91Ó°ÊÓ

We can make a substitution to simplify the integrand. Let's set \(u = \cot (x)\). Therefore, the integral becomes: $$\int\frac{4^u}{\sin^2x}dx$$ We also need to find the derivative of \(u\) with respect to \(x\). We have: $$\frac{du}{dx} = -\csc^2(x)$$ Now, we can rewrite \(\sin^2(x)\) in terms of \(u\): $$\sin^2(x) = \frac{1}{1 + \cot^2(x)} = \frac{1}{1 + u^2}$$ Substituting this expression into the integral, we get: $$\int\frac{4^u}{1 + u^2}dx$$ Next, let's replace \(dx\) with its expression in terms of \(du\): $$dx = -\frac{du}{\csc^2(x)} = -\frac{du}{1 + u^2}$$ Substituting \(dx\) in the integral: $$\int 4^u (-du)$$ Now we have a simpler integral to deal with: $$-\int 4^udu$$

In order to solve the integral, we will need to apply integration by parts. Let's set \(v = u\) and \(dw = 4^u du\). Now we can find the derivatives and integrals of these functions: $$\frac{dv}{du} = 1 \Rightarrow dv = du$$ $$w = \int 4^udu = \frac{1}{\ln 4}4^u$$ Now, applying integration by parts: $$-\int 4^udu = -(uvw - \int w dv) = -\frac{1}{\ln 4}4^uu + \frac{1}{\ln 4}\int\frac{1}{\ln 4}4^udu$$ Now, we have a constant factor \(\frac{1}{\ln 4}\) multiplying the integral, which can be taken out of the integral: $$-\frac{1}{\ln 4}\int4^udu + \frac{1}{(\ln 4)^2}\int4^udu$$ Combining these integrals and factoring out the common term \(\frac{1}{\ln 4}\int4^udu\), we get: $$\left(-1 + \frac{1}{\ln 4}\right)\frac{1}{\ln 4}\int4^udu$$ From this point, we can substitute back \(u = \cot(x)\) and find the integral with respect to \(x\): $$\left(-1 + \frac{1}{\ln 4}\right)\frac{1}{\ln 4}\int4^{\cot(x)}(-\csc^2(x)dx)$$ Finally, we take out the constant factors: $$\boxed{\left(-1 + \frac{1}{\ln 4}\right)\frac{1}{\ln 4}\int\frac{4^{\cot(x)}}{\sin^2(x)} dx}$$ And this is the final result of the integral.