(e) evaluate the definite integral\n int_{x = 1}^{x=pi/2} left(sqrt{x}-\frac{1}{x}-csc^{2}(x)\right) dx

Explanation:

Step1: Split the integral

By the property of definite - integrals $\int_{a}^{b}(f(x)-g(x)-h(x))dx=\int_{a}^{b}f(x)dx-\int_{a}^{b}g(x)dx-\int_{a}^{b}h(x)dx$. So, $\int_{1}^{\frac{\pi}{2}}(\sqrt{x}-\frac{1}{x}-\csc^{2}(x))dx=\int_{1}^{\frac{\pi}{2}}\sqrt{x}dx-\int_{1}^{\frac{\pi}{2}}\frac{1}{x}dx-\int_{1}^{\frac{\pi}{2}}\csc^{2}(x)dx$.

Step2: Integrate each term

1. For $\int_{1}^{\frac{\pi}{2}}\sqrt{x}dx$, since $\sqrt{x}=x^{\frac{1}{2}}$, and $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, then $\int_{1}^{\frac{\pi}{2}}x^{\frac{1}{2}}dx=\left[\frac{2}{3}x^{\frac{3}{2}}\right]_{1}^{\frac{\pi}{2}}=\frac{2}{3}(\frac{\pi}{2})^{\frac{3}{2}}-\frac{2}{3}(1)^{\frac{3}{2}}=\frac{2}{3}(\frac{\pi^{\frac{3}{2}}}{2\sqrt{2}})-\frac{2}{3}=\frac{\pi^{\frac{3}{2}}}{3\sqrt{2}}-\frac{2}{3}$.
2. For $\int_{1}^{\frac{\pi}{2}}\frac{1}{x}dx$, since $\int\frac{1}{x}dx=\ln|x|+C$, then $\int_{1}^{\frac{\pi}{2}}\frac{1}{x}dx=[\ln x]_{1}^{\frac{\pi}{2}}=\ln\frac{\pi}{2}-\ln1=\ln\frac{\pi}{2}$.
3. For $\int_{1}^{\frac{\pi}{2}}\csc^{2}(x)dx$, since $\int\csc^{2}(x)dx=-\cot x + C$, then $\int_{1}^{\frac{\pi}{2}}\csc^{2}(x)dx=[-\cot x]_{1}^{\frac{\pi}{2}}=-\cot\frac{\pi}{2}+\cot1 = 0+\cot1=\cot1$.

Step3: Combine the results

$\int_{1}^{\frac{\pi}{2}}(\sqrt{x}-\frac{1}{x}-\csc^{2}(x))dx=\frac{\pi^{\frac{3}{2}}}{3\sqrt{2}}-\frac{2}{3}-\ln\frac{\pi}{2}-\cot1$.

Answer:

$\frac{\pi^{\frac{3}{2}}}{3\sqrt{2}}-\frac{2}{3}-\ln\frac{\pi}{2}-\cot1$