Passage

Let $$\displaystyle f\left ( x \right )=\frac{\sin 2x \cdot \sin \left ( \dfrac{\pi }{2}\cos x \right )}{2x-\pi }$$

Then answer the following question.

Mathematics

$$\int_{0}^{\pi }xf\left ( x \right )dx=$$

$$0$$

$$\displaystyle \frac{8}{\pi }$$

$$\displaystyle \frac{16}{\pi ^{2}}$$

$$\displaystyle \frac{8}{\pi ^{2}}$$

ANSWER

$$\displaystyle \frac{8}{\pi ^{2}}$$

SOLUTION

Given $$\displaystyle =\int_{0}^{\pi }xf\left ( x \right )dx=\dfrac{1}{2}\int_{0}^{\pi }\dfrac{\left ( 2x-\pi +\pi \right )\sin 2x\sin \left ( \dfrac{\pi }{2}\cos x \right )}{2x-\pi }dx$$
   $$=\displaystyle \dfrac{1}{2}\int_{0}^{\pi }\sin 2x\sin \left ( \dfrac{\pi }{2}\cos x \right )dx+\dfrac{\pi }{2}\int_{0}^{\pi }f\left ( x \right )dx$$
   $$=\displaystyle \dfrac{1}{2}\int_{0}^{\pi }\sin 2x\sin \left ( \dfrac{\pi }{2}\cos x \right )dx+0$$          (As $$\int_{0}^{\pi }f\left ( x \right )dx=0$$)
   $$\displaystyle =\dfrac{2}{\pi }\int_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}\dfrac{2t}{\pi }\sin tdt=\dfrac{4}{\pi ^{2}}\int_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}t\sin tdt=\dfrac{8}{\pi ^{2}}\int_{0}^{\dfrac{\pi }{2}}t\sin tdt=\dfrac{8}{\pi ^{2}}$$
(Substituting $$\displaystyle \dfrac{\pi }{2}\cos x=t$$ then for $$x=0$$, $$\displaystyle t=\dfrac{\pi }{2}$$ and $$x=\pi $$, $$\displaystyle t=-\dfrac{\pi }{2}$$)

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Mathematics

$$\int_{0}^{\pi }f\left ( x \right )dx=$$

$$\displaystyle \frac{4}{\pi }$$

$$\displaystyle \frac{8}{\pi }$$

$$\displaystyle \frac{8}{\pi ^{2}}$$

$$0$$

ANSWER

$$0$$

SOLUTION

$$\displaystyle I=\int_{0}^{\pi }f\left ( x \right )dx=\int_{0}^{\pi }\dfrac{\sin 2x\sin \left ( \dfrac{\pi }{2}.\cos x \right )}{2x-\pi }dx$$
$$\therefore $$   $$\displaystyle I=\int_{0}^{\pi }\dfrac{\sin 2\left ( \pi -x \right )\sin \left ( \dfrac{\pi }{2}.\cos \left ( \pi -x \right ) \right )}{2\left ( \pi -x \right )-\pi }dx$$
     $$\displaystyle =\int_{0}^{\pi }\dfrac{\left ( -1 \right )\sin 2x.\sin \left ( \dfrac{\pi }{2}\cos x \right )}{\pi -2x}dx=-I$$
$$\therefore $$   $$2I=0$$     $$\therefore $$   $$I=0$$

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Questions 203525

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