This question was by Hadjer2015

But it didn't post properly so I am doing it here.

please help me

$$\int$$(1-cos(x/3))/sin(x/2)

$$\int( x^{2}) /((x^2+3)(x+1))$$

[address of original question]

Melody
Mar 22, 2015

#8**+5 **

I looked for a way to get the sin and cos terms the "same" in some sense, and I could see that if we had integer multiples of some common angle we could expand the result to get both in terms of the one angle.

There might well be a more straightforward way of doing the integral!

.

Alan
Mar 22, 2015

#1**+5 **

$$\\\displaystyle\int \frac{x^2}{(x^2+3)(x+1)}\;dx\\\\

Let \\

\frac{x^2}{(x^2+3)(x+1)}=\frac{Ax+B}{x^2+3}+\frac{C}{x+1}\\\\

\frac{x^2}{(x^2+3)(x+1)}=\frac{(Ax+B)(x+1)}{(x^2+3)(x+1)}+\frac{C(x^2+3)}{(x^2+3)(x+1)}\\\\

so\\

(Ax+B)(x+1)+C(x^2+3)=x^2\\\\

Ax^2+Ax+Bx+B+Cx^2+3C=x^2\\\\

(A+C)x^2+(A+B)x+(B+3C)=x^2\\\\$$

$$\\so\\

A+C=1 \;(1)\;\rightarrow \\

A+B=0\;(2)\; \rightarrow \\

B+3C=0\;\; \rightarrow\;\;B=-3C\;\; (3) \;\;$sub into (2)$\\

A-3C=0 (4)\\

(1)-(4)\\

4C=1\\

C=\frac{1}{4}=0.25 \\

A=1-0.25=0.75\\

B=-3*0.25=-0.75\\\\$$

$$\\so\\

\frac{x^2}{(x^2+3)(x+1)}=\frac{(0.75x-0.75)(x+1)}{(x^2+3)(x+1)}+\frac{0.25(x^2+3)}{(x^2+3)(x+1)}\\\\

\frac{x^2}{(x^2+3)(x+1)}=\frac{(0.75x-0.75)}{(x^2+3)}+\frac{0.25}{(x+1)}\\\\

\frac{x^2}{(x^2+3)(x+1)}=\frac{0.75x}{(x^2+3)}-\frac{0.75}{(x^2+3)}+\frac{0.25}{(x+1)}\\\\

\frac{x^2}{(x^2+3)(x+1)}=\frac{0.375*2x}{(x^2+3)}-\frac{0.75}{(x^2+3)}+\frac{0.25}{(x+1)}\\\\\\

\displaystyle\int\;\frac{x^2}{(x^2+3)(x+1)}\;dx\\\\\\

=\displaystyle\int\;\frac{0.375*2x}{(x^2+3)}\;dx-\displaystyle\int\;\frac{0.75}{(x^2+3)}\;dx+\displaystyle\int\;\frac{0.25}{(x+1)}\;dx\\\\\\

=0.375\;ln(x^2+3)+0.35\;ln(x+1)-\displaystyle\int\;\frac{0.75}{(x^2+3)}\;dx\\\\\\

$ Now I don't know what to do. BUMMER!$$$

**I guess I wasn't meant to do it that way afterall. LOL**

Melody
Mar 22, 2015

#2**+5 **

Melody....for that last integral.....let u = x , a = √3, du = dx, and we have the form.....

-0.75 ∫ 1 / [ u^2 + a^2] du =

-0.75 [ tan^{-1} (u/a)] / a + C =

-.0.75 [tan^{ -1} (x / √3)] / √3 + C

CPhill
Mar 22, 2015

#4**+5 **

Well...you got most of the difficult part done with the partial fraction stuff....!!!!

That's worth at least 3 points.........

P.S. - I don't want to appear as though I'm a genius.....I went to the integral tables to find the proper form for that last part.....LOL!!!

CPhill
Mar 22, 2015

#5**+5 **

For the first integral:

This should be carefully checked as I tend to be a little cavalier with constants sometimes!

.

Alan
Mar 22, 2015

#7**0 **

Nice "double-substitution," Alan.

P.S - How did it occur to you to let cos(x/3) = cos(2Θ) and sin(x/ 2) = sin(3Θ) ???

CPhill
Mar 22, 2015

#8**+5 **

Best Answer

I looked for a way to get the sin and cos terms the "same" in some sense, and I could see that if we had integer multiples of some common angle we could expand the result to get both in terms of the one angle.

There might well be a more straightforward way of doing the integral!

.

Alan
Mar 22, 2015