∫sin^3 (√x) dx


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Question Number 43861 by Tinkutara last updated on 16/Sep/18

$\int \sin^{3} \sqrt{x} d x$
Commented by Tinkutara last updated on 16/Sep/18

$A n s w e r g i v e n i s$ $- 3 x^{3 / 2} \cos^{3} \sqrt{x} + 6 x^{1 / 3} \sin^{3} \sqrt{x} +$ ${6 \cos}^{3} \sqrt{x} + C$ $P l e a s e t e l l i f t h e a n s w e r i s c o r r e c t$ $o r w r o n g!$
Commented by maxmathsup by imad last updated on 16/Sep/18
$changement (√x)=t give I = ∫ sin^3 (t)(2t)dt =2 ∫ t sin^3 (t)dt but sin^3 (t)=(((e^(it) −e^(−it) )/(2i)))^3 =(1/(−8i)){ e^(it) −e^(−it) }^3 =(i/8){ Σ_(k=0) ^3 C_3 ^k e^(ikt) (−1)^(3−k) e^(−i(3−k)t) } =(i/8) Σ_(k=0) ^3 (−1)^(3−k) C_3 ^k e^(i(2k−3)t) =(i/8){ −e^(−3it) +3 e^(−it) −3 e^(it) +e^(i3t) }=(i/8){ 2isin(3t) −6i sin(t)} =−(1/4)sin(3t) +(3/4)sin(t) ⇒ I = 2 ∫ t ((3/4)sint −(1/4)sin(3t))dt=(3/2) ∫ t sint dt −(1/2) ∫ t sin(3t)dt but =(3/2) H −(1/2) K by parte H =−t cost +∫ cost =−t cost +sint also K =−(t/3)cos(3t) + ∫ (1/3)cos(3t)dt =−(t/3)cos(3t) +(1/9)sin(3t) ⇒ I =−(3/2)t cost +(3/2)sint +(t/6)cos(3t) −(1/(18))sin(3t) +c =−(3/2)(√x)cos((√x)) +(3/2)sin((√x)) +((√x)/6)cos(3(√x))−(1/(18))sin(3(√x)) +c .$
$c h a n g e m e n t \sqrt{x} = t g i v e I = \int {s i n}^{3} (t) (2 t) d t$ $= 2 \int t {s i n}^{3} (t) d t b u t {s i n}^{3} (t) = {(\frac{e^{i t} - e^{- i t}}{2 i})}^{3} = \frac{1}{- 8 i} {e^{i t} - e^{- i t}}^{3}$ $= \frac{i}{8} {\sum_{k = 0}^{3} C_{3}^{k} e^{i k t} {(- 1)}^{3 - k} e^{- i (3 - k) t}} = \frac{i}{8} \sum_{k = 0}^{3} {(- 1)}^{3 - k} C_{3}^{k} e^{i (2 k - 3) t}$ $= \frac{i}{8} {- e^{- 3 i t} + 3 e^{- i t} - 3 e^{i t} + e^{i 3 t}} = \frac{i}{8} {2 i s i n (3 t) - 6 i s i n (t)}$ $= - \frac{1}{4} s i n (3 t) + \frac{3}{4} s i n (t) \Rightarrow$ $I = 2 \int t (\frac{3}{4} s i n t - \frac{1}{4} s i n (3 t)) d t = \frac{3}{2} \int t s i n t d t - \frac{1}{2} \int t s i n (3 t) d t b u t$ $= \frac{3}{2} H - \frac{1}{2} K b y p a r t e H = - t c o s t + \int c o s t = - t c o s t + s i n t a l s o$ $K = - \frac{t}{3} c o s (3 t) + \int \frac{1}{3} c o s (3 t) d t = - \frac{t}{3} c o s (3 t) + \frac{1}{9} s i n (3 t) \Rightarrow$ $I = - \frac{3}{2} t c o s t + \frac{3}{2} s i n t + \frac{t}{6} c o s (3 t) - \frac{1}{18} s i n (3 t) + c$ $= - \frac{3}{2} \sqrt{x} c o s (\sqrt{x}) + \frac{3}{2} s i n (\sqrt{x}) + \frac{\sqrt{x}}{6} c o s (3 \sqrt{x}) - \frac{1}{18} s i n (3 \sqrt{x}) + c .$
Commented by Tinkutara last updated on 16/Sep/18
Thank you very much Sir! I got the answer. ��
Commented by maxmathsup by imad last updated on 16/Sep/18

