The value of int frac{1}{1+cos 8x} dx is | vedclass

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(D) We know that $1 + \cos 2	heta = 2 \cos^2 	heta$. Applying this identity for $	heta = 4x$,we get $1 + \cos 8x = 2 \cos^2 4x$. Substituting this

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$\frac{	an 4x}{8}+c$

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(D) We know that $1 + \cos 2	heta = 2 \cos^2 	heta$. Applying this identity for $	heta = 4x$,we get $1 + \cos 8x = 2 \cos^2 4x$. Substituting this into the integral: $I = \int \frac{1}{2 \cos^2 4x} dx$ $I = \frac{1}{2} \int \sec^2 4x dx$ Using the standard integral $\int \sec^2(ax) dx = \frac{	an(ax)}{a} + c$,we get: $I = \frac{1}{2} \cdot \frac{	an 4x}{4} + c$ $I = \frac{	an 4x}{8} + c$

The value of $\int \frac{1}{1+\cos 8x} dx$ is

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