integration by substitution question (1 Viewer)

nrlwinner · May 1, 2010

Can somebody help me with this one.

$\int \frac{dx}{(4+x^2)^\frac{3}{2}}$
using $x=2tan\theta$

shaon0 · May 1, 2010

nrlwinner said:
Can somebody help me with this one.

$\int \frac{dx}{(4+x^2)^\frac{3}{2}}$
using $x=2tan\theta$

I=S dx/(4+x^2)^3/2
Let x=2tan@, dx=2sec^2@ d@
I=S 2sec^2 @ d@ / 4^(3/2).sec ^3 @
=1/4 S cos@ d@
=1/4.sin@ +C
= 1/4.sin(atan(x/2)) +C
= 1/4.x/sqrt(4+x^2) +C

nrlwinner · May 1, 2010

Thanks. I got another one here. I've tried so hard but I can't seem to get the answer which is 7.

$\int_{3}^{8}\frac{x}{(x+1)-\sqrt{x+1}}dx$

I've been using the substitution u^2 = x+1

shaon0 · May 1, 2010

nrlwinner said:
Thanks. I got another one here. I've tried so hard but I can't seem to get the answer which is 7.

$\int_{3}^{8}\frac{x}{(x+1)-\sqrt{x+1}}dx$

I've been using the substitution u^2 = x+1

I=S[3,8] (x+1-1)/sqrt(x+1)(sqrt(x+1)-1) dx
Let u^2=x+1, 2u du=dx:
I=S[2,3] (u^2-1)/u(u-1) 2u du
=2 S[2,3] (u+1) du
=2 [1/2u^2+u] {2,3}
=2 (9/2+3-2-2)
=2(1/2+3)
=7

nrlwinner · May 1, 2010

Can someone show me where to go next. I've done a question up to here.

$\frac{1}{2}\int_{0}^{1}\frac{1-t^2}{t^4+t^2+1}$

I've been using t=tanx/2 and I'm trying to prove for 1/4ln3. If it's impossible from here, please tell me.

Drongoski · May 1, 2010

Not very straightforward. I'll skip the details.

$\int \frac{1-t^{2}}{t^{4}+t^{2}+1}\ dt\ \ =\ \ \int \frac{1-t^{2}}{(t^{2}+t+1)(t^{2}-t+1)} \ dt \\ \\ \\ = \ \ \int \frac{t+\frac{1}{2}}{t^{2}+t+1}\ -\ \frac{t-\frac{1}{2}}{t^{2}-t+1}\ dt\\ \\ =\ \ \frac{1}{2}ln\left | t^{2}+t+1 \right |\ -\ \frac{1}{2}ln\left | t^{2}-t+1 \right |\ +\ C\\ \\ \therefore \int_{0}^{1}\ XXX\ dt\ \ =\ \ \frac{1}{2}\left [ ln\left | \frac{t^{2}+t+1}{t^{2}-t+1} \right | \right ]^{1}_{0}\ \ =\ \ \frac{1}{2}ln\ 3$

OK . . . 1/2 my integral = 1/4 ln 3.

nrlwinner · May 1, 2010

Drongoski said:
Not very straightforward. I'll skip the details.

$\int \frac{1-t^{2}}{t^{4}+t^{2}+1}\ dt\ \ =\ \ \int \frac{1-t^{2}}{(t^{2}+t+1)(t^{2}-t+1)} \ dt \\ \\ \\ = \ \ \int \frac{t+\frac{1}{2}}{t^{2}+t+1}\ -\ \frac{t-\frac{1}{2}}{t^{2}-t+1}\ dt\\ \\ =\ \ \frac{1}{2}ln\left | t^{2}+t+1 \right |\ -\ \frac{1}{2}ln\left | t^{2}-t+1 \right |\ +\ C\\ \\ \therefore \int_{0}^{1}\ XXX\ dt\ \ =\ \ \frac{1}{2}\left [ ln\left | \frac{t^{2}+t+1}{t^{2}-t+1} \right | \right ]^{1}_{0}\ \ =\ \ \frac{1}{2}ln\ 3$

OK . . . 1/2 my integral = 1/4 ln 3.

Brilliant. But how did you know to divide the denominator like that because it is not reducible. (1st line I mean)

shaon0 · May 2, 2010

nrlwinner said:
Brilliant. But how did you know to divide the denominator like that because it is not reducible. (1st line I mean)

Terms of odd powers cancel, so the middle term of the quadratic has to be minus in this case.

Drongoski · May 2, 2010

nrlwinner said:
Brilliant. But how did you know to divide the denominator like that because it is not reducible. (1st line I mean)

Yes! But not a soul on bos has hired me to tutor!

Drongoski · May 2, 2010

Re 1st line of my solution:

$t^{4}+t^{2}+1\ =\ t^{4}+2t^{2}+1\ -\ t^{2}\ =\ (t^{2}+1)^{2}-t^{2}\\ =\ \ (t^{2}+t+1)(t^{2}-t+1)\\ \\Convert \ to\ \ partial\ fractions:\ \ Let\ \ \frac{1-t^{2}}{t^{4}+t^{2}+1}\ =\frac {At+B}{t^{2}-t+1}+\frac{Ct+D}{t^{2}+t+1}\\ \\ \therefore \ \ 1-t^{2}\ =\ (At+B)(t^{2}+t+1)\ +\ (Ct+D)(t^{2}-t+1)\\t=0\ \ \Rightarrow\ B+D=1\\t=1\ \ \Rightarrow\ 3(A+B)+(C+D)=0\\t=-1\ \Rightarrow \ -A+B\ +\ 3(-C+D) =0\\t=i\ \ \Rightarrow \ \ (Ai+B)(-1+i+1)\ +\ (Ci+D)(-1-i+1)\\ \ \ \Rightarrow \ \ B-D=0\ \ and\ \ C-A=2\\ \\ Solving\ \Rightarrow \ \ A=-1\ \ \ B=\frac{1}{2}\ \ \ C=1\ \ \ D=\frac{1}{2}$

integration by substitution question (1 Viewer)

nrlwinner

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