First, your .
Then look at this.
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!
First, your .
Then look at this.
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.
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.
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!
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!
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
Mind youthat and now you follow the first point at this link(click).
Merry Christmas!