evaluate the definite integral. enter exact values.\n int_{0}^{1}-\frac{2cot^{-1}(x)}{1 + x^{2}}dx=

Explanation:

Step1: Use substitution

Let $u = \cot^{-1}(x)$. Then $du=-\frac{1}{1 + x^{2}}dx$.

Step2: Change the limits of integration

When $x = 0$, $u=\cot^{-1}(0)=\frac{\pi}{2}$. When $x = 1$, $u=\cot^{-1}(1)=\frac{\pi}{4}$.

Step3: Rewrite the integral

The integral $\int_{0}^{1}-\frac{2\cot^{-1}(x)}{1 + x^{2}}dx$ becomes $2\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}u\ du$.

Step4: Integrate $u$

Using the power - rule for integration $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), for $y = u$, $\int u\ du=\frac{u^{2}}{2}+C$. So $2\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}u\ du=2\left[\frac{u^{2}}{2}\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$.

Step5: Evaluate the definite integral

$2\left[\frac{u^{2}}{2}\right]{\frac{\pi}{4}}^{\frac{\pi}{2}}=\left[u^{2}\right]{\frac{\pi}{4}}^{\frac{\pi}{2}}=\left(\frac{\pi}{2}\right)^{2}-\left(\frac{\pi}{4}\right)^{2}=\frac{\pi^{2}}{4}-\frac{\pi^{2}}{16}=\frac{4\pi^{2}-\pi^{2}}{16}=\frac{3\pi^{2}}{16}$.

Answer:

$\frac{3\pi^{2}}{16}$