1. Solved : Help Calculas

Thanks guys for the help provided.

2. Hello, tester85!

I'll explain the first one for you.
It's quite challenging and uses a technique you may not have seen.

$I \;=\;\int e^{ax}\sin bx\,dx$
Integrate by parts . . .

. . $\begin{array}{ccccccc}u &=& \sin bx & & dv &=& e^{ax}dx \\ du &=& b\cos bx\,dx & & v &=&\frac{1}{a}e^{ax} \end{array}$

We have: . $I \;=\;\tfrac{1}{a}\,e^{ax}\sin bx - \tfrac{b}{a}\int\! e^{ax}\cos bx\,dx$

Integrate by parts (again):

. . $\begin{array}{ccccccc}u &=&\cos bx & & dv &=& e^{ax}dx \\ du &=& \text{-}b\sin bx\,dx & & v &=& \frac{1}{a}e^{ax} \end{array}$

We have: . $I \;=\;\tfrac{1}{a}\,e^{ax}\sin bx - \tfrac{b}{a}\bigg[\tfrac{1}{a}e^{ax}\cos bx + \tfrac{b}{a}\int\! e^{ax}\sin bx\,dx\bigg] + C$

. . $I \;=\;\tfrac{1}{a},e^{ax}\sin bx - \tfrac{b}{a^2}\,e^{ax}\cos bx - \tfrac{b^2}{a^2}\underbrace{\int\! e^{ax}\sin bx\,dx}_{\text{This is }I} + C$

. . $I \;=\;\tfrac{1}{a}\,e^{ax}\sin bx - \tfrac{b}{a^2}\,e^{ax}\cos bx - \tfrac{b^2}{a^2}\,I + C$

Multiply by $a^2\!:\;\;a^2I \;=\;ae^{ax}\sin bx - be^{ax}\cos bx - b^2I + C$

. . . . . . . . $a^2I + b^2I \;=\;ae^{ax}\sin bx - be^{ax}\cos bx + C$

Factor: . $(a^2+b^2)I \;=\;e^{ax}(a\sin bx - b\cos bx) + C$

. . . . . . . . . . . $I \;=\;\frac{e^{ax}}{a^2+b^2}(a\sin bx - b\cos bx) + C$

Therefore: . $\int e^{ax}\sin bx\,dx \;=\;\frac{e^{ax}}{a^2+b^2}(a\sin bx - b\cos bx) + C$

3. alternatively you can use the well known identity:

$
\sin bx = \frac{{e^{jbx} - e^{ - jbx} }}
{{2j}}$

so the integral becomes a simple combination of exponents:

$
\int {e^{ax} \sin bxdx = \frac{1}
{{2j}}\int {e^{x(a + jb)} - e^{x(a - jb)} dx} }
$

$\frac{1}
{{2j}}\int {e^{x(a + jb)} - e^{x(a - jb)} dx} = \frac{1}
{{2j}}\left[ {\frac{1}
{{a + jb}}e^{x(a + jb)} - \frac{1}
{{a - jb}}e^{x(a - jb)} } \right] =
$

$
= \frac{1}
{{2j}}e^{ax} \left[ {\frac{{ae^{jbx} - jbe^{jbx} - ae^{ - jbx} - jbe^{ - jbx} }}
{{a^2 + b^2 }}} \right] = \frac{{e^{ax} }} {{a^2 + b^2 }}\left( {a\sin bx - b\cos bx} \right)
$

4. Thanks Soroban and peritus. Can someone guide me with the rest, i have done it but unable to get the final solution.

5. $\int{\frac{x^2}{(x-1)(x+1)^2}\,dx}$

Use partial fractions.

$\frac{x^2}{(x-1)(x+1)^2} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{(x+1)^2}$

$= \frac{A(x+1)^2 + B(x-1)(x+1) + C(x-1)}{(x-1)(x+1)^2}$

Since the denominators are the same, the numerators are equal...

