1. ## Integration

1. Evaluate integral(e^(7x)cos(2x)dx)

I let u =e^(7x) and so du = 7e^(7x)dx. dv is cos(2x) so V= 1/2sin(2x). I did the procedure, but i cant get the answer. Please show me some intermediate steps please!

2. Evaluate I = integral(486cot(6x)/sin^2(6x)+18sin(6x)+81) dx

No exactly sure how to go about this problem. I can factor the bottom into (sin(6x)+9)^2. I am using partial fraction decomposition, but the numerator seems to confuse me.

Help is much appreciated!
Thanks to all who contribute.

2. Originally Posted by Belowzero78
1. Evaluate integral(e^(7x)cos(2x)dx)

I let u =e^(7x) and so du = 7e^(7x)dx. dv is cos(2x) so V= 1/2sin(2x). I did the procedure, but i cant get the answer. Please show me some intermediate steps please!
Your work is good up to here.

Now take note that $-\tfrac{7}{2}\int e^{7x}\sin\!\left(2x\right)\,dx$ appears in your answer. Apply integration by parts again to get some constant multiple of $\int e^{7x}\cos\!\left(2x\right)\,dx$ to reappear. At this stage, treat the integral as a variable, and solve the equation for the integral (and don't forget the +C at the end!).

Does this make sense? Can you continue where you left off?

2. Evaluate I = integral(486cot(6x)/sin^2(6x)+18sin(6x)+81) dx

No exactly sure how to go about this problem. I can factor the bottom into (sin(6x)+9)^2. I am using partial fraction decomposition, but the numerator seems to confuse me.

Help is much appreciated!
Thanks to all who contribute.
Note that $\int\frac{486\cot\!\left(6x\right)}{\sin^2\!\left( 6x\right)+18\sin\!\left(6x\right)+81}\,dx=\int\fra c{486\cos\!\left(6x\right)}{\sin\!\left(6x\right)\ left(\sin\!\left(6x\right)+9\right)^2}\,dx$

Now apply the substitution $u=\sin\!\left(6x\right)$ and the reduced integral can be solved using partial fractions.

Does this make sense? Can you try to take it from here?

3. Originally Posted by Belowzero78
1. Evaluate integral(e^(7x)cos(2x)dx)

I let u =e^(7x) and so du = 7e^(7x)dx. dv is cos(2x) so V= 1/2sin(2x). I did the procedure, but i cant get the answer. Please show me some intermediate steps please!

2. Evaluate I = integral(486cot(6x)/sin^2(6x)+18sin(6x)+81) dx

No exactly sure how to go about this problem. I can factor the bottom into (sin(6x)+9)^2. I am using partial fraction decomposition, but the numerator seems to confuse me.

Help is much appreciated!
Thanks to all who contribute.
1. $\int{e^{7x}\cos{2x}\,dx}$

Let $u = e^{7x}$ so that $du = 7e^{7x}$.

Let $dv = \cos{2x}$ so that $v = \frac{1}{2}\sin{2x}$.

So $\int{e^{7x}\cos{2x}\,dx} = \frac{1}{2}e^{7x}\sin{2x} - \int{\frac{7}{2}e^{7x}\sin{2x}\,dx}$

$= \frac{1}{2}e^{7x}\sin{2x} - \frac{7}{2}\int{e^{7x}\sin{2x}\,dx}$.

Now let $u = e^{7x}$ so that $du = 7e^{7x}$

Let $dv = \sin{2x}$ so that $v = -\frac{1}{2}\cos{2x}$.

So $\int{e^{7x}\cos{2x}\,dx} = \frac{1}{2}e^{7x}\sin{2x} - \int{\frac{7}{2}e^{7x}\sin{2x}\,dx}$

$= \frac{1}{2}e^{7x}\sin{2x} - \left[-\frac{1}{2}e^{7x}\cos{2x} - \int{-\frac{7}{2}e^{7x}\cos{2x}\,dx}\right]$

$= \frac{1}{2}e^{7x}\sin{2x} + \frac{1}{2}e^{7x}\cos{2x} - \frac{7}{2}\int{e^{7x}\cos{2x}\,dx}$.

So $\frac{9}{2}\int{e^{7x}\cos{2x}\,dx} = \frac{1}{2}e^{7x}\sin{2x} + \frac{1}{2}e^{7x}\cos{2x}$

$\frac{9}{2}\int{e^{7x}\cos{2x}\,dx} = \frac{1}{2}e^{7x}\left(\sin{2x} + \cos{2x}\right)$

$\int{e^{7x}\cos{2x}\,dx} = \frac{1}{9}e^{7x}\left(\sin{2x} + \cos{2x}\right) + C$.

4. Ok, that works for to the 2nd one, so i decomposed into fractions and i get A/u +b/(U+9) + C/(u+9)^2. How do i get them so i compare coefficient to solve for the constants? I also found the common denominator.

5. prove It, the answer given is e^(7x)(7cos(2x)+2sin(2x))/53 +C

6. Originally Posted by Belowzero78
prove It, the answer given is e^(7x)(7cos(2x)+2sin(2x))/53 +C
Then go over my work and find my mistake.

7. Im not understanding how they get the denominator as 53. The denominator im getting is 56, so my integral is = 2/7(1/2e^(7x)sin(2x)+7/4e^(7x)cos(2x)). How do i get a 53 from that?