# Math Help - integral of ∫√(1+(4/9)*x^2)dx

1. ## integral of ∫√(1+(4/9)*x^2)dx

Hello!

I need help integrating ∫√(1+(4/9)*x^2)dx

x=(3/2)tan θ dx=sec^2 θ dθ

∫√(1+(4/9)*x^2)dx=∫√(1+(4/9)(9/4)tan^2 θ)dx=∫√(1+tan^2 θ)dx=∫√(1+tan^2 θ)*sec^2 θ dθ=∫√(sec^2 θ)*sec^2 θ)dθ=∫sec^3 θ dθ

Then I marked:

u=sec θ dv=sec^2 θ
du=sec θ*tan θ v=tan θ

∫sec^3 θ dθ=sec θ*tan θ-∫sec θ*tan θ*tan θ dθ
∫sec^3 θ dθ=sec θ*tan θ-∫sec θ*tan^2 θ dθ |tan^2=sec^2 θ-1
∫sec^3 θ dθ=sec θ*tan θ-∫sec θ*(sec^2 θ-1) dθ
∫sec^3 θ dθ=sec θ*tan θ-∫sec^3 θ dθ+∫sec θ dθ
2*∫sec^3 θ dθ=sec θ*tan θ+ln|sec θ+tan θ|+C
∫sec^3 θ dθ=(1/2)sec θ*tan θ+(1/2)ln|sec θ+tan θ|+C

x=(3/2)tan θ --> tan θ=(2/3)*x --> (4/9)*x^2=tan^2 θ --> (4/9)*x^2+1=1+tan^2 θ

(4/9)*x^2+1=sec^2 θ --> sec θ=√((4/9)*x^2+1)

So:

∫sec^3 θ dθ=∫√(1+(4/9)*x^2)dx

∫√(1+(4/9)*x^2)dx=(1/2)√((4/9)*x^2+1)*(2/3)*x+(1/2)ln|√((4/9)*x^2+1)+(2/3)x|+C

=(1/2)(2/3)(1/3)√(4*x^2+9)*x+(1/2)ln|(√(4*x^2+9)+2*x))/3|+C

=(1/9)*x*√(4*x^2+9)+(1/2)ln|(√(4*x^2+9)+2*x))/3|+C

I checked with my TI-89 calculator, I typed exactly this:

∫(√(1+(4/9)*x^2),x)

which gave:

∫(√(1+(4/9)*x^2),x)=(9*ln(abs(√(4*x^2+9)+2*x))+2*x*√(4*x^2+ 9))/12

so my answer is wrong. It gives a different value with same x compared to calculator's answer. Was x=(3/2)tan θ the mistake? Tho I don't see anything wrong with that.

2. ## Re: integral of ∫√(1+(4/9)*x^2)dx

Originally Posted by flyingdog
I need help integrating ∫√(1+(4/9)*x^2)dx
Code:
x=(3/2)tan θ       dx=sec^2 θ dθ
!
First, your $dx=\tfrac{3}{2}\sec^2(\theta)d\theta$.

Then look at this.

3. ## Re: integral of ∫√(1+(4/9)*x^2)dx

Do you have to try it by Integration of parts? If not, the following integrals from a list of integrals may be easier. I think those are the form you need.

Assume (ax2 + bx + c) cannot be reduced to the following expression (px + q)2 for some p and q.

$R = \sqrt{ax^2+bx+c}$

$\int R\,dx= \frac{2ax+b}{4a} R+ \frac{4ac-b^{2}}{8a} \int \frac{dx}{ R}$

$\int\frac{dx}{R} = \frac{1}{\sqrt{a}}\ln\left|2\sqrt{a}R+2ax+b\right| \qquad \mbox{(for }a>0\mbox{)}$

First simplify $\sqrt{1+(4/9)x^2},$ which should be $(1/3)\sqrt{4x^2+9},$, pull out the 1/3 and save it until the end, then use the formulas with a = 4, b = 0, and c = 9.

4. ## Re: integral of ∫√(1+(4/9)*x^2)dx

Hello, flyingdog!

(But I don't understand why it wrote 9/12.)

