Key Concepts and Formulas

• Trigonometric Identities: $\tan x = \frac{\sin x}{\cos x}$
• Indefinite Integral Properties: $\int [f(x) + g(x)] \, dx = \int f(x) \, dx + \int g(x) \, dx$
• Substitution Method: If $\int f(g(x))g'(x) \, dx$, let $u = g(x)$, then $du = g'(x) \, dx$, and the integral becomes $\int f(u) \, du$.
• Logarithmic Integration: $\int \frac{1}{x} \, dx = \ln |x| + C$

Step-by-Step Solution

Step 1: Rewrite the integrand in terms of sine and cosine. We begin by expressing $\tan x$ as $\frac{\sin x}{\cos x}$ to simplify the expression. $\int \frac{5\tan x}{\tan x - 2} \, dx = \int \frac{5\frac{\sin x}{\cos x}}{\frac{\sin x}{\cos x} - 2} \, dx$

Step 2: Simplify the fraction within the integral. Multiply the numerator and denominator of the fraction by $\cos x$ to eliminate the inner fractions. $= \int \frac{5\frac{\sin x}{\cos x} \cdot \cos x}{(\frac{\sin x}{\cos x} - 2) \cdot \cos x} \, dx = \int \frac{5\sin x}{\sin x - 2\cos x} \, dx$

Step 3: Manipulate the numerator to create a term proportional to the derivative of the denominator. We want to express the numerator in terms of $(\sin x - 2\cos x)$ and its derivative $(\cos x + 2\sin x)$. We aim to write $5\sin x = A(\sin x - 2\cos x) + B(\cos x + 2\sin x)$. Expanding, we have $5\sin x = (A+2B)\sin x + (-2A+B)\cos x$. Equating coefficients gives $A+2B = 5$ and $-2A+B = 0$. From the second equation, $B = 2A$. Substituting into the first equation, $A+2(2A) = 5$, so $5A = 5$, which means $A = 1$ and $B = 2$. $\int \frac{5\sin x}{\sin x - 2\cos x} \, dx = \int \frac{(\sin x - 2\cos x) + 2(\cos x + 2\sin x)}{\sin x - 2\cos x} \, dx$

Step 4: Separate the integral into two parts. Split the fraction into two separate fractions, making it easier to integrate each part individually. $= \int \frac{\sin x - 2\cos x}{\sin x - 2\cos x} \, dx + \int \frac{2(\cos x + 2\sin x)}{\sin x - 2\cos x} \, dx$ $= \int 1 \, dx + 2 \int \frac{\cos x + 2\sin x}{\sin x - 2\cos x} \, dx$

Step 5: Evaluate the first integral. The integral of 1 with respect to $x$ is simply $x$. $= x + 2 \int \frac{\cos x + 2\sin x}{\sin x - 2\cos x} \, dx$

Step 6: Evaluate the second integral using substitution. Let $u = \sin x - 2\cos x$. Then, the derivative of $u$ with respect to $x$ is $du = (\cos x + 2\sin x) \, dx$. Substitute these into the integral. $= x + 2 \int \frac{du}{u}$ $= x + 2 \ln |u| + C$

Step 7: Substitute back for $u$. Replace $u$ with its original expression in terms of $x$. $= x + 2 \ln |\sin x - 2\cos x| + C$

Step 8: Compare with the given form. We are given that $\int \frac{5\tan x}{\tan x - 2} \, dx = x + a \ln |\sin x - 2\cos x| + k$. Comparing this with our result, $x + 2 \ln |\sin x - 2\cos x| + C$, we can see that $a = 2$.

Common Mistakes & Tips

• Sign Errors: Be very careful with signs, especially when differentiating and substituting.
• Absolute Values: Remember to include absolute values inside the logarithm, as the argument of the logarithm must be positive.
• Strategic Manipulation: The key to solving this problem is to strategically manipulate the integrand to create a term proportional to the derivative of the denominator. This often involves adding and subtracting terms.

Summary

We started by rewriting the integrand in terms of sine and cosine. Then, we manipulated the numerator to create a term proportional to the derivative of the denominator. This allowed us to separate the integral into two parts, one of which could be easily integrated directly, and the other which could be solved using substitution. Finally, we compared our result with the given form to find the value of $a$. The value of $a$ is $2$.

Final Answer

The final answer is \boxed{2}, which corresponds to option (D).