Question

Trigometric simplification help?

Answer 1

\[\int\limits \sin(8x) \cos(3x) dx\]

Answer 2

I think i'm close to the answer I need someone to check it!

Answer 3

\[=\frac{ 1 }{ 2 } (\frac{ 1 }{ 5 }\cos(5x) +\frac{ 1 }{ 11 }\cos(11x))+c\]

Answer 4

It's almost good.
You just need a negative sign before everything.

Answer 5

could you show me where?

Answer 6

\[-\frac{1}{10}\cos(5x)-\frac{1}{22}\cos(11x)+C\]

But you really ought to figure out why it is missing from your result...

Answer 7

is it because of the anti-derivative of sin would be -cos right?

Answer 8

the anti-derivative of sin is -cos, whether that's where you slipped, I can't say without seeing your work. Seems plausible, however!

Answer 9

can you help me with one more?

Answer 10

Only one way to find out :-)

Answer 11

\[\int\limits \sin ^{4}xdx\]

Answer 12

There is no odds in this case what do I do?

Answer 13

Can you integrate \(\sin^2x~ dx\) and \(\cos^2x~dx\)?

Answer 14

\[\frac{ 1 }{ 2 }(-\frac{ 1 }{ 5 }\cos ^{5}(x)\sin(x))+c\]

Answer 15

\[\sin^4x = \sin^2x(1-\cos^2x)= \sin^2x - \frac{1}{4}(2\sin x\cos x)^2 = \sin^2x - \frac{1}{4}\sin^2(2x)\]Can you integrate that?

Answer 16

not sure how to do this one

Answer 17

\[\sin^2x = \frac{1}{2}(1-\cos(2x))\]

Answer 18

You should be able to do it with that last identity

Answer 19

\[=-4cosx +c\]

Answer 20

Mmm, no.

If you had a 2x as the argument to the cos function, you're still going to have it in the anti-derivative.

\[\frac{1}{2}[\int dx - \int \cos(2x) dx]= \frac{1}{2}[x - \frac{1}{2}\sin(2x)]+C\]