Question

Questions 2 cos x + 3 sin x *(i) Evaluate dx 4 cos x + 5 sinx​

Answer 1

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$\displaystyle \int \frac{2 \cos x + 3 \sin x}{4 \cos x + 5 \sin x} \: dx$

$\sf \purple{here }$

• $\sf numertor = 2 \cos x + 3 sin x$
• $\sf denominator =4 \cos x + 5 \sin x$

$\sf numerator \: can \: be \: represented \: as \: follows$

$\small{ \sf numerator = a( \frac{d}{dx} (4\cos x+5\sin x))+ b(4\cos x+5\sin x)}$

$\small{ \implies\sf numerator = a( - 4\sin x+5\cos x) + b(4\cos x+5\sin x)}$

$\small{ \implies\sf numerator = ( 5b- 4a)\sin x+(5a + 4b)\cos x \: ...(1)}$

$\sf \purple{since}$

$\sf numertor = 2 \cos x + 3 sin x$

$\sf \purple{substituting \: in \: (1)}$

$\small{ \implies\sf 2 \cos x + 3 sin x = ( 5b- 4a)\sin x+(5a + 4b)\cos x}$

$\sf \purple{comparing \: coefficients}$

$\implies \sf 5b - 4a = 3 \: ...(2)$

$\implies \sf 5a + 4b= 2 \: ...(3)$

$\sf \purple{solving \: (2) \: and \: (3),we \: get}$

$\sf a = \frac{ - 2}{41}$

$\sf b = \frac{23}{41}$

$\small{ \implies\displaystyle \int \frac{ \frac{ - 2}{41} ( \frac{d}{dx} (4\cos x+5\sin x)) + \frac{23}{41} (4\cos x+5\sin x)}{4 \cos x + 5 \sin x} \: dx}$

$\tiny{ \implies\displaystyle \frac{ - 2}{41}\int \frac{ \frac{d}{dx}(4\cos x+5\sin x) }{4 \cos x + 5 \sin x} \: dx + \frac{23}{41} \int \frac{4\cos x+5\sin x}{4 \cos x + 5 \sin x} \: dx}$

$\tiny{ \implies\displaystyle \frac{ - 2}{41} \log|4 \cos x + 5 \sin x| + \frac{23}{41} \int dx}$

$\orange{ \small{ \therefore\displaystyle \frac{ - 2}{41} \log|4 \cos x + 5 \sin x| + \frac{23}{41} x \: + c}}$

HOPE IT HELPS!!