)+\frac{\sqrt{1-49x^{2}}}{7}+C\).

A good way to check if you gotten the proper answer, is to take the derivative of your answer which should give you the original integral, since integrals and derivatives are inverses of each other.