Q231 / 270 · NDA 2024 (Phase 1) · Mathematics

Given that $\int \frac{3 \cos x+4 \sin x}{2 \cos x+5 \sin x} d x=\frac{\alpha x}{29}+\frac{\beta}{29} \ln |2 \cos x+5 \sin x|+c$ What is the value of $\alpha$ ?

  • A
    7
  • B
    13
  • C
    17
  • Correct answer
    D
    26

Solution

Let, $\begin{aligned} & \mathrm{I}=\int \frac{3 \cos x+4 \sin x}{2 \cos x+5 \sin x} d x \\ & 3 \cos x+4 \sin x=\mathrm{A}(2 \cos x+5 \sin x) \\ & +\mathrm{B}\left[\frac{d}{d x}(2 \cos x+5 \sin x)\right] \\ & 3 \cos x+4 \sin x=\mathrm{A}(2 \cos x+5 \sin x) \\ & +\mathrm{B}(-2 \sin x+5 \cos x) \end{aligned}$ Comparing co-efficient of $\sin x$ and $\cos x$ $\begin{aligned} & 3=2 A+5 B \qquad ....(i)\\ & 4=5 A-2 B \qquad ....(ii) \end{aligned}$ Solving (i) and (ii) then we get, $\begin{aligned} & \mathrm{A}=\frac{26}{29}, \mathrm{~B}=\frac{7}{29} \\ & \mathrm{I}=\frac{\int \frac{26}{29}(2 \cos x+5 \sin x)+\frac{7}{29}(-2 \sin x+5 \cos x)}{2 \cos x+5 \sin x} d x \\ &=\int \frac{26}{29} d x+\frac{7}{29} \int \frac{(-2 \sin x+5 \cos x)}{2 \cos x+5 \sin x} d x \\ &=\frac{26}{29} x+\frac{7}{29} \text { In }(2 \cos x+5 \sin x)+c \\ & \alpha=26, \beta=7[\text { [on comparing] } \end{aligned}$ $\alpha=26$

Hence, the correct answer is option: 'D'.

Relevant Questions