What is the definite integral of 1/(36+x^2) with bounds [0, 6]

[Chandasouk]

Homework Statement

What is the definite integral of 1/(36+x^2) with bounds [0, 6]?

I've only been taught U substitution to handle problems like these. I let u = 36+x^2 and du=2xdx but I am stuck and don't know what to do. Te answer is pi/24 but I don't know how to obtain it.

Is there a better u to choose? if so, what?

 

[gabbagabbahey]

Hint: What is $\frac{d}{du}\tan^{-1}(u)$?

 

[Squeezebox]

U substitution isn't needed. Look at an integration table.

Hint: The function being integrated takes the shape of 1/(x²+a²)

 

[Chandasouk]

Okay, I used that formula u gave me although in my book it is written as

du/a^2+u^2 = 1/a*tan^-1(u/a)+C

But I had to evaluate it with a calculator [0, 6]

1/6*tan^-1(1)-1/6*tan^-1(0) = 0.1308 which is about pi/24

How would I solve it without a calculator?

 

[Squeezebox]

Okay, I used that formula u gave me although in my book it is written as

du/a^2+u^2 = 1/a*tan^-1(u/a)+C

But I had to evaluate it with a calculator [0, 6]

1/6*tan^-1(1)-1/6*tan^-1(0) = 0.1308 which is about pi/24

How would I solve it without a calculator?

Well, 1/6*tan^-1(1)-1/6*tan^-1(0)=1/6*tan^-1(1)-0=1/6*tan^-1(1). So I guess you would need to know that tan^-1(1)=$$\pi$$/4.

 

[gabbagabbahey]

Keep in mind that $u=\tan^{-1}(x)$ means that $\tan(u)=\frac{\sin(u)}{\cos(u)}=x$...so where does $\frac{\sin(u)}{\cos(u)}$ equal zero? Where does it equal one?

 

[physicsman2]

factor out a 36 in the denominator and in the denominator you'll get: 36(1+(x/6)^2)

let u = x/6 and when you take the derivative and substitute in the original, you should get:

(1/6)[1/(1 + u^2)] and integrate, remembering that 1/(1 + u^2) is tan^-1 u after you integrate, then you can go from there.

 

[SgtSniper90]

answer after you integrate is (1/6)arctan(x/6) + C..

then just evaluate it for 0 to 6

(1/6)arctan(6/6)-[(1/6)arctan(0/6)]

[(1/6)*(Pi/4)]-0= Pi/24