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Question Number 190927 by alcohol last updated on 14/Apr/23

	Answered by MikeH last updated on 15/Apr/23
	$I = ∫t^7 sin(2t^4 )dt let u =2 t^4 ⇒ du = 8t^3 dt ⇒ dt = (du/(8t^3 )) ⇒ I = ∫t^7 sin(u)(du/(8t^3 )) = (1/8)∫t^4 sin(u)du I = (1/(16))∫usin udu by parts { ((a = u ⇒ a′ = 1)),((b′ = sin u⇒b = −cos u)) :} I = −(1/(16))[−ucos u + ∫cos u du] I = −(1/(16))(−ucos u + sin u) + k I = −(1/(16)) (−2t^4 cos (2t^4 ) + sin (2t^4 )) + k welcome again to COT and maths 2 hehe$
	$I = \int t^{7} \sin (2 t^{4}) d t$ $let u = 2 t^{4} \Rightarrow d u = 8 t^{3} d t \Rightarrow d t = \frac{d u}{8 t^{3}}$ $\Rightarrow I = \int t^{7} \sin (u) \frac{d u}{8 t^{3}} = \frac{1}{8} \int t^{4} \sin (u) d u$ $I = \frac{1}{16} \int u \sin u d u$ $by parts {\begin{cases} a = u \Rightarrow a^{'} = 1 \\ b^{'} = \sin u \Rightarrow b = - \cos u \end{cases}$ $I = - \frac{1}{16} [- u \cos u + \int \cos u d u]$ $I = - \frac{1}{16} (- u \cos u + \sin u) + k$ $I = - \frac{1}{16} (- 2 t^{4} \cos (2 t^{4}) + \sin (2 t^{4})) + k$ $welcome again to COT and maths 2 hehe$
	Answered by MikeH last updated on 15/Apr/23

	$2 . 2.1 g (x, y, z) = e^{3 y + 4 z} \sin (5 x)$ $\frac{\partial g}{\partial x} = 5 e^{3 y + 4 z} \cos (5 x) \Rightarrow \frac{\partial^{2} g}{\partial x^{2}} = - 25 e^{3 y + 4 z} \sin (5 x)$ $\frac{\partial g}{\partial y} = 3 e^{3 y + 4 z} \sin (5 x) \Rightarrow \frac{\partial^{2} g}{\partial y^{2}} = 9 e^{3 y + 4 z} \sin (5 x)$ $\frac{\partial g}{\partial z} = 4 e^{3 y + 4 z} \sin (5 x) \Rightarrow \frac{\partial^{2} g}{\partial z^{2}} = 16 e^{3 y + 4 z} \sin (5 x)$ $\frac{\partial^{2} g}{\partial x^{2}} + \frac{\partial^{2} g}{\partial y^{2}} + \frac{\partial^{2} g}{\partial z^{2}} = - 25 e^{3 y + 4 z} \sin (5 x) + 9 e^{3 y + 4 z} \sin (5 x) + 16 e^{3 y + 4 z} \sin (5 x) = 0$ $Since \frac{\partial^{2} g}{\partial x^{2}} + \frac{\partial^{2} g}{\partial y^{2}} + \frac{\partial^{2} g}{\partial z^{2}} = 0 g (x, y, z) satisfies laplace equation$ $w e l c o m e t o C O T, w e l c o m e t o m a t h s 2$
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