Question

Help me solve this please!!!!
f (e^-2x)sinxdx

f (cos(ax))/(rad(9+sinax)) dx

Answer 1

u substitution?

Answer 2

can you show me step by step ?

Answer 3

\[u = 9+\sin(ax)~,~ du = a\cos(ax)dx\]

Answer 4

Oh there are 2 problems here.

Answer 5

yeah i'm having trouble with them i haven't taken calc 1 in like a year so i'm trying to refresh my memory

Answer 6

\[\int e^{-2x}\sin(x)dx\]Use LIATE for determining the u substitution:
1. Log 2. Inverse 3. Algebraic (polynomials) 4. Trig 5. exponentials\[u=\sin(x)~,~ du = \cos(x)dx~,~ dv = e^{-2x}dx~,~v = \int e^{-2x}dx = -\frac{1}{2}e^{-2x}dx \]

Answer 7

how do i solve the second one?

Answer 8

Integration by parts: \[\int udv = uv - \int vdu\]

Answer 9

For the second one.. look at my first post.

\(\color{blue}{\text{Originally Posted by}}\) @Jhannybean
\[u = 9+\sin(ax)~,~ du = a\cos(ax)dx\]
\(\color{blue}{\text{End of Quote}}\)

Answer 10

is -1/2e^-2xdx the final form?

Answer 11

No, you have to plug it into the format for integration by parts.

\(\color{blue}{\text{Originally Posted by}}\) @Jhannybean
Integration by parts: \[\int udv = uv - \int vdu\]
\(\color{blue}{\text{End of Quote}}\)

Answer 12

@Jhannybean can you show it to me plugged in?

Answer 13

I've already identified what your u, v , du, and dv are, all you have to do is plug them into the formula and solve

Answer 14

Simplify it*

Answer 15

@Jhannybean how did you get the u, v, du, and dv