use a change of variables or the table to evaluate the following definite integral\n int_{\frac{pi}{2}}^{\fra…

Explanation:

Step1: Set substitution variable

Let $u = \sin x$, then $du=\cos xdx$.

Step2: Change the limits of integration

When $x = \frac{\pi}{2}$, $u=\sin(\frac{\pi}{2}) = 1$. When $x=\frac{3\pi}{4}$, $u=\sin(\frac{3\pi}{4})=\frac{\sqrt{2}}{2}$.

Step3: Rewrite the integral

The integral $\int_{\frac{\pi}{2}}^{\frac{3\pi}{4}}\frac{\cos x}{\sin^{2}x}dx$ becomes $\int_{1}^{\frac{\sqrt{2}}{2}}\frac{du}{u^{2}}=\int_{1}^{\frac{\sqrt{2}}{2}}u^{- 2}du$.

Step4: Integrate using power - rule

The antiderivative of $u^{-2}$ is $-u^{-1}+C=-\frac{1}{u}+C$.

Step5: Evaluate the definite integral

$-\left[\frac{1}{u}\right]_{1}^{\frac{\sqrt{2}}{2}}=-\left(\frac{1}{\frac{\sqrt{2}}{2}}-\frac{1}{1}\right)=-( \sqrt{2}-1)=1 - \sqrt{2}$.

Answer:

$1-\sqrt{2}$