∫5cos x−4sin x/2cos x+sin x dx


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Question Number 54529 by tarun kunar last updated on 05/Feb/19

$\int 5 \cos x - 4 \sin x / 2 \cos x + \sin x d x$
Commented by maxmathsup by imad last updated on 05/Feb/19

$l e t I = \int \frac{5 c o s x - 4 s i n x}{2 c o s x + s i n x} d x l e t u s e t h e c h a n g e m e n t t a n (\frac{x}{2}) = t \Rightarrow$ $I = \int \frac{\frac{5 (1 - t^{2})}{1 + t^{2}} - \frac{8 t}{1 + t^{2}}}{2 \frac{1 - t^{2}}{1 + t^{2}} + \frac{2 t}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = \int \frac{- 5 t^{2} - 8 t + 5}{(t^{2} + 1) (2 - 2 t^{2} + 2 t)} d t$ $= \int \frac{5 t^{2} + 8 t - 5}{(t^{2} + 1) (t^{2} - t - 1)} d t l e t d e c o m p o s e F (t) = \frac{5 t^{2} + 8 t - 5}{(t^{2} + 1) (t^{2} - t - 1)} d t$ $r o o t s o f t^{2} - t - 1 \to Δ = 1 + 4 = 5 \Rightarrow t_{1} = \frac{1 + \sqrt{5}}{2} a n d t_{2} = \frac{1 - \sqrt{5}}{2}$ $F (t) = \frac{a}{t - t_{1}} + \frac{b}{t - t_{2}} + \frac{c t + d}{t^{2} + 1}$ $a = {l i m}_{t \to t_{1}} (t - t_{1}) F (t) = \frac{5 t_{1}^{2} + 8 t_{1} - 5}{(t_{1}^{2} + 1) \sqrt{5}}$ $b = {l i m}_{t \to t_{z}} (t - t_{2}) F (t) = \frac{5 t_{2}^{2} + 8 t_{2} - 5}{(t_{2}^{2} + 1) (- \sqrt{5})}$ ${l i m}_{t \to + \infty} t F (t) = 0 = a + b + c \Rightarrow c = - a - b$ $F (0) = 5 = - \frac{a}{t_{1}} - \frac{b}{t_{2}} + d \Rightarrow d = 5 + \frac{a}{t_{1}} + \frac{b}{t_{2}} \Rightarrow$ $\int F (t) d t = a l n ∣ t - t_{1} ∣ + b l n ∣ t - t_{2} ∣ + \frac{c}{2} l n (t^{2} + 1) + d a r c t a n t + λ$
Commented by maxmathsup by imad last updated on 05/Feb/19

$\Rightarrow I = a l n ∣ t a n (\frac{x}{2}) - t_{1} ∣ + b l n ∣ t a n (\frac{x}{2}) - t_{2} ∣ + \frac{c}{2} l n (1 + {t a n}^{2} (\frac{x}{2})) + \frac{d x}{2} + λ .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 05/Feb/19

	$\int \frac{5 c o s x - 4 s i n x}{2 c o s x + s i n x} d x$ $\int \frac{a (2 c o s x + s i n x) + b \frac{d}{d x} (2 c o s x + s i n x)}{2 c o s x + s i n x} d x$ $a \int d x + b \int \frac{d (2 c o s x + s i n x)}{2 c o s x + s i n x}$ $a x + b l n (2 c o s x + s i n x) + c$ $n o w w e c a l c u l a t e a a n d b$ $5 c o s x - 4 s i n x = a (2 c o s x + s i n x) + b \frac{d}{d x} (2 c o s x + s i n x)$ $5 c o s x - 4 s i n x = a (2 c o s x + s i n x) + b (- 2 s i n x + c o s x)$ $5 c o s x - 4 s i n x = c o s x \times (2 a + b) + s i n x \times (a - 2 b)$ $c o m p a r i n g c o e f f i c i e n t o f c o s x a n d s i n x b o t s i d e$ $2 a + b = 5 \times 1 [s o l v e i t]$ $a - 2 b = - 4 \times 2$ $2 a + b = 5$ $2 a - 4 b = - 8$ $5 b = 13 s o b = \frac{13}{5}$ $2 a + b = 5$ $2 a = 5 - \frac{13}{5} \to 2 a = \frac{12}{5}$ $s o a = \frac{6}{5}$ $s o a n s w e r i s$ $= \frac{6}{5} (x) + (\frac{13}{5}) l n (2 c o s x + s i n x) + c$
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