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$\int \cos^2(x) \csc^2(2x) \, dx$

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Gaussian integral

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Q.

\int \cos^2(x) \csc^2(2x) \, dx

Simplify using trigonometric identities: Simplify the integrand using trigonometric identities. $\newline$ We know that $\csc(\theta) = \frac{1}{\sin(\theta)}$ and $\cos^2(\theta) = \frac{(1 + \cos(2\theta))}{2}$ (from the double angle formula). $\newline$ So, $\cos^2(x)\csc^2(2x) = \frac{(1 + \cos(2x))}{2} \cdot \frac{1}{\sin^2(2x)}$ .
Further simplify the integrand: Further simplify the integrand. $\newline$ We can write $\frac{1}{\sin^2(2x)}$ as $\left(\frac{1}{\sin(2x)}\right)^2$ , which is $\csc^2(2x)$ . $\newline$ So, $\cos^2(x)\csc^2(2x) = \left(\frac{1}{2} + \frac{\cos(2x)}{2}\right) * \csc^2(2x)$ .
Break into two parts: Break the integral into two parts. \int \cos^\(2(x)\csc^ $2$ ( $2$ x)\,dx = \int (\frac{ $1$ }{ $2$ })\csc^ $2$ ( $2$ x)\,dx + \int (\frac{\cos( $2$ x)}{ $2$ })\csc^ $2$ ( $2$ x)\,dx.
Integrate first part: Integrate the first part $\int(\frac{1}{2})\csc^2(2x)\,dx$ . We know that $\int\csc^2(\theta)\,d\theta = -\cot(\theta)$ , so: $\int(\frac{1}{2})\csc^2(2x)\,dx = -(\frac{1}{2})\cot(2x)/2$ .
Integrate second part: Integrate the second part $\int(\cos(2x)/2)\csc^2(2x)\,dx$ . This integral is more complex and requires a substitution or another method to solve. However, we notice that there is no straightforward substitution or simplification that can be applied here. This suggests that there might be an error in the problem or the approach we are taking.

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