Evaluate the definite integral: int_0^{pi /8} | vedclass

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Question

(A) We are given the integral $I = \int_0^{\pi /8} \frac{\sec^2 2x}{2} \, dx$. Taking the constant factor out,we get $I = \frac{1}{2} \int_0^{\pi /8}

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(A) We are given the integral $I = \int_0^{\pi /8} \frac{\sec^2 2x}{2} \, dx$. Taking the constant factor out,we get $I = \frac{1}{2} \int_0^{\pi /8} \sec^2 2x \, dx$. We know that the integral of $\sec^2(ax)$ is $\frac{	an(ax)}{a}$. Applying this,we get $I = \frac{1}{2} \left[ \frac{	an 2x}{2} ight]_0^{\pi /8}$. $I = \frac{1}{4} [	an 2x]_0^{\pi /8}$. Substituting the limits,$I = \frac{1}{4} [	an(2 	imes \frac{\pi}{8}) - 	an(0)]$. $I = \frac{1}{4} [	an(\frac{\pi}{4}) - 0]$. Since $	an(\frac{\pi}{4}) = 1$,we have $I = \frac{1}{4} [1] = \frac{1}{4}$.

Evaluate the definite integral: $\int_0^{\pi /8} \frac{\sec^2 2x}{2} \, dx$

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