how to solve this definite integral 6/(1+x^2) from -1 to 1?

Answer 1

∫[from -1 to 1 of] 6 dx/(1+ x²) =
6 * arctan(x) | from x = -1 to 1 =
6 (arctan(1) - arctan(-1)) =
6 (π/4 - (-π/4)) =
3π

Answer 2

This is a trick "out of the blue" that you come to recognize over time as useful when you see (1+x^2)^(-1):

Write x = tan t. Then dx = (1 + tan^2(t))dt, so dt = dx/(1+tan^2(t)). Since x = tan t, this says that dt = dx/(1+x^2).

In doing a substitution, we need to change the bounds of integration. I think it's easiest to understand what interval for t to use if you look at the graph of the tangent function. The interval you want is [-π/4, π/4] since tan(-π/4) = -1 and tan(π/4) = 1, and tan t goes through all the values between -1 and 1 as t goes from -π/4 to π/4.

So, using "int(a,b)f(x) dx" to mean "the integral of f(x) with respect to x from a to b," we have the following:

int(-1,1)6/(1+x^2) dx
= 6*int(-1,1)dx/(1+x^2)
= 6*int(-π/4,π/4) dt
= 6(π/4 - (-π/4))
= 3π

Answer 3

I = 6. tan^(-1)x between -1 and 1
I = 6[(tan^(-1).1 - tan^(-1).(- 1)]
I = 6[π/4 + π/4 ]
I = 6π/2 = 3π