0-3-3-2-x-3-4x-2-9-dx-

Question Number 13364 by tawa tawa last updated on 19/May/17

\int_{0}^{\frac{3 \sqrt{3}}{2}} \frac{x^{3}}{\sqrt{{4 x}^{2} - 9}} dx

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 19/May/17

y o u s h o u l d c h a n g e t h e l i m i t s o f i n t e g r a l .

f u n c t i o n h a v e n o t r e a l v a l v e s i n i n t e r v a l : [0, \frac{3 \sqrt{3}}{2}] .

Answered by ajfour last updated on 19/May/17

c o n s i d e r i n g I = \int \frac{x^{3}}{\sqrt{4 x^{2} - 9}} d x

l e t t = 4 x^{2} - 9

d t = 8 x d x

t d t = 32 x^{3} d x - 9 \times 8 x d x

32 x^{3} d x = (t + 9) d t

s o,

I = \frac{1}{32} \int \frac{(t + 9) d t}{\sqrt{t}}

= \frac{1}{32} (\frac{3}{2} t^{3 / 2} + 18 \sqrt{t}) + C

= \frac{3}{64} {(4 x^{2} - 9)}^{3 / 2} + \frac{9}{16} (4 x^{2} - 9) + C

i f t h e l i m i t s w e r e a p p r o p r i a t e

i w o u l d h a v e t r i e d t o g e t a n a n s w e r .

Commented by tawa tawa last updated on 19/May/17

God bless you sir

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 19/May/17

\frac{2 x}{3} = i . t g t \Rightarrow d x = \frac{3}{2} i (1 + {t g}^{2} t) d t \Rightarrow

c o s t = \frac{1}{\sqrt{1 - \frac{4 x^{2}}{9}}} = \frac{3}{\sqrt{9 - 4 x^{2}}} = \frac{- 3 i}{\sqrt{4 x^{2} - 9}}

4 x^{2} - 9 = - 4 . \frac{9}{4} {t g}^{2} t - 9 = - 9 ({t g}^{2} t + 1) =

= \frac{- 9}{{c o s}^{2} t} (i^{2} = - 1)

I = \int \frac{\frac{27}{8} i^{3} . {t g}^{3} t . \frac{3}{2} i . (1 + {t g}^{2} t) d t}{3 i / c o s t} =

= - \frac{27}{16} i \int \frac{c o s t . {t g}^{3} t}{{c o s}^{2} t} d t = - \frac{27 i}{16} \int \frac{{s i n}^{3} t}{{c o s}^{4} t} d t =

= - \frac{27 i}{16} \int \frac{s i n t (1 - {c o s}^{2} t) d t}{{c o s}^{4} t} =

= \frac{- 27 i}{16} \int [\frac{s i n t}{{c o s}^{4} t} - \frac{s i n t}{{c o s}^{2} t}] d t =

= \frac{- 27 i}{16} (- \frac{1}{5 {c o s}^{5} t} + \frac{1}{{c o s}^{3} t}) + C . =

= \frac{- 27 i}{16} (- \frac{1}{5} \frac{{(4 x^{2} - 9)}^{2} \sqrt{4 x^{2} - 9}}{- 3^{5} i} + \frac{(4 x^{2} - 9) \sqrt{4 x^{2} - 9}}{- 27 i}) + C =

= \frac{- (4 x^{2} - 9) \sqrt{4 x^{2} - 9}}{16 \times 45} (4 x^{2} - 54) + C = F (x)

F (\frac{3 \sqrt{3}}{2}) = \frac{- 18 \sqrt{18}}{16 \times 45} (27 - 54) = + \frac{81}{40} \sqrt{2}

F (0) = \frac{27 i}{16 \times 45} (- 54) = - \frac{81}{40} i

\Rightarrow I = \frac{81}{40} (\sqrt{2} - i) .