Solve Integral with Integration by Parts: 3xcos(x/2)dx

[Telemachus]

Homework Statement

Hi there. I'm confused about this exercise. It asks me to solve the integral using integration by parts. And the integral is:

\displaystyle\int_{}^{}3x\cos(\displaystyle\frac{x}{2})dx

The Attempt at a Solution

What I did:

u=3x
du=3dx
dv=cos(\displaystyle\frac{x}{2})
v=\sin(\displaystyle\frac{x}{2})

Then:

\displaystyle\int_{}^{}3x\cos(\displaystyle\frac{x}{2})dx=3x\sin(\displaystyle\frac{x}{2})-\displaystyle\int_{}^{}\sin(\displaystyle\frac{x}{2})3dx=3x\sin(\displaystyle\frac{x}{2})-3\cos(\displaystyle\frac{x}{2})

And derive gives me:

\displaystyle\int_{}^{}3x\cos(\displaystyle\frac{x}{2})dx=12\cos(\displaystyle\frac{x}{2})+6x\sin(\displaystyle\frac{x}{2})

So I think I'm doing something wrong, but I don't know what. And actually in my first attempt to a solution I got something like: \displaystyle\int_{}^{}3x\cos(\displaystyle\frac{x}{2})dx=\displaystyle\frac{1}{3}[x\sin(\displaystyle\frac{x}{2})-\cos(\displaystyle\frac{x}{2})], I arrived to this result by first taking the 3 out of the integral, and passing it to the other side at the end.

Any help will be thanked.

Bye there.

 

[stevenb]

Homework Statement

Hi there. I'm confused about this exercise. It asks me to solve the integral using integration by parts. And the integral is:

\displaystyle\int_{}^{}3x\cos(\displaystyle\frac{x}{2})dx

The Attempt at a Solution

What I did:

u=3x
du=3dx
dv=cos(\displaystyle\frac{x}{2})
v=\sin(\displaystyle\frac{x}{2})

Then:

\displaystyle\int_{}^{}3x\cos(\displaystyle\frac{x}{2})dx=3x\sin(\displaystyle\frac{x}{2})-\displaystyle\int_{}^{}\sin(\displaystyle\frac{x}{2})3dx=3x\sin(\displaystyle\frac{x}{2})-3\cos(\displaystyle\frac{x}{2})

And derive gives me:

\displaystyle\int_{}^{}3x\cos(\displaystyle\frac{x}{2})dx=12\cos(\displaystyle\frac{x}{2})+6x\sin(\displaystyle\frac{x}{2})

So I think I'm doing something wrong, but I don't know what. And actually in my first attempt to a solution I got something like: \displaystyle\int_{}^{}3x\cos(\displaystyle\frac{x}{2})dx=\displaystyle\frac{1}{3}[x\sin(\displaystyle\frac{x}{2})-\cos(\displaystyle\frac{x}{2})], I arrived to this result by first taking the 3 out of the integral, and passing it to the other side at the end.

Any help will be thanked.

Bye there.

You have the right approach but you made at least 3 silly mistakes. Your expression for "v" is off by a factor of 2 and you are off by a minus sign when you integrate sine. The final step is really silly when you factor out a 3 and get 1/3.

 

[HallsofIvy]

Homework Statement

Hi there. I'm confused about this exercise. It asks me to solve the integral using integration by parts. And the integral is:

\displaystyle\int_{}^{}3x\cos(\displaystyle\frac{x}{2})dx

The Attempt at a Solution

What I did:

u=3x
du=3dx
dv=cos(\displaystyle\frac{x}{2})
v=\sin(\displaystyle\frac{x}{2})

This is incorrect. Let u= x/2. Then du= (1/2)dx so dx= 2du.
\int cos(x/2)dx= 2\int cos(u)du= 2 sin(u)+ C= 2sin(x/2)+ C
If dv= cos(x/2), then v= 2sin(x/2), not just sin(x/2).

Then:

\displaystyle\int_{}^{}3x\cos(\displaystyle\frac{x}{2})dx=3x\sin(\displaystyle\frac{x}{2})-\displaystyle\int_{}^{}\sin(\displaystyle\frac{x}{2})3dx=3x\sin(\displaystyle\frac{x}{2})-3\cos(\displaystyle\frac{x}{2})

And derive gives me:

\displaystyle\int_{}^{}3x\cos(\displaystyle\frac{x}{2})dx=12\cos(\displaystyle\frac{x}{2})+6x\sin(\displaystyle\frac{x}{2})

So I think I'm doing something wrong, but I don't know what. And actually in my first attempt to a solution I got something like: \displaystyle\int_{}^{}3x\cos(\displaystyle\frac{x}{2})dx=\displaystyle\frac{1}{3}[x\sin(\displaystyle\frac{x}{2})-\cos(\displaystyle\frac{x}{2})], I arrived to this result by first taking the 3 out of the integral, and passing it to the other side at the end.

Any help will be thanked.

Bye there.

 

[Telemachus]

Thanks :)