The teacher hasn't uploaded a solution for this question yet.
We are given the integral: $\int\frac{\cos x}{(4+\sin^{2}x)(5-4\cos^{2}x)}dx$ We can rewrite $\cos^2 x$ as $1 - \sin^2 x$. Thus, the integral becomes: $\int\frac{\cos x}{(4+\sin^{2}x)(5-4(1-\sin^{2}x))}dx = \int\frac{\cos x}{(4+\sin^{2}x)(5-4+4\sin^{2}x)}dx = \int\frac{\cos x}{(4+\sin^{2}x)(1+4\sin^{2}x)}dx$
Let $t = \sin x$. Then, $dt = \cos x \, dx$. The integral becomes: $\int\frac{1}{(4+t^{2})(1+4t^{2})}dt$
We need to decompose the fraction $\frac{1}{(4+t^{2})(1+4t^{2})}$ into partial fractions. Let $\frac{1}{(4+t^{2})(1+4t^{2})} = \frac{A}{4+t^{2}} + \frac{B}{1+4t^{2}}$ Multiplying both sides by $(4+t^{2})(1+4t^{2})$, we get: $1 = A(1+4t^{2}) + B(4+t^{2})$ $1 = A + 4At^{2} + 4B + Bt^{2}$ $1 = (A+4B) + (4A+B)t^{2}$ Comparing the coefficients, we have the following system of equations: $A+4B = 1$ $4A+B = 0$ From the second equation, $B = -4A$. Substituting this into the first equation: $A + 4(-4A) = 1$ $A - 16A = 1$ $-15A = 1$ $A = -\frac{1}{15}$ Then, $B = -4A = -4(-\frac{1}{15}) = \frac{4}{15}$ So, $\frac{1}{(4+t^{2})(1+4t^{2})} = \frac{-1/15}{4+t^{2}} + \frac{4/15}{1+4t^{2}} = \frac{1}{15}\left(\frac{4}{1+4t^{2}} - \frac{1}{4+t^{2}}\right)$
Now we integrate: $\int\frac{1}{(4+t^{2})(1+4t^{2})}dt = \frac{1}{15}\int\left(\frac{4}{1+4t^{2}} - \frac{1}{4+t^{2}}\right)dt = \frac{1}{15}\left(4\int\frac{1}{1+(2t)^{2}}dt - \int\frac{1}{4+t^{2}}dt\right)$ We know that $\int\frac{1}{a^{2}+x^{2}}dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right) + C$. So, $\int\frac{1}{1+(2t)^{2}}dt = \frac{1}{2}\arctan(2t) + C_{1}$ and $\int\frac{1}{4+t^{2}}dt = \int\frac{1}{2^{2}+t^{2}}dt = \frac{1}{2}\arctan\left(\frac{t}{2}\right) + C_{2}$ Therefore, the integral becomes: $\frac{1}{15}\left(4\cdot\frac{1}{2}\arctan(2t) - \frac{1}{2}\arctan\left(\frac{t}{2}\right)\right) + C = \frac{1}{15}\left(2\arctan(2t) - \frac{1}{2}\arctan\left(\frac{t}{2}\right)\right) + C$
Substitute $t = \sin x$ back into the expression: $\frac{1}{15}\left(2\arctan(2\sin x) - \frac{1}{2}\arctan\left(\frac{\sin x}{2}\right)\right) + C$
Final Answer: $\frac{1}{15}\left(2\arctan(2\sin x) - \frac{1}{2}\arctan\left(\frac{\sin x}{2}\right)\right) + C$
AI generated content. Review strictly for academic accuracy.