Question

I need to find the integral of cos(ax)sin(bx) dx without using the product to sum formula. Any help is greatly appreciated. This is turning into a nightmare. I need to find the integral of cos(ax)sin(bx) dx without using the product to sum formula. Any help is greatly appreciated. This is turning into a nightmare. @Mathematics

Answer 1

Write J = integral of cos ax . sin bx.

Integrate twice by parts and you'll land on an expression

J = (expression in cos and sin) + (constant) J

Then solve for J.

Answer 2

For example, integrating once,
\[ J = \int \cos ax . \sin bx \ dx = 1/a . \sin ax . \sin bx - b/a \int \sin ax . \cos bx \ dx\]

Answer 3

Now repeat that until the integral you get back is J.

Answer 4

\[\int\limits \cos(ax)sing(bx) = -1/b(\cos(ax)\cos(bx)) - a/b ^{2}\sin(ax)\sin(bx) - a ^{2}/b ^{2} \int\limits \cos(ax)\sin(bx) dx\]
I got this far and now I am supposed to use the cyclic trick to finish it but I cant just add the Int of cos(ax)sin(bx) to both sides b/c it is attached to the a squared over b squared.

Answer 5

You're almost there. Moving both integrals the the left-hand side, LHS, we have:

\[ (1 + a^2/b^2) J = -1/b . \cos ax . \cos bx - a/b^2 . \sin ax . \sin bx \]

Now finish solving for J.

Answer 6

But double-check your work. I'm pretty sure you've made a sign error.

Answer 7

the first integration is u= cos(ax) v= -1/b cos(bx)
du= -asin(ax)dx dv= sin(bx)dx
then the second integration is u= sin(ax) v= 1/bsin(bx)
du=acos(ax)dx dv=cos(bx)dx
which after that leaves me with:
int of cos(ax)sin(bx)dx =
-1/bcos(bx)cos(ax) - a/bsquared sin(ax)sin(bx)
- asquared/bsquared int of cos(ax)sin(bx)dx [which is what
I started with so I have to add it to both sides for the cyclic trick]
How do I do that since it is hooked by multiplication to the asquared/bsquared???

Answer 8

If J = g(x) + c J, where c is a constant, then

J = g(x) / (1-c)

That's what you do. I don't understand why that's so mysterious.

Answer 9

Hence if
\[ J = \int cos(ax) sin(bx) dx \]
then
\[ J = 1/a . sin(ax) sin(bx) - b/a \int sin(ax) cos(bx) dx \ \ \ \ --- (*)\]
and
\[ \int sin(ax) cos(bx) dx = -1/a . cos(ax) cos(bx) - b/a \int cos(ax).sin(bx) dx \]
\[ = -1/a . cos(ax) cos(bx) - b/a . J \]
Recombining this expression with (*),
\[ J = 1/a^2 . ( a . sin(ax).sin(bx) + b . cos(ax).cos(bx) ) + b^2/a^2 . J\]

Solving now for J:
\[ \frac{a^2-b^2}{a^2} J = \frac{1}{a^2} ( a . sin(ax).sin(bx) + b . cos(ax).cos(bx) ) \]
hence
\[ J = \frac{1}{a^2-b^2} ( a . sin(ax).sin(bx) + b . cos(ax).cos(bx) ) \]