$\int \sqrt{1 + \tfrac{4}{9}x^2}\,dx$

Formula: . $\int\sec^3\!\theta\,d\theta \:=\:\tfrac{1}{2}\left(\sec\theta\tan\theta + \ln|\sec\theta + \tan\theta|\right) + C$

Let $\tfrac{2}{3}x \,=\,\tan\theta \quad\Rightarrow\quaddx \,=\,\tfrac{3}{2}\sec^2\!\theta\,d\theta$

. . $\sqrt{1 + \left(\tfrac{2}{3}x\right)^2} \:=\:\sqrt{1 + \tan^2\!\theta} \:=\:\sqrt{\sec^2\!\theta} \:=\:\sec\theta$

Substitute: . $\int \sec\theta\cdot\tfrac{3}{2}\sec^2\!\theta\,d\theta \:=\: \tfrac{3}{2}\int \sec^3\!\theta\,d\theta$

. . . . . . . . $=\;\tfrac{3}{2}\cdot\tfrac{1}{2}\big(\sec\theta \tan\theta + \ln|\sec\theta + \tan\theta|\big) + C$

Back-substitute: . $\tan\theta \,=\,\frac{2x}{3} \quad\Rightarrow\quad \sec\theta \,=\,\frac{\sqrt{4x^2+9}}{3}$

We have: . $\frac{3}{4}\left(\frac{\sqrt{4x^2+9}}{3}\cdot\frac {2x}{3} + \ln\left|\frac{4x^2+9}{3} + \frac{2x}{3}\right|\right) + C$

. . . . . . $=\;\frac{3}{4}\left(\frac{2x\sqrt{4x^2+9}}{9} + \ln\left|\frac{2x + \sqrt{4x^2+9}}{3}\right|\right) + C$

That final "3" vanishes in the final answer.
Someone can explain that for you.

5. ## Re: integral of ∫√(1+(4/9)*x^2)dx

Originally Posted by flyingdog
Hello!

I need help integrating ∫√(1+(4/9)*x^2)dx

x=(3/2)tan θ dx=sec^2 θ dθ

∫√(1+(4/9)*x^2)dx=∫√(1+(4/9)(9/4)tan^2 θ)dx=∫√(1+tan^2 θ)dx=∫√(1+tan^2 θ)*sec^2 θ dθ=∫√(sec^2 θ)*sec^2 θ)dθ=∫sec^3 θ dθ

Then I marked:

u=sec θ dv=sec^2 θ
du=sec θ*tan θ v=tan θ

∫sec^3 θ dθ=sec θ*tan θ-∫sec θ*tan θ*tan θ dθ
∫sec^3 θ dθ=sec θ*tan θ-∫sec θ*tan^2 θ dθ |tan^2=sec^2 θ-1
∫sec^3 θ dθ=sec θ*tan θ-∫sec θ*(sec^2 θ-1) dθ
∫sec^3 θ dθ=sec θ*tan θ-∫sec^3 θ dθ+∫sec θ dθ
2*∫sec^3 θ dθ=sec θ*tan θ+ln|sec θ+tan θ|+C
∫sec^3 θ dθ=(1/2)sec θ*tan θ+(1/2)ln|sec θ+tan θ|+C

x=(3/2)tan θ --> tan θ=(2/3)*x --> (4/9)*x^2=tan^2 θ --> (4/9)*x^2+1=1+tan^2 θ

(4/9)*x^2+1=sec^2 θ --> sec θ=√((4/9)*x^2+1)

So:

∫sec^3 θ dθ=∫√(1+(4/9)*x^2)dx

∫√(1+(4/9)*x^2)dx=(1/2)√((4/9)*x^2+1)*(2/3)*x+(1/2)ln|√((4/9)*x^2+1)+(2/3)x|+C

=(1/2)(2/3)(1/3)√(4*x^2+9)*x+(1/2)ln|(√(4*x^2+9)+2*x))/3|+C

=(1/9)*x*√(4*x^2+9)+(1/2)ln|(√(4*x^2+9)+2*x))/3|+C

I checked with my TI-89 calculator, I typed exactly this:

∫(√(1+(4/9)*x^2),x)

which gave:

∫(√(1+(4/9)*x^2),x)=(9*ln(abs(√(4*x^2+9)+2*x))+2*x*√(4*x^2+ 9))/12

so my answer is wrong. It gives a different value with same x compared to calculator's answer. Was x=(3/2)tan θ the mistake? Tho I don't see anything wrong with that.

