AP DSC School Assistant Math Model Test 2

Concept:Integration using substitution $t = \tan\frac{x}{2}$ and partial fractions to match the given form.Explanation:First, factor the denominator: $(\sin x +4)(\sin x -1)$. Decompose into partial fractions:$$\frac{1}{(\sin x+4)(\sin x-1)} = \frac{1}{5}\left(\frac{1}{\sin x-1} - \frac{1}{\sin x+4}\right).$$Integrate each term separately. Use the substitution $t = \tan\frac{x}{2}$. Then $\sin x = \frac{2t}{1+t^2}$ and $dx = \frac{2\,dt}{1+t^2}$.For $\int\frac{dx}{\sin x-1}$:$$\int\frac{dx}{\sin x-1} = \int\frac{\frac{2}{1+t^2}}{\frac{2t}{1+t^2} - 1}\,dt = \int\frac{2\,dt}{2t - (1+t^2)} = \int\frac{2\,dt}{-(t-1)^2} = -2\int\frac{dt}{(t-1)^2}.$$Similarly, for $\int\frac{dx}{\sin x+4}$:$$\int\frac{dx}{\sin x+4} = \int\frac{\frac{2}{1+t^2}}{\frac{2t}{1+t^2}+4}\,dt = \int\frac{2\,dt}{2t+4(1+t^2)} = \int\frac{2\,dt}{4t^2+2t+4}.$$Simplify the denominator: $4t^2+2t+4 = 4\left(t^2+\frac{t}{2}+1\right) = 4\left[\left(t+\frac14\right)^2 + \frac{15}{16}\right] = \frac{(4t+1)^2+15}{4}.$Thus the integral becomes:$$\int\frac{2\,dt}{\frac{(4t+1)^2+15}{4}} = 8\int\frac{dt}{(4t+1)^2+15} = \frac{8}{15}\int\frac{dt}{\left(\frac{4t+1}{\sqrt{15}}\right)^2+1}.$$Substitute $u = \frac{4t+1}{\sqrt{15}}$, $du = \frac{4}{\sqrt{15}}dt$, giving:$$\frac{8}{15}\cdot\frac{\sqrt{15}}{4}\int\frac{du}{u^2+1} = \frac{2}{\sqrt{15}}\tan^{-1}u = \frac{2}{\sqrt{15}}\tan^{-1}\left(\frac{4t+1}{\sqrt{15}}\right).$$Now combine with the factor $\frac15$ and the sign from the partial fractions:$$\frac15\left[ -2\int\frac{dt}{(t-1)^2} \;-\; \frac{2}{\sqrt{15}}\tan^{-1}\left(\frac{4t+1}{\sqrt{15}}\right) \right].$$Integrate: $\int\frac{dt}{(t-1)^2} = -\frac{1}{t-1} + C$. So the expression becomes:$$\frac15\left[ -2\left(-\frac{1}{t-1}\right) \;-\; \frac{2}{\sqrt{15}}\tan^{-1}\left(\frac{4t+1}{\sqrt{15}}\right) \right] = \frac{2}{5(t-1)} - \frac{2}{5\sqrt{15}}\tan^{-1}\left(\frac{4t+1}{\sqrt{15}}\right) + C.$$Replace $t$ with $\tan\frac{x}{2}$: $$\frac{2}{5\left(\tan\frac{x}{2}-1\right)} - \frac{2}{5\sqrt{15}}\tan^{-1}\left(\frac{4\tan\frac{x}{2}+1}{\sqrt{15}}\right) + C.$$Comparing with $\frac{A}{\tan\frac{x}{2}-1} + B\tan^{-1}f(x) + C$, we identify:$A = \frac{2}{5},\quad B = -\frac{2}{5\sqrt{15}},\quad f(x) = \frac{4\tan\frac{x}{2}+1}{\sqrt{15}}.$Answer:The correct option is C: $A=\frac{2}{5},\ B=-\frac{2}{5\sqrt{15}},\ f(x)=\frac{4\tan\frac{x}{2}+1}{\sqrt{15}}$.