Evaluate ∫9cos⁡x−sin⁡x5cos⁡x+4sin⁡xdx

∫9cos⁡x−sin⁡x5cos⁡x+4sin⁡xdxLet Nr≡Addx(Dr)+B(Dr)Take 9cos⁡x−sin⁡x≡Addx(5cos⁡x+4sin⁡x)+B(5cos⁡x+4sin⁡x)−(1)9cos⁡x−sin⁡x≡A(−5sin⁡x+4cos⁡x)+B(5cos⁡x+4sin⁡x) Comparing the coefficients of sinx terms and cosx terms on both sides we get−5A+4B=−1 ...(1)4A+5B=9 ...(2)Soving (1) & (2) we get A B 1 4 1 -5 4 5 -9 4 5 ⇒A -36-5=B4-45=1-25-16 ⇒A-41 =B-41=1-41 A = 1 B = 1 in equation (1)9cos⁡x−sin⁡x=1. ddx(5cos⁡x+4sin⁡x)+1(5cos⁡x+4sin⁡x) ∫9cos⁡x−sin⁡x5cos⁡x+4sin⁡xdx=∫ddx(5cos⁡x+4sin⁡x)+(5cos⁡x+4sin⁡x)5cos⁡x+4sin⁡xdx∴=∫ddx(5cos⁡x+4sin⁡x)5cos⁡x+4sin⁡xdx+∫5cos⁡x+4sin⁡x5cos⁡x+4sin⁡xdx=log⁡|5cos⁡x+4sin⁡x|+x+c∴∫9cos⁡x−sin⁡x5cos⁡x+4sin⁡xdx=log⁡(5cos⁡x+4sin⁡x)+x+c

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Evaluate $\int \frac{9 \cos x - \sin x}{5 \cos x + 4 \sin x} dx$

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Detailed Solution

$\int \frac{9 \cos x - \sin x}{5 \cos x + 4 \sin x} dx$
Let $Nr \equiv A \frac{d}{dx} (Dr) + B (Dr)$
Take $9 \cos x - \sin x$
$\begin{array}{l} \equiv A \frac{d}{dx} (5 \cos x + 4 \sin x) + B (5 \cos x + 4 \sin x) - (1) \\ 9 \cos x - \sin x \\ \equiv A (- 5 \sin x + 4 \cos x) + B (5 \cos x + 4 \sin x) \end{array}$
Comparing the coefficients of sinx terms and cosx terms on both sides we get
$\begin{array}{l} - 5 A + 4 B = - 1 . . . (1) \\ 4 A + 5 B = 9 . . . (2) \end{array}$
Soving (1) & (2) we get
$A B 1 4 1 - 5 4 5 - 9 4 5 \Rightarrow \frac{A}{- 36 - 5} = \frac{B}{4 - 45} = \frac{1}{- 25 - 16} \Rightarrow \frac{A}{- 41} = \frac{B}{- 41} = \frac{1}{- 41}$
A = 1 B = 1 in equation (1)
$\begin{array}{l} 9 \cos x - \sin x = 1. \frac{d}{dx} (5 \cos x + 4 \sin x) + 1 (5 \cos x + 4 \sin x) \\ \int \frac{9 \cos x - \sin x}{5 \cos x + 4 \sin x} dx \\ = \int \frac{\frac{d}{dx} (5 \cos x + 4 \sin x) + (5 \cos x + 4 \sin x)}{5 \cos x + 4 \sin x} dx \\ ∴= \int \frac{\frac{d}{dx} (5 \cos x + 4 \sin x)}{5 \cos x + 4 \sin x} dx + \int \frac{5 \cos x + 4 \sin x}{5 \cos x + 4 \sin x} dx \\ = \log | 5 \cos x + 4 \sin x | + x + c \\ ∴ \int \frac{9 \cos x - \sin x}{5 \cos x + 4 \sin x} dx \\ = \log (5 \cos x + 4 \sin x) + x + c \end{array}$