Final answer:
To find the integral of ∫(1/(1+x²))dx from negative infinity to 0, we can use the trigonometric substitution method. Let's substitute x = tan(t). The integral becomes ∫(1/(1+tan²(t)))sec²(t) dt. Using the trigonometric identity sec²(t) = 1 + tan²(t), we can simplify the integral to ∫dt. The solution is π/2.
Step-by-step explanation:
To find the integral of ∫(1/(1+x²))dx from negative infinity to 0, you can use the trigonometric substitution method. Let's substitute x = tan(t). The integral becomes ∫(1/(1+tan²(t)))sec²(t) dt.
Using the trigonometric identity sec²(t) = 1 + tan²(t), we can simplify the integral to ∫dt. Since t = arctan(x), the limits of integration change to arctan(0) = 0 and arctan(-∞) = -π/2.
Therefore, the integral is simply the difference of the limits: 0 - (-π/2) = π/2.