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Question Number 46499 by Tawa1 last updated on 27/Oct/18

	Answered by MJS last updated on 27/Oct/18

	$\int \frac{6 \sin x \cos^{2} x + \sin 2 x - 23 \sin x}{{(1 - \cos x)}^{2} (5 - \sin^{2} x)} d x =$ $= \int \frac{{6 \cos}^{2} x + 2 \cos x - 23}{{(1 - \cos x)}^{2} (4 + \cos^{2} x)} \sin x d x =$ $[t = \cos x \to d x = - \frac{d t}{\sin x}]$ $= - \int \frac{6 t^{2} + 2 t - 23}{{(t - 1)}^{2} (t^{2} + 4)} d t = \int (\frac{4 t - 5}{t^{2} + 4} + \frac{3}{{(t - 1)}^{2}} - \frac{4}{t - 1}) d t =$ $= 4 \int \frac{t}{t^{2} + 4} d t - 5 \int \frac{d t}{t^{2} + 4} + 3 \int \frac{d t}{{(t - 1)}^{2}} - 4 \int \frac{d t}{t - 1} =$ $= 2 \ln (t^{2} + 4) - \frac{5}{2} \arctan \frac{t}{2} - \frac{3}{t - 1} - 4 \ln (t - 1) =$ $= 2 \ln \frac{t^{2} + 4}{{(t - 1)}^{2}} - \frac{5}{2} \arctan \frac{t}{2} + \frac{3}{1 - t} =$ $= 2 \ln ∣ \frac{4 + \cos^{2} x}{{(1 - \cos x)}^{2}} ∣ - \frac{5}{2} \arctan \frac{\cos x}{2} + \frac{3}{1 - \cos x} + C$
	Commented by Tawa1 last updated on 27/Oct/18