So $x^2 = A(x+1)^2 + B(x-1)(x+1) + C(x-1)$

Expand and you get

$x^2 + 0x + 0 = (A + B)x^2 + (2A + C)x + A - B - C$

Equating like powers of x gives

$A + B = 1, 2A + C = 0, A - B - C = 0$.

Solving the equations simultaneously gives

$A = \frac{1}{4}, B = \frac{3}{4}, C = -\frac{1}{2}$.

So $\frac{x^2}{(x-1)(x+1)^2} = \frac{1}{4} \left(\frac{1}{x-1}\right) + \frac{3}{4} \left(\frac{1}{x+1}\right) - \frac{1}{2} \left(\frac{1}{(x+1)^2}\right)$

$\int{\frac{x^2}{(x-1)(x+1)^2}\,dx} = \frac{1}{4} \int{\frac{1}{x-1}\,dx} + \frac{3}{4} \int{\frac{1}{x+1}\,dx} - \frac{1}{2} \int{\frac{1}{(x+1)^2}\,dx}$

$= \frac{1}{4} \ln |x-1| + \frac{3}{4} \ln |x + 1| + \frac{1}{2(x+1)} + c$

6. Hi prove it.

I got the same answer as what you got, but the answer given is

1/4 ln|(x-1)(x+1)^3|+1/2(x+1)+C

7. Originally Posted by tester85
Hi prove it.

I got the same answer as what you got, but the answer given is

1/4 ln|(x-1)(x+1)^3|+1/2(x+1)+C
They've used the log laws...

$\log m + \log n = \log mn$ and $p\log m = \log{m^p}$

8. Ok now i get it thanks.

Therefore, 3/4ln(x+1)=1/4ln(x+1)^3

9. For the last one, you need to set up the partial fractions as

$\frac{A}{x+1} + \frac{Bx+C}{x^2 + 1}$.

You should find $A = \frac{1}{2}, B = -\frac{1}{2}, C = \frac{1}{2}$.

Knowledge that $\frac{-x + 1}{x^2 + 1} = \frac{-x}{x^2 + 1} + \frac{1}{x^2 + 1}$ should also help you out.

10. I am able to do the partial fractions part but the final answer is wierd, it is given as 1/4ln2 + pi/8 . Where did the pi come from ?

11. Originally Posted by tester85
I am able to do the partial fractions part but the final answer is wierd, it is given as 1/4ln2 + pi/8 . Where did the pi come from ?

12. Ok noted with thanks prove it. Will try again later. Does anyone has any tips for the second integration question ? The answer given is 4 and i am not able to get it.

13. That depends.

Is it...

$\sqrt{1 - \cos{\frac{x}{2}}}$ or $\sqrt{1 - \frac{\cos{x}}{2}}$ or $\sqrt{\frac{1 - \cos{x}}{2}}$?

14. Originally Posted by tester85
integrate from 2pi to 0 sqrt(1-cosx/2)
I'd like to help, but I'm not sure if my interpretation is correct. Do you mean

$\int_0^{2\pi}\sqrt{\frac{1-\cos x}{2}}\,dx$ or

$\int_0^{2\pi}\sqrt{1-\frac{\cos x}{2}}\,dx$ or

$\int_0^{2\pi}\sqrt{1-\cos \left(\frac{x}{2}\right)}\,dx$??

--Chris

15. Originally Posted by Chris L T521
I'd like to help, but I'm not sure if my interpretation is correct. Do you mean

$\int_0^{2\pi}\sqrt{\frac{1-\cos x}{2}}\,dx$ or

$\int_0^{2\pi}\sqrt{1-\frac{\cos x}{2}}\,dx$ or

$\int_0^{2\pi}\sqrt{1-\cos \left(\frac{x}{2}\right)}\,dx$??

--Chris
It is this bro. $\int_0^{2\pi}\sqrt{\frac{1-\cos x}{2}}\,dx$

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