Question

Need help with integration. the integral of
cos(ax)sin(bx)dx by using the cyclic trick. I cannot use the product to sum formula. @Mathematics

Answer 1

You can use integration by parts

Answer 2

\[\frac{cos((b-a))x}{2(b-a)}-\frac{cos((b+a)x)}{2(b+a)}\]

Answer 3

i gots no idea what a cyclic trick is tho

Answer 4

It's when you get the same integral on both sides when using integration by parts and then you add it to both sides. Something like that the OP was saying

Answer 5

ohh ....

Answer 6

we just did a set of stuff like this for a group project in class which is why i remember the formula ...

Answer 7

or at least i hope i remembered it lol

Answer 8

\[\int x^n\ ln(x)\ dx=\frac{x^{n+1}}{n+1}ln(x)-\frac{x^{n+1}}{2(n+1)}+C\]

i think

Answer 9

nah, thats (n+1)^2 , i knew there was a 2 in it

Answer 10

The integration by parts trick goes as follows...
\[ I = \int cos(ax)sin(bx) dx =\frac{- cos(ax)cos(bx)}{b} - \int\frac{a*sin(ax)cos(bx)}{b}dx \]

\[ = \frac{-cos(ax)cos(bx)}{b} - \frac{a*sin(ax)sin(bx)}{b^2} + \int\frac{a^2cos(ax)sin(bx)}{b^2}dx\]
\[= \frac{-cos(ax)cos(bx)}{b} - \frac{a*sin(ax)sin(bx)}{b^2} + \frac{a^2}{b^2}I \]
so
\[(1-\frac{a^2}{b^2})I = \frac{-cos(ax)cos(bx)}{b} - \frac{a*sin(ax)sin(bx)}{b^2} \]
and, being careful to recognize that we need some constant C,
\[I = \frac{ \frac{-cos(ax)cos(bx)}{b} - \frac{a*sin(ax)sin(bx)}{b^2}}{1-\frac{a^2}{b^2}} +C \]
or equivalently,
\[ I = \frac{b*cos(ax)cos(bx) + a*sin(ax)sin(bx)}{a^2-b^2} +C \]

Answer 11

Someone clearly has a lot of free time on their hands

Answer 12

It's either this or revising my statement of purpose for the billionth time...