1. ## Integration

It is given that . Find the values of the constants and .

How would you solve this?

2. Let $I = \int_0^{\frac{\pi}{2}} e^{4y} \sin (3y) dy$

Use integration by parts twice.

For the first round, use: $u = e^{4y}$ and $dv =\sin (3y) dy$

To get: $I = -\frac{1}{3}e^{4y}\cos (3y) \bigg|_0^{\frac{\pi}{2}} + \frac{4}{3} {\color{blue}\int_0^{\frac{\pi}{2}} e^{4y} \cos (3y) dy}$

Use integrationg by parts again on the blue using: $u = e^{4y}$ and $dv = \cos (3y) dy$

To get: $I = -\frac{1}{3}e^{4y}\cos (3y)\bigg|_0^{\frac{\pi}{2}} + \frac{4}{3} \left(\frac{1}{3} e^{4y} \sin (3y) \bigg|_0^{\frac{\pi}{2}} - \frac{4}{3} \int_0^{\frac{\pi}{2}} e^{4y} \sin (3y) dy \right)$

You should notice that you get $I$ on the right hand side again. So juse use a little bit of algebra to solve for $I$:

${\color{red}I} = -\frac{1}{3}e^{4y}\cos (3y)\bigg|_0^{\frac{\pi}{2}} + \frac{4}{3} \left(\frac{1}{3} e^{4y} \sin (3y) \bigg|_0^{\frac{\pi}{2}} - \frac{4}{3} {\color{red}I} \right)$

3. Hello, Haris!

It is given that: . $\int^{\frac{\pi}{2}}_0 e^{4x}\sin(3x)\,dx \;=\; pe^{2\pi} + q$

Find the values of the constants $p$ and $q.$

Obviously, we do the integration first . . .

. . $\int^{\frac{\pi}{2}}_0e^{4x}\sin(3x)\,dx\;=\;\frac {e^{4x}}{25}\bigg[4\sin(3x) - 3\cos(3x)\bigg]^{\frac{\pi}{2}}_0$ ..
This takes more work than I want to show here

Evaluate: . $\frac{e^{4(\frac{\pi}{2})}}{25} \bigg[4\sin\tfrac{3\pi}{2} -3\cos\tfrac{3\pi}{2}\bigg] - \frac{e^0}{25}\bigg[4\sin0 - 3\cos0 \bigg]$

. . . . . . $= \;\frac{e^{2\pi}}{25}\bigg[4(-1) - 3(0)\bigg] - \frac{1}{25}\bigg[4(0) - 3(1)\bigg]$

. . . . . . $= \;\frac{e^{2\pi}}{25}(-4) - \frac{1}{25}(-3)$

. . . . . . $=\;-\frac{4}{25}e^{2\pi} + \frac{3}{25}$

Therefore: . $p \:=\:-\frac{4}{25},\;\;q \:=\:\frac{3}{25}$

4. $\int^\frac{\pi}{2}_0 e^{4x}sin(3x) dx$

but $sin(3x)=Im[e^{3ix}]$ so,

$\int^\frac{\pi}{2}_0 e^{4x}e^{3ix} dx$

$\int^\frac{\pi}{2}_0 e^{(4+3i)x} dx$

$I=\bigg[\frac{1}{4+3i} e^{(4+3i)x}\bigg]^\frac{\pi}{2}_0$

$I=\bigg[\frac{1}{25} (4-3i)e^{(4+3i)x}\bigg]^\frac{\pi}{2}_0$

$I=\bigg[\frac{1}{25} (4-3i)e^{4x}e^{3ix}\bigg]^\frac{\pi}{2}_0$

$I=\bigg[\frac{1}{25}e^{4x} (4-3i)(cos(3x)+isin(3x))\bigg]^\frac{\pi}{2}_0$

$I=\bigg[\frac{1}{25}e^{4x}(4cos(3x)+4isin(3x)-3icos(3x)+3sin(3x))\bigg]^\frac{\pi}{2}_0$

Now take the imaginary part of the result:

$I=\bigg[\frac{1}{25}e^{4x}(4sin(3x)-3cos(3x))\bigg]^\frac{\pi}{2}_0$

Is this method correct?

5. Yes, that's another way of doing it.

When (in particular) studying the improper integral $\int_0^\infty e^{-\alpha x}\sin(\beta x)\,dx$ its calculation is very easy by using that method.