Use the substitution u = arctan(x) to evaluate int0^1 arctan(x)/1+x^2\,dx. Find the exact value of the integral.

Let u = arctan(x), so du = 1/1+x^2\,dx. This matches exactly the integrand, since arctan(x)/1+x^2\,dx = u\,du. Convert the bounds: when x=0, u = arctan(0) = 0; when x=1, u = arctan(1) = π/4. The integral becomes int0^π/4 u\,du = left[u^2/2right]0^π/4 = (π/4)^2/2 - 0 = π^2/32. So the value of the original integral is π^2/32.

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Definite integral via u-substitution with arctan

Evaluate the definite integral $\int_0^1 \frac{\arctan(x)}{1+x^2}\,dx$ using the substitution $u = \arctan(x)$ .

Use the substitution $u = \arctan(x)$ to evaluate $\int_0^1 \frac{\arctan(x)}{1+x^2}\,dx.$ Find the exact value of the integral.

Hint

Compute $du$ in terms of $dx$ : since $u = \arctan(x)$ , $du = \frac{1}{1+x^2}\,dx$ . This means the integral becomes $\int u\,du$ after substitution. Don't forget to convert the bounds: when $x=0$ , $u=\arctan(0)$ ; when $x=1$ , $u=\arctan(1)$ .

Step-by-step solution

Let $u = \arctan(x)$ , so $du = \frac{1}{1+x^2}\,dx$ . This matches exactly the integrand, since $\frac{\arctan(x)}{1+x^2}\,dx = u\,du$ . Convert the bounds: when $x=0$ , $u = \arctan(0) = 0$ ; when $x=1$ , $u = \arctan(1) = \pi/4$ .

The integral becomes $\int_0^{\pi/4} u\,du = \left[\frac{u^2}{2}\right]_0^{\pi/4} = \frac{(\pi/4)^2}{2} - 0 = \frac{\pi^2}{32}.$ So the value of the original integral is $\frac{\pi^2}{32}$ .

Definite integral via u-substitution with arctan

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