In this case a hyperbolic substitution might be easier than a trigonometric substitution. Try \displaystyle \begin{align*} x = \frac{3}{2}\sinh{(t)} \implies dx = \frac{3}{2}\cosh{(t)}\,dt \end{align*}.

6. ## Re: integral of ∫√(1+(4/9)*x^2)dx

Originally Posted by Plato
First, your $dx=\tfrac{3}{2}\sec^2(\theta)d\theta$.

Then look at this.
I added dx=(3/2)sec^2 θ dθ to my equation. The factor (3/2) was missing from it and I got the correct answer. At start I needed to do:

∫√(1+(4/9)*x^2)dx=∫√(1+(4/9)(9/4)tan^2 θ)dx=∫√(1+tan^2 θ)dx=∫√(1+tan^2 θ)*(3/2)*sec^2 θ dθ=(3/2)*∫√(sec^2 θ)*sec^2 θ)dθ=(3/2)*∫sec^3 θ dθ

Then I intergrated normally

.
.
.

∫√(1+(4/9)*x^2)dx=(1/2)√((4/9)*x^2+1)*(2/3)*x+(1/2)ln|√((4/9)*x^2+1)+(2/3)x|+C

=(1/2)(2/3)(1/3)√(4*x^2+9)*x+(1/2)ln|(√(4*x^2+9)+2*x))/3|+C

=(1/9)*x*√(4*x^2+9)+(1/2)ln|(√(4*x^2+9)+2*x))/3|+C

now the factor (3/2):

(3/2)*∫√(1+(4/9)*x^2)dx=(3/2)*((1/9)*x*√(4*x^2+9)+(1/2)ln|(√(4*x^2+9)+2*x))/3|+C)

--> (1/6)*x*√(4*x^2+9)+(3/4)ln|(√(4*x^2+9)+2*x))/3|+C

I used the link you showed to me and WolframAlpha simplifies it with one hyperbolic function:

(1/6)*x*√(4*x^2+9)+(3/4)ln|(√(4*x^2+9)+2*x))/3|+C=(1/6)*x*√(4*x^2+9)+(3/4)ln|√(((2/3)*x)^2+9)+(2/3)*x)|+C

=(1/6)*x*√(4*x^2+9)+(3/4)(sinh((2/3)*x))^(-1)+C but I still like (1/6)*x*√(4*x^2+9)+(3/4)ln|(√(4*x^2+9)+2*x))/3|+C more.

7. ## Re: integral of ∫√(1+(4/9)*x^2)dx

Originally Posted by Spacemoss
Do you have to try it by Integration of parts? If not, the following integrals from a list of integrals may be easier. I think those are the form you need.

Assume (ax2 + bx + c) cannot be reduced to the following expression (px + q)2 for some p and q.

$R = \sqrt{ax^2+bx+c}$

$\int R\,dx= \frac{2ax+b}{4a} R+ \frac{4ac-b^{2}}{8a} \int \frac{dx}{ R}$

$\int\frac{dx}{R} = \frac{1}{\sqrt{a}}\ln\left|2\sqrt{a}R+2ax+b\right| \qquad \mbox{(for }a>0\mbox{)}$

First simplify $\sqrt{1+(4/9)x^2},$ which should be $(1/3)\sqrt{4x^2+9},$, pull out the 1/3 and save it until the end, then use the formulas with a = 4, b = 0, and c = 9.
Intergrating by parts is not necessary.

I used your formula and I got:

∫√(4*x^2+9)dx=((2*4*x+0)/4*4)* √(4*x^2+9)+((4*4*9-0^2)/(8*4))*(1/√4)*ln|(2*√4*√(4*x^2+9)+2*4*x)+C
=((8*x)/16)*√(4*x^2+9)+(144/32)*(1/2)*ln|(4*√(4*x^2+9)+8*x)|+C
=(1/2)*x*√(4*x^2+9)+(9/4)*ln|4*(√(4*x^2+9)+2*x)|+C

now the missing factor (1/3):

(1/3)* ∫√(4*x^2+9)dx=(1/3)*((1/2)*x*√(4*x^2+9)+(9/4)*ln|4*(√(4*x^2+9)+2*x)|)+C
--> (1/6)*x*√(4*x^2+9)+(3/4)*ln|4*(√(4*x^2+9)+2*x)|+C

Looks like it worked well! Fast and quite simple. But is this now possible?