	$God bless you sir$
	Answered by tanmay.chaudhury50@gmail.com last updated on 27/Oct/18
	$∫((sinx{6cos^2 x+2cosx−23})/((cosx−1)^2 (5−1+cos^2 x)))dx t=cosx dt=−sinxdx ∫((−(6t^2 +2t−23)dt)/((t−1)^2 (4+t^2 ))) ((6t^2 +2t−23)/((t−1)^2 (4+t^2 )))=(a/((t−1)))+(b/((t−1)^2 ))+((ct+d)/(4+t^2 )) 6t^2 +2t−23=a(t−1)(4+t^2 )+b(4+t^2 )+(ct+d)(t−1)^2 6t^2 +2t−23=a(4t+t^3 −4−t^2 )+(4b+bt^2 )+(ct+d)(t^2 −2t+1) 6t^2 +2t−23=a(t^3 −t^2 +4t−4)+(4b+bt^2 )+(ct^3 −2ct^2 +ct+dt^2 −2dt+d) 6t^2 +2t−23=t^3 (a+c)+t^2 (−a+b−2c+d)+t(4a+c−2d)+(−4a+4b+d) a+c=0 −a+b−2c+d=6 4a+c−2d^ =2 −4a+4b+d=−23 −a+b+2a+d=6←from eqn 1(because a=−c) 4a−a+2d=2←( 3a=2−2d a=((2−2d)/3) a+b+d=6 ((2−2d)/3)+b+d=6 2−2d+3b+3d=18 3b=16−d b=((16−d)/3) 4×((16−d)/3)−4×((2−2d)/3)+d=23 ((64−4d−8+8d+3d=)/)69 7d=69−56=13 d=((13)/7) a=((2−2×((13)/7))/3)=((−12)/(21))=((−4)/7) c=(4/7) b=((16−((13)/7))/3)=((112−13)/(21))=((99)/(21))=((33)/7) ∫(a/(t−1))dt+∫(b/((t−1)^2 ))dt+∫((ct+d)/(4+t^2 ))dt (((−4)/7))ln(t−1)+(((33)/7))×((−1)/((t−1)))+∫(((4/7)t+((13)/7))/(4+t^2 ))dt (((−4)/7))ln(t−1)+(((−33)/7))×(1/(t−1))+(2/7)∫((d(4+t^2 ))/(4+t^2 ))+((13)/7)∫(dt/(4+t^2 )) (((−4)/7))ln(t−1)+(((−33)/7))×(1/(t−1))+((2/7))ln(4+t^2 )+(((13)/7))×(1/2)tan^(−1) ((t/2))+C (((−4)/7))ln(cosx−1)+(((−33)/7))×(1/(−1+cosx))+((2/7))ln(4+cos^2 x)+((13)/(14))tan^(−1) (((cosx)/2))+C$
	$\int \frac{s i n x {6 {c o s}^{2} x + 2 c o s x - 23}}{{(c o s x - 1)}^{2} (5 - 1 + {c o s}^{2} x)} d x$ $t = c o s x d t = - s i n x d x$ $\int \frac{- (6 t^{2} + 2 t - 23) d t}{{(t - 1)}^{2} (4 + t^{2})}$ $\frac{6 t^{2} + 2 t - 23}{{(t - 1)}^{2} (4 + t^{2})} = \frac{a}{(t - 1)} + \frac{b}{{(t - 1)}^{2}} + \frac{c t + d}{4 + t^{2}}$ $6 t^{2} + 2 t - 23 = a (t - 1) (4 + t^{2}) + b (4 + t^{2}) + (c t + d) {(t - 1)}^{2}$ $6 t^{2} + 2 t - 23 = a (4 t + t^{3} - 4 - t^{2}) + (4 b + {b t}^{2}) + (c t + d) (t^{2} - 2 t + 1)$ $6 t^{2} + 2 t - 23 = a (t^{3} - t^{2} + 4 t - 4) + (4 b + {b t}^{2}) + ({c t}^{3} - 2 {c t}^{2} + c t + {d t}^{2} - 2 d t + d)$ $6 t^{2} + 2 t - 23 = t^{3} (a + c) + t^{2} (- a + b - 2 c + d) + t (4 a + c - 2 d) + (- 4 a + 4 b + d)$ $a + c = 0$ $- a + b - 2 c + d = 6$ $4 a + c - 2 \overset{}{d} = 2$ $- 4 a + 4 b + d = - 23$ $- a + b + 2 a + d = 6 \leftarrow f r o m e q n 1 (b e c a u s e a = - c)$ $4 a - a + 2 d = 2 \leftarrow ($ $3 a = 2 - 2 d$ $a = \frac{2 - 2 d}{3}$ $a + b + d = 6$ $\frac{2 - 2 d}{3} + b + d = 6$ $2 - 2 d + 3 b + 3 d = 18$ $3 b = 16 - d$ $b = \frac{16 - d}{3}$ $4 \times \frac{16 - d}{3} - 4 \times \frac{2 - 2 d}{3} + d = 23$ $\frac{64 - 4 d - 8 + 8 d + 3 d =}{} 69$ $7 d = 69 - 56 = 13 d = \frac{13}{7}$ $a = \frac{2 - 2 \times \frac{13}{7}}{3} = \frac{- 12}{21} = \frac{- 4}{7} c = \frac{4}{7}$ $b = \frac{16 - \frac{13}{7}}{3} = \frac{112 - 13}{21} = \frac{99}{21} = \frac{33}{7}$ $\int \frac{a}{t - 1} d t + \int \frac{b}{{(t - 1)}^{2}} d t + \int \frac{c t + d}{4 + t^{2}} d t$ $(\frac{- 4}{7}) l n (t - 1) + (\frac{33}{7}) \times \frac{- 1}{(t - 1)} + \int \frac{\frac{4}{7} t + \frac{13}{7}}{4 + t^{2}} d t$ $(\frac{- 4}{7}) l n (t - 1) + (\frac{- 33}{7}) \times \frac{1}{t - 1} + \frac{2}{7} \int \frac{d (4 + t^{2})}{4 + t^{2}} + \frac{13}{7} \int \frac{d t}{4 + t^{2}}$ $(\frac{- 4}{7}) l n (t - 1) + (\frac{- 33}{7}) \times \frac{1}{t - 1} + (\frac{2}{7}) l n (4 + t^{2}) + (\frac{13}{7}) \times \frac{1}{2} {t a n}^{- 1} (\frac{t}{2}) + C$ $(\frac{- 4}{7}) l n (c o s x - 1) + (\frac{- 33}{7}) \times \frac{1}{- 1 + c o s x} + (\frac{2}{7}) l n (4 + {c o s}^{2} x) + \frac{13}{14} {t a n}^{- 1} (\frac{c o s x}{2}) + C$
	Commented by tanmay.chaudhury50@gmail.com last updated on 27/Oct/18

	$p l s c h e c k c a l c u l a t i o n t o f i n d t h e v a l u e o f a b c d$
	Commented by Tawa1 last updated on 27/Oct/18

	$God bless you sir$
	Commented by MJS last updated on 27/Oct/18

	$error in line above the first blue line :$ $4 a - a - 2 d = 2$
	Commented by tanmay.chaudhury50@gmail.com last updated on 27/Oct/18

	$y e s s i r t h a n k y o u s i r . . . i s h a l r e c t i f y a n d r e p o s t i t . . .$
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