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\(\int_{0}^{\pi} x.f(\sin x)\; dx=\)
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A. \(x\cdot \int_{0}^{\pi }f \left ( \sin\: x \right )\: dx\) |
| B. \(\pi \cdot \int_{0}^{\pi }f \left ( \sin\: x \right )\: dx\) |
| C. \(2\pi \cdot \int_{0}^{\pi }f \left ( \sin\: x \right )\: dx\) |
| D. \(\frac{\pi}{2} \cdot \int_{0}^{\pi }f \left ( \sin\: x \right )\: dx\) |
| E. \(\int_{0}^{\pi }f \left ( \cos\: x \right )\: dx\) |
[ 6-8728 - op net sinds 5.4.2021-(E)-16.10.2026 ]
Translation in E N G L I S H
Oplossing - Solution
\(I=\int_{0}^{\pi }\left [x\! \cdot f\left (\sin\: x \right ) \right ]dx \\ Stel \; x=\pi -t \; \Rightarrow \; dx=\!-dt \\ {\color{Yellow} .}\qquad \qquad \qquad \qquad x=0\;\; k.o.m.\; \; t=\pi \\ {\color{Yellow} .}\qquad \qquad \qquad \qquad x=\pi \;\; k.o.m.\; \; t=0 \\ I=\int_{\pi }^{0} \left [(\pi -t).f(\sin\: t) \right ]\cdot (-dt)\\ = -\pi \int_{\pi }^{0}f(\sin\: t)dt+\int_{\pi }^{0}\left [t\cdot f(sin\: t) \right ]dt\\ =\pi \int_{0}^{\pi }f(\sin\: t)\: dt - \int_{0}^{\pi }\left [ t\cdot f(\sin\: t) \right ]dt\\ \Rightarrow I=\pi \int_{0}^{\pi }f(\sin\: t)\: dt\: -\: I \\ \Rightarrow I=\frac{\pi }{2}\int_{0}^{\pi }f(\sin\: t)\: dt={\color{Red} (!)\: }=\frac{\pi }{2}\int_{0}^{\pi }f(\sin\: x)\: dx\)