$y o u a r e w e l c o m e s i r .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 16/Sep/18
	$t^2 =x ∫sin^3 t×2tdt sin3t=3sint−4sin^3 t ∫(((3sint−sin3t)/4))×2t dt (3/2)∫sint.tdt−(1/2)∫sin3t×t dt (3/2)I_1 −(1/2)I_2 ∫sint ×t dt t×−cost−∫[(dt/dt)∫sint dt]dt −tcost−∫−costdt =−tcost+sint (3/2)(−tcost+sint) (3/2)((√x) cos(√x) +sin(√x) ) ∫tsin3tdt =t×((−cos3t)/3)−∫[(dt/dt)∫sin3tdt]dt =((−tcos3t)/3)+∫((cos3t)/3) =((−tcos3t)/3)+((sin3t)/9) =((3sint−4sin^3 t)/9)+(((−t)(4cos^3 t−3cost))/3) =((3sin(√x) −4sin^3 (√x) )/9)+(((−(√x) )(4cos^3 (√x) −3cos(√x) ))/3) so (1/2)I_2 ((3sin(√x) −4sin^3 (√x))/(18))+(((−(√x) )(4cos^3 (√x) −3cos(√x) ))/6) answer is (3/2)I_1 −(1/2)I_2 (3/2)((√x) cos(√x) +sin(√x) )−{((3sin(√x) −4sin^3 (√x) )/(18))+(((−(√x) )(4cos^3 (√x) −3cos(√x) ))/6)}$
	$t^{2} = x$ $\int {s i n}^{3} t \times 2 t d t$ $s i n 3 t = 3 s i n t - 4 {s i n}^{3} t$ $\int (\frac{3 s i n t - s i n 3 t}{4}) \times 2 t d t$ $\frac{3}{2} \int s i n t . t d t - \frac{1}{2} \int s i n 3 t \times t d t$ $\frac{3}{2} I_{1} - \frac{1}{2} I_{2}$ $\int s i n t \times t d t$ $t \times - c o s t - \int [\frac{d t}{d t} \int s i n t d t] d t$ $- t c o s t - \int - c o s t d t$ $= - t c o s t + s i n t$ $\frac{3}{2} (- t c o s t + s i n t)$ $\frac{3}{2} (\sqrt{x} c o s \sqrt{x} + s i n \sqrt{x})$ $\int t s i n 3 t d t$ $= t \times \frac{- c o s 3 t}{3} - \int [\frac{d t}{d t} \int s i n 3 t d t] d t$ $= \frac{- t c o s 3 t}{3} + \int \frac{c o s 3 t}{3}$ $= \frac{- t c o s 3 t}{3} + \frac{s i n 3 t}{9}$ $= \frac{3 s i n t - 4 {s i n}^{3} t}{9} + \frac{(- t) (4 {c o s}^{3} t - 3 c o s t)}{3}$ $= \frac{3 s i n \sqrt{x} - 4 {s i n}^{3} \sqrt{x}}{9} + \frac{(- \sqrt{x}) (4 {c o s}^{3} \sqrt{x} - 3 c o s \sqrt{x})}{3}$ $s o$ $\frac{1}{2} I_{2}$ $\frac{3 s i n \sqrt{x} - 4 {s i n}^{3} \sqrt{x}}{18} + \frac{(- \sqrt{x}) (4 {c o s}^{3} \sqrt{x} - 3 c o s \sqrt{x})}{6}$ $a n s w e r i s \frac{3}{2} I_{1} - \frac{1}{2} I_{2}$ $\frac{3}{2} (\sqrt{x} c o s \sqrt{x} + s i n \sqrt{x}) - {\frac{3 s i n \sqrt{x} - 4 {s i n}^{3} \sqrt{x}}{18} + \frac{(- \sqrt{x}) (4 {c o s}^{3} \sqrt{x} - 3 c o s \sqrt{x})}{6}}$
	Commented by Tinkutara last updated on 16/Sep/18
	Thank you very much Sir! I got the answer. ��
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