(1/6)*x*√(4*x^2+9)+(3/4)*ln|4*(√(4*x^2+9)+2*x)|+C
=(1/6)*x*√(4*x^2+9)+(3/4)*ln|√(4*x^2+9)+2*x|+(3/4)*ln|4|+C
=(1/6)*x*√(4*x^2+9)+(3/4)*ln|√(4*x^2+9)+2*x|+C

to hide (3/4)*ln|4| behind C?

8. ## Re: integral of ∫√(1+(4/9)*x^2)dx

Hello and thank you Soroban for answering!

I did not know that I needed to add the factor (2/3) also to dx=sec^2 θ dθ.

So the final "3" can be just considered a part of C? Meaning (3/4)*ln|(1/3)|+C=C?

9. ## Re: integral of ∫√(1+(4/9)*x^2)dx

Originally Posted by Prove It
In this case a hyperbolic substitution might be easier than a trigonometric substitution. Try \displaystyle \begin{align*} x = \frac{3}{2}\sinh{(t)} \implies dx = \frac{3}{2}\cosh{(t)}\,dt \end{align*}.
This looks like a good way to go, but I got stuck..

x=(3/2)sinh t dx=(3/2)cosh t dt

∫√(1+(4/9)*x^2)dx=∫√(1+sinh^2 t)*(3/2)*cosh t dt
=(3/2)∫√(cosh^2 t)*cosh t dt=(3/2)*∫cosh^2 t dt
=(3/2)∫(1/2)*(cosh 2t+1)dt=(3/4)*(∫cosh 2t dt+∫1dt)
=(3/4)*((1/2)*sinh 2t+t+C)=(3/4)*(1/2)*2*sinh t*cosh t+(3/4)*t+C
=(3/4)*sinh t*cosh t+(3/4)*t+C

I looked alternative answers too, but the problem is the "t" which I do not seem to be able to convert.

10. ## Re: integral of ∫√(1+(4/9)*x^2)dx

Originally Posted by flyingdog
This looks like a good way to go, but I got stuck..

x=(3/2)sinh t dx=(3/2)cosh t dt

∫√(1+(4/9)*x^2)dx=∫√(1+sinh^2 t)*(3/2)*cosh t dt
=(3/2)∫√(cosh^2 t)*cosh t dt=(3/2)*∫cosh^2 t dt
=(3/2)∫(1/2)*(cosh 2t+1)dt=(3/4)*(∫cosh 2t dt+∫1dt)
=(3/4)*((1/2)*sinh 2t+t+C)=(3/4)*(1/2)*2*sinh t*cosh t+(3/4)*t+C
=(3/4)*sinh t*cosh t+(3/4)*t+C

I looked alternative answers too, but the problem is the "t" which I do not seem to be able to convert.

First of all, convert \displaystyle \begin{align*} \cosh{(t)} = \sqrt{1 + \sinh^2{(t)}} \end{align*} and also note you originally made the substitution \displaystyle \begin{align*} x = \frac{3}{2}\sinh{(t)} \end{align*}, so that means \displaystyle \begin{align*} t = \textrm{arsinh}\,{\left( \frac{2x}{3} \right) } \end{align*}...

11. ## Re: integral of ∫√(1+(4/9)*x^2)dx

Originally Posted by flyingdog
Looks like it worked well! Fast and quite simple. But is this now possible?

(1/6)*x*√(4*x^2+9)+(3/4)*ln|4*(√(4*x^2+9)+2*x)|+C
=(1/6)*x*√(4*x^2+9)+(3/4)*ln|√(4*x^2+9)+2*x|+(3/4)*ln|4|+C
=(1/6)*x*√(4*x^2+9)+(3/4)*ln|√(4*x^2+9)+2*x|+C

to hide (3/4)*ln|4| behind C?

Mind youthat $\int \sqrt{1 + \frac{4}{9}x^2}\,dx = \frac{2}{3}\int\sqrt{x^2+\left(\frac{3}{2}\right)^ 2}\,dx$ and now you follow the first point at this link(click).