If int frac{1+cos 8 x}{tan 2 x-cot 2 x} d x=f | vedclass

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(D) We are given the integral $I = \int \frac{1+\cos 8 x}{	an 2 x-\cot 2 x} d x$. Using the identity $1+\cos 8x = 2\cos^2 4x$, we have: $I = \int \fr

Vedclass · Accepted Answer

$\frac{33}{16}$

Vedclass · Answer

(D) We are given the integral $I = \int \frac{1+\cos 8 x}{	an 2 x-\cot 2 x} d x$. Using the identity $1+\cos 8x = 2\cos^2 4x$, we have: $I = \int \frac{2 \cos ^2 4 x}{\frac{\sin 2 x}{\cos 2 x}-\frac{\cos 2 x}{\sin 2 x}} d x$. Simplifying the denominator: $\frac{\sin 2 x}{\cos 2 x}-\frac{\cos 2 x}{\sin 2 x} = \frac{\sin^2 2x - \cos^2 2x}{\sin 2x \cos 2x} = \frac{-\cos 4x}{\frac{1}{2}\sin 4x} = -2\cot 4x$. Thus, $I = \int \frac{2 \cos^2 4x}{-2\cot 4x} d x = -\int \frac{\cos^2 4x}{\frac{\cos 4x}{\sin 4x}} d x = -\int \cos 4x \sin 4x d x$. Multiplying and dividing by $2$: $I = -\frac{1}{2} \int 2 \sin 4x \cos 4x d x = -\frac{1}{2} \int \sin 8x d x$. Integrating: $I = -\frac{1}{2} \left( -\frac{\cos 8x}{8} ight) + c = \frac{1}{16} \cos 8x + c$. Comparing with $f(x) \cdot \cos(g(x)) + c$, we get $f(x) = \frac{1}{16}$ and $g(x) = 8x$. Therefore, $f\left(\frac{1}{4}ight) + g\left(\frac{1}{4}ight) = \frac{1}{16} + 8\left(\frac{1}{4}ight) = \frac{1}{16} + 2 = \frac{33}{16}$.

If $\int \frac{1+\cos 8 x}{\tan 2 x-\cot 2 x} d x=f(x) \cdot \cos (g(x))+c$, then $f\left(\frac{1}{4}\right)+g\left(\frac{1}{4}\right)=$

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