The value of limm→∞ ∫$$0$$π$$/2$$ $$sin2m$$⁡xdx∫$$0$$π$$/2$$ $$sin2m+1$$⁡xdx

Question

The value of limm→∞ ∫$$0$$π$$/2$$ $$sin2m$$⁡xdx∫$$0$$π$$/2$$ $$sin2m+1$$⁡xdx

Infinity Learn · Accepted Answer

We know that $$I2n=$$∫$$0$$π$$/2$$ $$sin2n$$⁡$$xdx=2n$$−$$12n$$×$$2n$$−$$32n$$−$$2$$×…×$$12$$×π$$2$$,$$I2n+1=$$∫$$0$$π$$/2$$ $$sin2n+1$$⁡$$xdx=2n2n+1$$×$$2n$$−$$22n$$−$$1$$×…×$$23$$ andAlso, $$I2m+1=2m2m+1I2m$$−$$1For$$ all x∈(0$$,π$$/2),$$sin2m$$−$$1$$⁡x>$$sin2m$$⁡x>$$sin2m+1$$⁡xIntegrating from $$0$$ to π$$/2$$, we get $$I2m$$−$$1$$≥$$I2m$$≥$$I2m+1hence$$ $$I2m$$−$$1I2m+1$$≥$$I2mI2m+1$$≥$$1$$                (i)Also$$   $$I2m$$−$$1I2m+1=2m+12mHence$$ limm→∞ $$I2m$$−$$1I2m+1=limm$$→∞ $$2m+12m=1From$$ $$(i) and using sandwitch theorem we havelimm→∞ $$I2mI2m+1=1$$

The value of $lim_{m \to \infty} \frac{\int_{0}^{π / 2} \sin^{2 m} x d x}{\int_{0}^{π / 2} \sin^{2 m + 1} x d x}$

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The value of limm→∞ ∫0π/2 sin2m⁡xdx∫0π/2 sin2m+1⁡xdx

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The value of $lim_{m \to \infty} \frac{\int_{0}^{π / 2} \sin^{2 m} x d x}{\int_{0}^{π / 2} \sin^{2 m + 1} x d x}$