+0

This question was by Hadjer2015 Difficult Integration

0
230
8
+91435

This question was by Hadjer2015

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

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

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

[address of original question]

http://web2.0calc.com/questions/nbsp_20

Melody  Mar 22, 2015

#8
+26399
+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
Sort:

#1
+91435
+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
+80935
+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
#3
+91435
0

Thanks Chris, I should have recognised that. :)

Melody  Mar 22, 2015
#4
+80935
+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
+26399
+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
#6
+91435
0

Thanks Alan. :)

Melody  Mar 22, 2015
#7
+80935
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
+26399
+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

17 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details