If \int_{0}^{2 \pi} \frac{dx}{2 + sin 4 x} = \lambda \int_{0}^{\pi / 2} \frac{dx}{3 + \left(cos\right)^{2} x} then λ equals

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Introduction to Definite Integral

Question

If $\int_{0}^{2 π} \frac{dx}{2 + \sin 4 x} = λ \int_{0}^{π / 2} \frac{dx}{3 + \cos^{2} x}$ then λ equals

$I = \int_{0}^{2 π} \frac{dx}{2 + \sin 4 x}$
Applying king, $I = \int_{0}^{2 π} \frac{dx}{2 - \sin 4 x}$   ⇒ $2 I = 4 \int_{0}^{2 π} \frac{dx}{4 - \sin^{2} 4 x}$ Put 4x = t
⇒     $2 I = \frac{4}{4} \int_{0}^{8 π} \frac{dt}{4 - \sin^{2} t} = \int_{0}^{8 π} \frac{dt}{3 + \cos^{2} t}$
∴    $2 I = \int_{0}^{8 π} \frac{dx}{\underset{\begin{array}{l} ↓ \\ periodicwithperiod π \end{array}}{3 + \cos^{2} x}} = 8 \int_{0}^{π} \frac{dx}{3 + \cos^{2} x}$
$I = 4 \int_{0}^{π} \frac{dx}{3 + \cos^{2} x}$ ⇒    $I = 8 \int_{0}^{π / 2} \frac{dx}{3 + \cos^{2} x}$

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