∫(dt/(3sint+4cost))


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Question Number 92910 by s.ayeni14@yahoo.com last updated on 09/May/20

$\int \frac{dt}{3 sint + 4 cost}$
Commented by msup by abdo last updated on 09/May/20

$I = \int \frac{d t}{3 s i n t + 4 c o s t} v h a n g e m e n t$ $t a n (\frac{t}{2}) = x g i v e$ $I = \int \frac{2 d x}{(1 + x^{2}) (3 \times \frac{2 x}{1 + x^{2}} + 4 \times \frac{1 - x^{2}}{1 + x^{2}})}$ $= 2 \int \frac{d x}{6 x + 4 - 4 x^{2}} = \int \frac{d x}{- 2 x^{2} + 3 x + 2}$ $= - \int \frac{d x}{2 x^{2} - 3 x - 2}$ $Δ = 9 - 4 (- 4) = 25 \Rightarrow$ $x_{1} = \frac{3 + 5}{4} = 2 a n d x_{2} = \frac{3 - 5}{4} = - \frac{1}{2}$ $I = - \int \frac{d x}{2 (x - 2) (x + \frac{1}{2})}$ $= - \frac{1}{2} \times \frac{2}{5} \int (\frac{1}{x - 2} - \frac{1}{x + \frac{1}{2}}) d x$ $= - \frac{1}{5} l n ∣ \frac{x - 2}{x + \frac{1}{2}} ∣ + C$ $= - \frac{1}{5} l n ∣ \frac{2 x - 4}{2 x + 1} ∣ + C$ $= - \frac{1}{5} l n ∣ \frac{2 t a n (\frac{x}{2}) - 4}{2 t a n (\frac{x}{2}) + 1} ∣ + C$
Commented by i jagooll last updated on 10/May/20

$\frac{1}{5} \int \frac{dt}{\cos (t - θ)}, where \tan θ = \frac{3}{4}$ $= \frac{1}{5} \int \sec (t - θ) d (t - θ)$ $= \frac{1}{5} \ln ∣ \sec (t - θ) + \tan (t - θ) ∣ + c$ $= \frac{1}{5} \ln ∣ \frac{1}{\cos tcos θ + \sin tsin θ} + \frac{\tan t - \tan θ}{1 + \tan t . \tan θ} ∣ + c$ $= \frac{1}{5} \ln ∣ \frac{1}{\frac{4}{5} \cos t + \frac{3}{5} \sin t} + \frac{\tan t - \frac{3}{4}}{1 + \frac{3}{4} \tan t} ∣ + c$
	Answered by mr W last updated on 09/May/20

	$= \frac{1}{5} \int \frac{d t}{\frac{3}{5} \sin t + \frac{4}{5} \cos t}$ $= \frac{1}{5} \int \frac{d t}{\sin α \sin t + \cos α \cos t}$ $= \frac{1}{5} \int \frac{d (t - α)}{\cos (t - α)}$ $= \frac{1}{5} \int \frac{d u}{\cos u} w i t h u = t - α$ $= \frac{1}{5} \int \frac{d (\sin u)}{1 - \sin^{2} u}$ $= \frac{1}{5} \int \frac{d v}{1 - v^{2}} w i t h v = \sin u = \sin (t - α)$ $= \frac{1}{10} \int (\frac{1}{1 - v} + \frac{1}{1 + v}) d v$ $= \frac{1}{10} \ln ∣ \frac{1 + v}{1 - v} ∣ + C$ $= \frac{1}{10} \ln ∣ \frac{1 + \sin (t - α)}{1 - \sin (t - α)} ∣ + C$ $= \frac{1}{10} \ln ∣ \frac{1 + \sin t \cos α - \sin α \cos t}{1 - \sin t \cos α + \sin α \cos t} ∣ + C$ $= \frac{1}{10} \ln ∣ \frac{5 + 4 \sin t - 3 \cos t}{5 - 4 \sin t + 3 \cos t} ∣ + C$
	Commented by s.ayeni14@yahoo.com last updated on 09/May/20

	$sir how ?$
	Commented by mr W last updated on 09/May/20

	$h o w a r e y o u ?$
	Commented by s.ayeni14@yahoo.com last updated on 09/May/20

	$by the integrals of the form \int \frac{1}{a + bsint + ccost} dt$ $then set x = \frac{t}{2}$
	Commented by mr W last updated on 09/May/20

	$w h a t h a p p e n s i f o n e {d o e s n}^{'} t d o t h a t$ $w a y b u t s o l v e s t h e q u e s t i o n n e v e r t h e l e s s ?$
	Commented by mr W last updated on 09/May/20

	$c a n y o u t e l l m e w h e r e t h e e r r o r i s i n m y$ $w a y ?$
	Commented by Ar Brandon last updated on 09/May/20
	��It wasn't gonna make a great difference if Mr W used the same method everyone's currently using to arrive at the answer. Thanks Mr W for drawing my attention to your method.��
	Commented by Ar Brandon last updated on 09/May/20

	$I understood your method and now feel$ $I could arrive at the answer at level$ $\frac{1}{5} \int \frac{1}{\cos u} du = \frac{1}{5} \ln ∣ \sec u + \tan u ∣$
	Answered by s.ayeni14@yahoo.com last updated on 09/May/20
	$x=tan(t/2) 3sint+4cost=((6x)/(1+x^2 ))+((4(1−3^2 ))/(1+x^2 )) =((4+6x−4x^2 )/(1+x^2 )) ∫(dt/(3sint+4cost))=∫((1+x^2 )/(4+6x−4x^2 ))(((2dx)/(1+x^2 ))) =∫(1/(2+3x−2x^2 ))dx =(1/2)∫(1/(1+(3/2)−x^2 ))dx 1+(3/2)x−x^2 =((5/4))^2 −(x−(3/4))^2 (1/2)∫(1/(((5/4))^2 −(x−(3/4))^2 ))dx=(1/5)ln{((1+2x)/(4−2x))}+C =(1/5)ln{((1+2tan(t/2))/(4−2tan(t/2)))}+C$
	$x = \tan \frac{t}{2}$ $3 sint + 4 cost = \frac{6 x}{1 + x^{2}} + \frac{4 (1 - 3^{2})}{1 + x^{2}}$ $= \frac{4 + 6 x - {4 x}^{2}}{1 + x^{2}}$ $\int \frac{dt}{3 sint + 4 cost} = \int \frac{1 + x^{2}}{4 + 6 x - {4 x}^{2}} (\frac{2 dx}{1 + x^{2}})$ $= \int \frac{1}{2 + 3 x - {2 x}^{2}} dx$ $= \frac{1}{2} \int \frac{1}{1 + \frac{3}{2} - x^{2}} dx$ $1 + \frac{3}{2} x - x^{2} = {(\frac{5}{4})}^{2} - {(x - \frac{3}{4})}^{2}$ $\frac{1}{2} \int \frac{1}{{(\frac{5}{4})}^{2} - {(x - \frac{3}{4})}^{2}} dx = \frac{1}{5} \ln {\frac{1 + 2 x}{4 - 2 x}} + C$ $= \frac{1}{5} \ln {\frac{1 + 2 \tan \frac{t}{2}}{4 - 2 \tan \frac{t}{2}}} + C$
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