sin-3-x-dx-

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

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

\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. ��