Integral of 1/(x^2 + 36)dx When and how to draw triangle?

[Lo.Lee.Ta.]

1. ∫1/(x² + 36)dx


2. I started by trying a trig substitution.

The normal form, "a² + x², x=(a)tan(θ)," I thought could be reversed here:

x² + 6²
x = 6tan(θ)
dx= 6sec²θdθ


∫1/[(6tanθ)² + 36] = ∫1/[36(tan²θ + 1)]*6sec²θdθ

= ∫1/[36sec²θ]*6sec²θdθ

= ∫1/6dθ

= (1/6)θ + C

From before, θ= arctan(1/6(x))

1/6(arctan(1/6(x))) + C

In my class, sometimes the professor says we need to draw a triangle to figure out the value...

I'm a little confused about that. Would I need to draw a triangle here to figure out an exact value?

Since tan(1/6(x))=θ, do I draw a triangle and say the side opposite θ equals 1, and the adjacent side equals 6?

And then do I figure out from that what arctanθ equals...?
Would you help me with the triangle thing?

Thank you so much! :D

 

[voko]

What value do you need to figure out?

 

[Jazzermazzer]

1. ∫1/(x² + 36)dx2. I started by trying a trig substitution.

The normal form, "a² + x², x=(a)tan(θ)," I thought could be reversed here:

x² + 6²
x = 6tan(θ)
dx= 6sec²θdθ∫1/[(6tanθ)² + 36] = ∫1/[36(tan²θ + 1)]*6sec²θdθ

= ∫1/[36sec²θ]*6sec²θdθ

= ∫1/6dθ

= (1/6)θ + C

From before, θ= arctan(1/6(x))

1/6(arctan(1/6(x))) + C

In my class, sometimes the professor says we need to draw a triangle to figure out the value...

I'm a little confused about that. Would I need to draw a triangle here to figure out an exact value?

Since tan(1/6(x))=θ, do I draw a triangle and say the side opposite θ equals 1, and the adjacent side equals 6?

And then do I figure out from that what arctanθ equals...?
Would you help me with the triangle thing?

Thank you so much! :D

I used partial fractions... But it's a bit odd

X^2+36=(x+6i)(x-6i)

1=A(x+6i) + B(x-6i)
A= -1/12i , B = 1/12i
1/i= -i
i/12 Integ 1/(x+6i) -1/(x-6i) dx

i/12 ln |(x+6i)/(x-6i)| + c