solve ∫ (dx/(3−5sin x))


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Question Number 118338 by benjo_mathlover last updated on 17/Oct/20

$s o l v e \int \frac{d x}{3 - 5 \sin x}$
	Answered by Dwaipayan Shikari last updated on 17/Oct/20

	$\int \frac{d x}{3 - 5 s i n x}$ $= 2 \int \frac{d t}{3 - \frac{10 t}{(1 + t^{2})}} . \frac{1}{1 + t^{2}} t = t a n \frac{x}{2}$ $= 2 \int \frac{d t}{3 + 3 t^{2} - 10 t} = \frac{2}{3} \int \frac{d t}{{(t - \frac{5}{3})}^{2} + 1 - \frac{25}{9}}$ $= \frac{2}{3} \int \frac{d t}{{(t - \frac{5}{3})}^{2} - {(\frac{4}{3})}^{2}}$ $= \frac{1}{4} l o g (\frac{t - \frac{5}{3} - \frac{4}{3}}{t - \frac{5}{3} + \frac{4}{3}}) = \frac{1}{4} l o g (\frac{3 t - 9}{3 t - 1}) = \frac{1}{4} l o g (\frac{3 t a n \frac{x}{2} - 9}{3 t a n \frac{x}{2} - 1}) + 4$
	Commented by som(math1967) last updated on 17/Oct/20

	$Are you ex - student of$ $Baranagar R . K . Mission$
	Commented by Dwaipayan Shikari last updated on 17/Oct/20

	$N o s i r . I a m c u r r e n t l y s t u d y i n g o n J a d a v p u r V i d y a p i t h$
	Commented by som(math1967) last updated on 17/Oct/20

	$Ok$
	Commented by Dwaipayan Shikari last updated on 17/Oct/20

	$A r e y o u f r o m B e n g a l s i r ?$
	Commented by som(math1967) last updated on 17/Oct/20

	$yes, kolkata dunlop$
	Answered by som(math1967) last updated on 17/Oct/20

	$\int \frac{\sec^{2} \frac{x}{2} dx}{{3 \sec}^{2} \frac{x}{2} - 5 \times 2 \sin \frac{x}{2} \cos \frac{x}{2} \sec^{2} \frac{x}{2}}$ $\int \frac{\sec^{2} \frac{x}{2} dx}{3 + {3 \tan}^{2} \frac{x}{2} - 10 \tan \frac{x}{2}}$ $let \tan \frac{x}{2} = z$ $\sec^{2} \frac{x}{2} \times \frac{1}{2} dx = dz$ $\sec^{2} \frac{x}{2} dx = 2 dz$ $2 \int \frac{dz}{{3 z}^{2} - 10 z + 3}$ $\frac{2}{3} \int \frac{dz}{z^{2} - \frac{10 z}{3} + 1}$ $\frac{2}{3} \int \frac{dz}{z^{2} - 2 . z . \frac{5}{3} + \frac{25}{9} + 1 - \frac{25}{9}}$ $\frac{2}{3} \int \frac{dz}{{(z - \frac{5}{3})}^{2} - {(\frac{4}{3})}^{2}}$ $\frac{2}{3} \times \frac{3}{2 \times 4} \ln ∣ \frac{z - \frac{5}{3} - \frac{4}{3}}{z - \frac{5}{3} + \frac{4}{3}} ∣ + c$ $\frac{1}{4} \ln ∣ \frac{z - \frac{9}{3}}{z - \frac{1}{3}} ∣ + c ans$ $z = \tan \frac{x}{2}$
	Answered by 1549442205PVT last updated on 17/Oct/20
	$Put t=tan(x/2)⇒dt=(1/2)(1+t^2 )dx F=∫(( dt)/((t^2 +1)(3−((10t)/(1+t^2 )))))=∫((2dt)/((3t^2 −10t+3))) =(2/(3t^2 −10t+3))=(a/(3t−1))+(b/(t−3)) ⇔2=(a+3b)t−3a−b⇔ { ((3a+b=−2)),((a+3b=0)) :} ⇔b=1/4,a=−3/4 F=((−3)/4)∫(dt/(3t−1))+(1/4)∫(dt/(t−3))= (1/4)ln∣((t−3)/(3t−1))∣+C =(1/4)ln∣((tan(x/2)−3)/(3tan(x/2)−1))∣+C$
	$Put t = \tan \frac{x}{2} \Rightarrow dt = \frac{1}{2} (1 + t^{2}) dx$ $F = \int \frac{dt}{(t^{2} + 1) (3 - \frac{10 t}{1 + t^{2}})} = \int \frac{2 dt}{({3 t}^{2} - 10 t + 3)}$ $= \frac{2}{{3 t}^{2} - 10 t + 3} = \frac{a}{3 t - 1} + \frac{b}{t - 3}$ $\Leftrightarrow 2 = (a + 3 b) t - 3 a - b \Leftrightarrow {\begin{cases} 3 a + b = - 2 \\ a + 3 b = 0 \end{cases}$ $\Leftrightarrow b = 1 / 4, a = - 3 / 4$ $F = \frac{- 3}{4} \int \frac{dt}{3 t - 1} + \frac{1}{4} \int \frac{dt}{t - 3} = \frac{1}{4} \ln ∣ \frac{t - 3}{3 t - 1} ∣ + C$ $= \frac{1}{4} \ln ∣ \frac{\tan \frac{x}{2} - 3}{3 \tan \frac{x}{2} - 1} ∣ + C$
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