Tag Info

New answers tagged

0

Wolfram Alpha shows only the whole number in the result field. If you push the spacebar, after you have input the term, Wolfram Alpha provides a result with more digits, which is $1311.339133974715$


0

Wolfram alpha is confused by your use of degrees, I know not why. If you replace "30 deg" by "pi/6" then everything is fine. Also, if you replace "0.5" by "(1/2)" then it allows you to look at the exact form, while still reporting the misleading integral result.


0

Try including to x significant digits (see here for example).


1

I think the problem comes from using degrees. I'm not sure exactly how Wolfram Alpha does its calculations on back-end, but compare your results with mine, where all I did was replace your 30 deg with pi/6 in both trig functions. Then click on 'more digits' a few times... My best guess is that adding the deg unit specification makes Wolfram Alpha consider ...


0

Try This. Don't include the deg in the command.


2

They are both the same. Note $$\left(1 + \frac{\pi^2}{x^2}\right)^{3/2} = \left(\frac{x^2 + \pi^2}{x^2}\right)^{3/2} = \frac{(x^2 + \pi^2)^{3/2}}{x^3}$$ So $$\frac{\pi}{\left(1 + \frac{\pi^2}{x^2}\right)^{3/2}x^2} = \frac{\pi}{\frac{(x^2 + \pi^2)^{3/2}}{x}} = \frac{\pi x}{(x^2 + \pi^2)^{3/2}}$$


2

TheĆ³phile's comment is a good answer. I will give another. The $\sqrt{4-x^2}$ and indeed anything of the form $\sqrt{a^2-b^2}$ --- think $\sqrt{x^2-b^2}$ or $\sqrt{a^2-x^2}$ --- or $x^2+b^2$ can be considered --- via Pythagoras --- as sides of a right-angled triangle. In this example we have $\sqrt{2^2-x^2}=b\Rightarrow b^2+x^2=2^2$ so that we have a ...


3

This is based on the fact that the usual parametric equation of the curve $x^2 + y^2 = R^2$ is $X = R\sin t, Y = R\cos t$. You want to make a change of variables, to make the integrand look simpler (that is the point of the change of variables). So you would like to simplify $$ \sqrt{4 - x^2} $$and in particular get rid of the $\sqrt.$. So you look for a ...


5

This is a common substitution technique known as Trig substitution, you substitute $x$ with a trigonometric function. Typically, when something is in the form $\sqrt{a-x^2}$, you substitute $x=\sqrt a\sin u$


2

In Mathematica, $f(\#1, \#2, \dots)\&$ is an abbreviation of the function $(x_1, x_2, ...) \to f(x_1, x_2, \dots)$, and Root[$f$, $k$] returns the $k$th zero of a polynomial function $f(x)$. (I'm not sure how they're ordered.) Thus as a function of time, $y(t)$ satisifes some fifth-degree polynomial with coefficients depending on $t$. To get an explicit ...


12

Since $$ \frac{1}{\cos x \sin x} =\frac{\cos x }{\sin x}+\frac{ \sin x}{\cos x} $$ you easily deduce that $$ \int\frac{1}{\cos x \sin x} {\rm d} x=\ln |\sin x|-\ln |\cos x|+C=\ln |\tan x|+C $$ for any constant $C$.


18

Rewrite it as: $$\frac 1{(\sin x)(\cos x)} = \frac 1 {\frac{\sin x}{\cos x}\cos^2x} = \frac 1 {\tan x \cdot \cos^2x} = \frac 1 {\tan x} \cdot \frac 1 {\cos^2x}$$ Now, since you know the derivative of the tangent, you substitute $u = \tan x$. You have that $\mathrm du = \frac 1 {\cos^2x} \mathrm dx$. I'm confident that you can continue from here.


2

Is the standard change when the integrand is a rational function of $\sin^2$, $\cos^2$ and $\sin\cos$. Check yourself that $\sin^2$, $\cos^2$ and $\sin\cos$ can be written in function of $\tan$ without square roots starting from $1+\tan^2=\sec^2$.


1

Your work agrees exactly with Wolfram Alpha's. Try expanding the denominator of your expression and the denominator of Wolfram Alpha's expression.


0

It only proves that you cannot rely on calculators and computer algebra systems and internet queries to solve problems for you... and that none of it is a substitute for the human brain. Im glad you found the right function to use... but had you not thought about it and blindly went along, like most people do, youd be wrong and everything that stems from it ...


1

It's applying the exponential function element-wise, rather than computing the matrix exponential.


3

The equation is separable $$\frac{dt}{dh}=h^{-j}$$ So $$t+c_1=\frac{h^{1-j}}{1-j}$$ Solving for $h$ gives in the most general case $$h=\Big((j-1) (-c_1-t)\Big)^{\frac{1}{1-j}}=\Big((j-1) (c_2-t)\Big)^{\frac{1}{1-j}}$$ which is what Wolfram Alpha returns.


1

You can simplify your life (and WA's too) taking into account the fact that $x,y,a,b$ are constants for the problem. So, defining some intermediate numbers, your equations write $$A=s+p$$ $$B=q+t$$ $$C=\sqrt{p^2+q^2}+\sqrt{s^2+t^2}$$ $$D=p^2+q^2+s^2+t^2$$ and the result is just a monster ! You can eliminate variables $s$ and $t$ using the first equations ...


0

Type Solve [{ x+a==s+p && y+b==q+t && Sqrt[x^2+y^2]+Sqrt[a^2+b^2]==Sqrt[s^2+t^2]+Sqrt[p^2+q^2] && x^2+y^2+a^2+b^2=(4/5)*(s^2+t^2+p^2+q^2) }, {s, t, p, q}] Then wolfram mathematica works correctly. Evaluation takes a few minutes... The answer is too long to display. It can print the answer and this takes also a few ...



Top 50 recent answers are included