integral of ∫√(1+(4/9)*x^2)dx

Hello!

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

I already tried this:

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.

Please help, thanks!

Re: integral of ∫√(1+(4/9)*x^2)dx

Quote:

Originally Posted by

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

I already tried this:

Code:

`x=(3/2)tan θ dx=sec^2 θ dθ`

!

First, your .

Then look at this.

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.

First simplify which should be , pull out the 1/3 and save it until the end, then use the formulas with a = 4, b = 0, and c = 9.

Re: integral of ∫√(1+(4/9)*x^2)dx

Hello, flyingdog!

The calculator's answer is correct.

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

Formula: .

Let

. .

Substitute: .

. . . . . . . .

Back-substitute: .

We have: .

. . . . . .

That final "3" vanishes in the final answer.

Someone can explain that for you.

Re: integral of ∫√(1+(4/9)*x^2)dx

Quote:

Originally Posted by

**flyingdog** Hello!

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

I already tried this:

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.

Please help, thanks!

In this case a hyperbolic substitution might be easier than a trigonometric substitution. Try .

Re: integral of ∫√(1+(4/9)*x^2)dx

Quote:

Originally Posted by

**Plato**

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.

Thanks for your answer!

Re: integral of ∫√(1+(4/9)*x^2)dx

Quote:

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.

First simplify

which should be

, 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?

Thank you for answering!

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?

Re: integral of ∫√(1+(4/9)*x^2)dx

Quote:

Originally Posted by

**Prove It** In this case a hyperbolic substitution might be easier than a trigonometric substitution. Try

.

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.

Thanks for answering

Re: integral of ∫√(1+(4/9)*x^2)dx

Quote:

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.

Thanks for answering

First of all, convert and also note you originally made the substitution , so that means ...

Re: integral of ∫√(1+(4/9)*x^2)dx

Quote:

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?

Thank you for answering!

Yep, constant + constant = constant, so it's still just c. Now just add the parts with a common denominator and you get what the calculator said.

Re: integral of ∫√(1+(4/9)*x^2)dx

Mind youthat and now you follow the first point at this link(click).

Merry Christmas!