Question

If $\int\frac{2\sin x+a\cos x}{b\sin x+4\cos x}dx=\frac{2}{5}x-\frac{1}{5}\log(b\sin x+4\cos x)+c$, then $a+b=$

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

Shortcut: In $\int \frac{N}{D} dx = Ax + B\log|D|$, you can always find the parameters by directly equating $N = A \cdot D + B \cdot D'$.

Answer 1

The Correct Option is C

Step 1: Concept
For integrating an expression of the form $\int \frac{p\sin x + q\cos x}{r\sin x + s\cos x} dx$, we express the numerator as $A(\text{Denominator}) + B(\text{Derivative of Denominator})$. The result is always $Ax + B\log|r\sin x + s\cos x| + c$.

Step 2: Meaning
From the given answer, the coefficient of $x$ is $A = \frac{2}{5}$ and the coefficient of the log term is $B = -\frac{1}{5}$. The denominator is $b\sin x + 4\cos x$, so its derivative is $b\cos x - 4\sin x$.

Step 3: Analysis
Equating the numerator: $2\sin x + a\cos x = \frac{2}{5}(b\sin x + 4\cos x) - \frac{1}{5}(b\cos x - 4\sin x)$. Compare the coefficients of $\sin x$ and $\cos x$ on both sides: 1. For $\sin x$: $2 = \frac{2}{5}b + \frac{4}{5} \implies 10 = 2b + 4 \implies 2b = 6 \implies b = 3$. 2. For $\cos x$: $a = \frac{8}{5} - \frac{1}{5}b$. Substitute $b = 3$: $a = \frac{8}{5} - \frac{3}{5} = \frac{5}{5} = 1$.

Step 4: Conclusion
Now calculate the requested sum: $a + b = 1 + 3 = 4$. This matches option (C).

Final Answer: (C)