Question

$\int \left[{\sin }^2(\frac{9\pi }{8}+\frac{x}{4})-{\sin }^2(\frac{7\pi }{8}+\frac{x}{4})\right]dx$

Answer 1

$\begin{array}{c}\int \left[{\sin }^2\left(\frac{9\pi }{8}+\frac{\pi }{4}\right)-{\sin }^2\left(\frac{7\pi }{8}+\frac{x}{4}\right)\right]dx\\ \int \left[{\sin }^2\left(\pi +\frac{\pi }{8}+\frac{\pi }{4}\right)-{\sin }^2\left(\pi -\frac{\pi }{8}+\frac{\pi }{4}\right)\right]dx\\ =\int \left[{\sin }^2\left(\frac{\pi }{8}+\frac{x}{4}\right)-{\sin }^2\left(\frac{\pi }{8}-\frac{x}{4}\right)\right]dx\\ \frac{1}{2}\int \left[2{\sin }^2\left(\frac{\pi }{8}+\frac{x}{4}\right)-2{\sin }^2\left(\frac{\pi }{8}-\frac{x}{4}\right)\right]dx\\ =\frac{1}{2}\int \left[1-\cos \left(\frac{\pi }{4}+\frac{x}{2}\right)-\left\{1-\cos \left(\frac{\pi }{4}-\frac{x}{2}\right)\right\}\right]dx\\ =\frac{1}{2}\int \left[1-\cos \left(\frac{\pi }{4}+\frac{x}{2}\right)-1+\cos \left(\frac{\pi }{4}-\frac{x}{2}\right)\right]dx\\ =\frac{1}{2}\int \left[\cos \left(\frac{\pi }{4}-\frac{x}{2}\right)-\cos \left(\frac{\pi }{4}+\frac{x}{2}\right)\right]dx\end{array}$

Putting

$\begin{array}{c}\frac{\pi }{4}-\frac{x}{2}=t\\ \Rightarrow \frac{-dx}{2}=dt\\ \Rightarrow dx=-2dt\end{array}$

$\begin{array}{c}=\frac{-2}{2}\int \left[\cos t-\cos \left(\frac{\pi }{4}+\frac{\pi }{4}-t\right)\right]dt\\ =-\int \left[\cos t-\cos \left(\frac{\pi }{2}-t\right)\right]dt\\ =-\int \left(\cos t-\sin t\right)dt\end{array}$

$\begin{array}{c}=-\left[\sin t+\cos t\right]+c\\ =-\left[\sin \left(\frac{\pi }{4}-\frac{x}{2}\right)+\cos \left(\frac{\pi }{4}-\frac{x}{2}\right)\right]+c\end{array}$