Solve int(dx)/(4sin^(2)x+8sin 2x-3cos^(2)x)

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Simplify using trigonometric identities: We need to simplify the expression inside the integral before we can integrate. Let's start by using trigonometric identities to rewrite the terms involving sine and cosine. We know that sin(2x) = 2sin(x)cos(x) and that sin^2(x) + cos^2(x) = 1. We can use these identities to rewrite the integral in terms of sin(x) only.Express cos^2(x) in terms of sin^2(x): First, let's express cos^2(x) in terms of sin^2(x) using the Pythagorean identity: cos^2(x) = 1 - sin^2(x). Then we can substitute this into the integral. The integral becomes: int(dx)/(4sin^2(x) + 8sin(2x) - 3(1 - sin^2(x))) Simplify the expression: int(dx)/(4sin^2(x) + 8sin(2x) - 3 + 3sin^2(x)) Combine like terms: int(dx)/(7sin^2(x) + 8sin(2x) - 3)Address the term 8sin(2x): Now, let's address the term 8sin(2x). We can use the double-angle identity for sine, which we've already mentioned: sin(2x) = 2sin(x)cos(x). Substituting this into the integral gives us: int(dx)/(7sin^2(x) + 8(2sin(x)cos(x)) - 3) Simplify the expression: int(dx)/(7sin^2(x) + 16sin(x)cos(x) - 3)Explore integration options: At this point, we need to find a way to integrate this expression. It's not immediately clear how to proceed, so let's try to simplify it further. One approach could be to use a substitution, but it's not obvious what substitution would work here. Another approach could be to rewrite the expression in a more familiar form, such as a sum of squares or a product of trigonometric functions. However, this expression does not lend itself easily to such simplifications. We may have made a mistake in our approach.

Solve $\int \frac{dx}{4\sin^2(x)+8\sin(2x)-3\cos^2(x)}$

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Solve $\int \frac{dx}{4\sin^2(x)+8\sin(2x)-3\cos^2(x)}$