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$$
\int_0^\pi\left(\cos ^2\left(\frac{3 \pi}{8}-\frac{x}{4}\right)-\cos ^2\left(\frac{11 \pi}{8}+\frac{x}{4}\right)\right) d x=
$$
Options:
\int_0^\pi\left(\cos ^2\left(\frac{3 \pi}{8}-\frac{x}{4}\right)-\cos ^2\left(\frac{11 \pi}{8}+\frac{x}{4}\right)\right) d x=
$$
Solution:
2336 Upvotes
Verified Answer
The correct answer is:
$\sqrt{2}$
$\begin{aligned} & \text { } I=\int_0^\pi\left(\cos ^2\left(\frac{3 \pi}{8}-\frac{x}{4}\right)-\cos ^2\left(\frac{11 \pi}{8}+\frac{x}{4}\right)\right) \cdot d x \\ & I=\int_0^\pi\left(\cos \left(\frac{3 \pi}{8}-\frac{x}{4}\right)-\cos \left(\frac{11 \pi}{8}+\frac{x}{4}\right)\right) \\ & \left(\cos \left(\frac{3 \pi}{8}-\frac{x}{4}\right)+\cos \left(\frac{11 \pi}{8}+\frac{x}{4}\right)\right) \cdot d x \\ & I=\int_0^\pi\left(2 \cos \left(\frac{14 \pi}{8 \times 2}\right) \cos \left(-\pi-\frac{x}{2}\right)\right)\end{aligned}$
$\begin{gathered}\left(-2 \sin \left(\frac{14 \pi}{8 \times 2}\right) \sin \left(-\pi-\frac{\pi}{2}\right)\right) d x \\ I=\int_0^\pi 2 \cos \left(\frac{7 \pi}{8}\right)\left(-\cos \frac{x}{2}\right)\left(=2 \sin \left(\frac{14 \pi}{8}\right)\left(\sin \frac{x}{2}\right)\right) d x \\ I=\int_0^\pi\left(2 \cos \frac{7 \pi}{8} \sin \frac{7 \pi}{8}\right)\left(-2 \sin \frac{x}{2} \cos \frac{x}{2}\right) d x \\ I=-\int_0^\pi \sin \frac{7 x}{4} \sin x d x \\ I=-\sin \left(\frac{\pi}{4}\right) \int_0^\pi \sin x d x \\ I=\frac{1}{\sqrt{2}}(-\cos x) \int_0^\pi=-\frac{1}{\sqrt{2}}(\cos \pi-\cos 0) \\ I=-\frac{1}{\sqrt{2}}(-1-1)=\frac{-2}{\sqrt{2}}=\sqrt{2}\end{gathered}$
$\begin{gathered}\left(-2 \sin \left(\frac{14 \pi}{8 \times 2}\right) \sin \left(-\pi-\frac{\pi}{2}\right)\right) d x \\ I=\int_0^\pi 2 \cos \left(\frac{7 \pi}{8}\right)\left(-\cos \frac{x}{2}\right)\left(=2 \sin \left(\frac{14 \pi}{8}\right)\left(\sin \frac{x}{2}\right)\right) d x \\ I=\int_0^\pi\left(2 \cos \frac{7 \pi}{8} \sin \frac{7 \pi}{8}\right)\left(-2 \sin \frac{x}{2} \cos \frac{x}{2}\right) d x \\ I=-\int_0^\pi \sin \frac{7 x}{4} \sin x d x \\ I=-\sin \left(\frac{\pi}{4}\right) \int_0^\pi \sin x d x \\ I=\frac{1}{\sqrt{2}}(-\cos x) \int_0^\pi=-\frac{1}{\sqrt{2}}(\cos \pi-\cos 0) \\ I=-\frac{1}{\sqrt{2}}(-1-1)=\frac{-2}{\sqrt{2}}=\sqrt{2}\end{gathered}$
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