Question

Solve the definite integral.

${\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}xcs{c}^{2}xdx$

Short answer

The solution is$\frac{1}{4}\left(\mathrm{\pi }+2\log 2\right)$.

Step-by-step solution

Step 1. Given information.

The given integral is${\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}xcs{c}^{2}xdx$.

Step 2. First, solve the indefinite integral.

$\begin{array}{rcl}I& =& \int xcs{c}^{2}xdx\\ & =& x\int cs{c}^{2}xdx-\int \left[\frac{d}{dx}\left(x\right)\int cs{c}^{2}xdx\right]dx\\ & =& x(-cotx)-\int 1(-cotx)dx\\ & =& -xcotx+\int cotxdx\\ & =& -xcotx+\log \left|\sin x\right|\dots \dots \dots \dots \left\{\int cotxdx=\log \left|\sin x\right|\right\}\end{array}$

Step 3. Now apply the limit of integration.

$\begin{array}{rcl}{\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}xcs{c}^{2}xdx& =& {\left[-xcotx+\log \left|\sin x\right|\right]}_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}\\ & =& \left[-\left(\frac{\mathrm{\pi }}{2}\right)cot\left(\frac{\mathrm{\pi }}{2}\right)+\log \left|\sin \left(\frac{\mathrm{\pi }}{2}\right)\right|\right]-\left[-\left(\frac{\mathrm{\pi }}{4}\right)cot\left(\frac{\mathrm{\pi }}{4}\right)+\log \left|\sin \left(\frac{\mathrm{\pi }}{4}\right)\right|\right]\\ & =& \frac{1}{4}\left(\mathrm{\pi }+2\log 2\right)\end{array}$

Step 4. Simplified answer.

Hence, the required value is$\frac{1}{4}\left(\mathrm{\pi }+2\log 2\right)$.