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!