∫((2sin(x)+3cos(x))/(3sin(x)+4cos(x)))dx


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 62266 by aliesam last updated on 18/Jun/19

$\int \frac{2 s i n (x) + 3 c o s (x)}{3 s i n (x) + 4 c o s (x)} d x$
Commented by maxmathsup by imad last updated on 19/Jun/19

$l e t A = \int \frac{2 s i n x + 3 c o s x}{3 s i n x + 4 c o s x} d x c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e$ $A = \int \frac{\frac{4 t}{1 + t^{2}} + 3 \frac{1 - t^{2}}{1 + t^{2}}}{\frac{6 t}{1 + t^{2}} + 4 \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = \int \frac{4 t + 3 - 3 t^{2}}{(6 t + 4 - 4 t^{2}) (1 + t^{2})} \frac{2 d t}{1 + t^{2}}$ $= \int \frac{- 3 t^{2} + 4 t + 3}{(1 + t^{2}) (- 2 t^{2} + 3 t + 2)} d t = \int \frac{3 t^{2} - 4 t - 3}{(t^{2} + 1) (2 t^{2} - 3 t - 2)} d t l e t d e c o m p o s e$ $F (t) = \frac{3 t^{2} - 4 t - 3}{(t^{2} + 1) (2 t^{2} - 3 t - 2)}$ $2 t^{2} - 3 t - 2 = 0 \to Δ = 9 - 4 (- 4) = 9 + 16 = 25 \Rightarrow t_{1} = \frac{3 + 5}{4} = 2, t_{2} = \frac{3 - 5}{4} = - \frac{1}{2} \Rightarrow$ $F (t) = \frac{3 t^{2} - 4 t - 3}{(t - 2) (t + \frac{1}{2}) (t^{2} + 1)} = \frac{2 (3 t^{2} - 4 t - 3)}{(t - 2) (2 t + 1) (t^{2} + 1)} = \frac{a}{t - 2} + \frac{b}{2 t + 1} + \frac{c t + d}{t^{2} + 1}$ $a = {l i m}_{t \to 2} (t - 2) F (t) = \frac{2 (12 - 8 - 3)}{5.5} = \frac{2}{25}$ $b = {l i m}_{t \to - \frac{1}{2}} (2 t + 1) F (t) = \frac{2 (\frac{3}{4} + 2 - 3)}{- \frac{5}{2} . \frac{5}{4}} = \frac{2 (3 + 8 - 12)}{4 (- \frac{5}{2}) . \frac{5}{4}} = \frac{4}{25} \Rightarrow$ $F (t) = \frac{2}{25 (t - 2)} + \frac{4}{25 (2 t + 1)} + \frac{c t + d}{t^{2} + 1}$ ${l i m}_{t \to + \infty} t F (t) = 0 = a + \frac{b}{2} + c \Rightarrow c = - a - \frac{b}{2} = - \frac{2}{25} - \frac{2}{25} = - \frac{4}{25} \Rightarrow$ $F (t) = \frac{2}{25 (t - 2)} + \frac{4}{25 (2 t + 1)} + \frac{- \frac{4}{25} t + d}{t^{2} + 1}$ $F (0) = \frac{3}{2} = - \frac{1}{25} + \frac{4}{25} + d = \frac{3}{25} + d \Rightarrow d = \frac{3}{2} - \frac{3}{25} = 3 (\frac{1}{2} - \frac{1}{25}) = 3 . \frac{23}{50} = \frac{69}{50} \Rightarrow$ $\int F (t) d t = \frac{2}{25} \int \frac{d t}{t - 2} + \frac{4}{25} \int \frac{d t}{2 t + 1} - \frac{1}{50} \int \frac{8 t - 69}{t^{2} + 1} d t + c$ $= \frac{2}{25} l n ∣ t - 2 ∣ + \frac{4}{50} l n ∣ 2 t + 1 ∣ - \frac{4}{50} \int \frac{2 t}{t^{2} + 1} d t + \frac{69}{50} \int \frac{d t}{t^{2} + 1} d t + c$ $= \frac{2}{25} l n ∣ t - 2 ∣ + \frac{4}{50} l n ∣ 2 t + 1 ∣ - \frac{4}{50} l n (t^{2} + 1) + \frac{69}{50} a r c t a n (t) + c \Rightarrow$ $A = \frac{2}{25} l n ∣ t a n (\frac{x}{2}) - 2 ∣ + \frac{4}{50} l n ∣ 2 t a n (\frac{x}{2}) + 1 ∣ - \frac{4}{50} l n (1 + {t a n}^{2} (\frac{x}{2})) + \frac{69}{100} x + c .$
	Answered by tanmay last updated on 18/Jun/19

	$N_{r} = a (D_{r}) + b (\frac{{d D}_{r}}{d x})$ $\int \frac{a (D_{r}) + b (\frac{{d D}_{r}}{d x})}{D_{r}} d x$ $a \int d x + b \int \frac{d (D_{r})}{D_{r}}$ $a x + b l n (D_{r}) + c$ $a x + b l n (3 s i n x + 4 c o s x) + c$ $n o w f i n d i n g v a l u e o f a a n d b$ $2 s i n x + 3 c o s x = a (3 s i n x + 4 c o s x) + b \frac{d}{d x} (3 s i n x + 4 c o s x)$ $2 s i n x + 3 c o s x = a (3 s i n x + 4 c o s x) + b (3 c o s x - 4 s i n x)$ $2 s i n x + 3 c o s x = (3 a - 4 b) s i n x + (4 a + 3 b) c o s x$ $3 a - 4 b = 2 \times 3$ $4 a + 3 b = 3 \times 4$ $9 a - 12 b = 6$ $16 a + 12 b = 12$ $25 a = 18 \to a = \frac{18}{25}$ $3 b = 3 - 4 a$ $3 b = 3 - \frac{72}{25} \to b = \frac{1}{25}$ $s o a n s w e r i s$ $\frac{18}{25} x + \frac{1}{25} l n (3 s i n x + 4 c o s x) + c$
	Commented by aliesam last updated on 18/Jun/19

	$t h a n k y o u s i r$
	Commented by tanmay last updated on 18/Jun/19

	$m o s t w e l c o m e s i r$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum