Question-213139

Question Number 213139 by Spillover last updated on 31/Oct/24

Answered by MrGaster last updated on 31/Oct/24

l e t u = \sqrt{x} + \sqrt{y} + z \Rightarrow d y = \frac{1}{2 \sqrt{x}} d x + \frac{1}{2 \sqrt{y}} d y + \frac{1}{2 \sqrt{z}} d x

\Rightarrow 0 \leq u \leq \frac{3 \sqrt{π}}{4}

\Rightarrow d x d y d z = 8 u^{2} d u

\Leftrightarrow \int_{0}^{\frac{3 \sqrt{π}}{4}} \sin (u) \cdot 8 u^{2} d u \Rightarrow 8 \int_{0}^{\frac{3 \sqrt{π}}{4}} u^{2} \sin (u) d u

l e t u = u^{2}, d u = 2 u d u, d w = \sin (u) d u, w = - \cos (u)

\Rightarrow \int u^{2} \sin (u) d u = - u^{2} \cos (u) + 2 \int u \cos (u)

l e t v = u, d v = d u, d w = \cos (u) d u, w = \sin (u)

\Rightarrow \int u \cos (u) d u = u \sin (u) d u = u \sin (u) + \cos (u)

\Rightarrow \int u^{2} \sin (u) d u = - u^{2} \cos (u) + 2 (u \sin (u) + \cos (u))

8 - [- (\frac{3 \sqrt{π}}{4}) \cos (\frac{3 \sqrt{π}}{4}) + \frac{3 \sqrt{π}}{2} \sin (\frac{3 \sqrt{π}}{4}) + 2 \cos (\frac{3 \sqrt{π}}{4}) + 2 \cos \frac{3 \sqrt{π}}{4}]

\Rightarrow 8 [- \frac{9 π}{16} \cos (\frac{3 \sqrt{π}}{4}) + \frac{3 \sqrt{π}}{2} \sin (\frac{3 \sqrt{π}}{4}) + 2 \cos (\frac{3 \sqrt{π}}{4})]

\Leftrightarrow 8 [(2 - \frac{9 π}{16}) \cos (\frac{3 \sqrt{π}}{4}) + \frac{3 \sqrt{π}}{2} \sin (\frac{3 \sqrt{π}}{4})]

Commented by Frix last updated on 31/Oct/24

No .

Commented by MathematicalUser2357 last updated on 02/Nov/24

= 21.097872

Answered by Frix last updated on 31/Oct/24

Using

(1)

\int \sin (α + \sqrt{t}) d t \overset{[u = \sqrt{t}]}{=} 2 \int u \sin (α + u) d u \overset{[by parts .]}{=}

= 2 (\sin (α + u) - u \cos (α + u)) =

= 2 (\sin (α + \sqrt{t}) - \sqrt{t} \cos (α + \sqrt{t}))

(2) Similar

\int \cos (α + \sqrt{t}) d t = 2 (\cos (α + \sqrt{t}) + \sqrt{t} \sin (α + \sqrt{t}))

We get

6 (4 \sqrt{π} \cos \sqrt{π} + (π - 4) \sin \sqrt{π}) -

- 12 (\sqrt{π} \cos \frac{\sqrt{π}}{2} - 2 \sin \frac{\sqrt{π}}{2}) +

+ \sqrt{π} (π - 12) \cos \frac{3 \sqrt{π}}{2} - 2 (3 π - 4) \sin \frac{3 \sqrt{π}}{2}

Answered by lepuissantcedricjunior last updated on 01/Nov/24

k = \int_{0}^{\frac{π}{4}} \int_{0}^{\frac{π}{4}} \int_{0}^{\frac{π}{4}} s i n (\sqrt{x} + \sqrt{y} + \sqrt{z}) d x d y d z

= I m \int_{0}^{\frac{π}{4}} \int_{0}^{\frac{π}{4}} \int_{0}^{\frac{π}{4}} e^{i (\sqrt{x} + \sqrt{y} + \sqrt{z})} d x d y d z

= I m (\int_{0}^{\frac{π}{4}} e^{i \sqrt{x}} d x \int_{0}^{\frac{π}{4}} e^{i \sqrt{y}} d y \int_{0}^{\frac{π}{4}} e^{i \sqrt{z}} d z)

p o s o n s x = n^{2}; y = p^{2}; z = q^{2}

= I m [{{\int_{0}}^{\sqrt{\frac{π}{4}}} 2 {n e}^{i n} d n} {\int_{0}^{\sqrt{\frac{π}{4}}} 2 {p e}^{i p} d p} {\int_{0}^{\sqrt{\frac{π}{4}}} 2 {q e}^{i q} d q}]

= I m {{[\frac{2 {n e}^{i n}}{i} + 2 e^{i n} + \frac{2 {p e}^{i p}}{i} + 2 e^{i p} + \frac{2 {q e}^{i q}}{i} + 2 e^{i q}]}_{0}^{\frac{\sqrt{π}}{2}}

= I m [3 \frac{\sqrt{π}}{i} e^{i \frac{\sqrt{π}}{2}} - 6] = I m (- 6 + 3 \sqrt{π} e^{i (\frac{\sqrt{π} - π}{2})})

i = 3 \sqrt{π} s i n (\frac{\sqrt{π}}{2} - \frac{π}{2}) = - 3 \sqrt{π} c o s (\frac{\sqrt{π}}{2})

Commented by Frix last updated on 01/Nov/24

But {it}^{'} s wrong .

But {it}^{'} s wrong .

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