If $$f(x)$$ is an integrable function in π$$6$$,π$$3$$ and $$I1=$$∫π$$/6$$π$$/3sec2$$θ$$f(2sin2$$θ$$)d$$θ and $$I2=$$∫π$$/6$$π$$/3cosec2$$θ$$f(2sin2$$θ$$)d$$θ, then find $$I12I2$$

$$I1=$$∫π$$/6$$π$$/3sec2$$$$π$$$$2-$$θ$$f2sin2$$$$π$$$$2-$$θdθ $$=$$∫π$$/6$$π$$/3cosec2$$θ$$f(2sin2$$θ$$)d$$θ$$=I2$$ ∴$$I12I2=12=0.5$$

Evaluation of definite integrals

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Question

$If f (x) is an integrable function in [\frac{π}{6}, \frac{π}{3}] and I_{1} = \int_{π / 6}^{π / 3} \sec^{2} θ f (2 \sin 2 θ) d θ and$ $I_{2} = \int_{π / 6}^{π / 3} {cosec}^{2} θ f (2 \sin 2 θ) d θ, then find \frac{I_{1}}{2 I_{2}}$

Moderate

Solution

$I_{1} = \int_{π / 6}^{π / 3} \sec^{2} (\frac{π}{2} - θ) f (2 \sin 2 (\frac{π}{2} - θ)) d θ$
$= \int_{π / 6}^{π / 3} {cosec}^{2} θ f (2 \sin 2 θ) d θ = I_{2}$
$∴ \frac{I_{1}}{2 I_{2}} = \frac{1}{2} = 0.5$

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