# Tag Info

0

As suggested in the comments, the apparently complex solutions stem from expressions presumably found using Cardano's Formula to solve the cubic equation. If these solutions are manipulated not via algebraic manipulations, but rather using numerical methods, rounding errors may arise leading to results having small complex parts. I tried, however, to enter ...

0

Let us see. We have to evaluate: $$I=\int_{-6}^{-3}\frac{\sqrt{x^2-9}}{x}\mathrm{d}x.$$ You suggest substituting $u=\sqrt{x^2-9}$, thus getting $\mathrm{d}u=\frac{x}{\sqrt{x^2-9}}\mathrm{d}x\implies\mathrm{d}u=\frac{u}{-\sqrt{u^2+9}}\mathrm{d}u$. The minus is due to the fact that we are integrating over $[-6,-3]$ where $x<0$ and the root would be ...

3

WolframAlpha distinguishes between x^(1/3) and cbrt(x) - the first returns the principal root while the second returns the real root. For example, cbrt(-8) returns $-2$ while (-8)^(1/3) returns $1 + 1.73205i$. Thus, you can use cbrt in your WolframAlpha query. integrate (1/(3*sqrt(x)*cbrt(log(x))), x=0..2) Note that the output states that the ...

2

The problem is that the logarithm has a branch point in $0$, as well as $\sqrt[3]{z}$, so we must be careful when defining the branches we are taking. If we simply set $x=e^t$ we have: $$I = \int_{-\infty}^{\log 2}\frac{e^{t/2}}{3 t^{1/3}}\,dt = \int_{0}^{\log 2}3t^{-1/3}e^{t/2}\,dt+\int_{-\infty}^{0}3t^{-1/3}e^{t/2}\,dt$$ where the first integral can be ...

4

If $\cos1$ is algebraic, so is $\sin1=\sqrt{1-\cos^21}$. Thus $e^i=\cos1+i\sin1$ is algebraic. But, by Lindemann's theorem, $e^\alpha$ is transcendental whenever $\alpha$ is a nonzero algebraic number.

3

Since $i$ is a non zero algebraic number then $\lbrace i , 2i ,0\rbrace$ is a set of distinct algebraic numbers in $\mathbb{C}$. By the Lindemann-Weierstrass theorem for any non-zero algebraic numbers $\beta_{1},\beta_{2},\beta_{3}$ we have $\beta_1 e^{i} + \beta_2 e^{2i}+\beta_3 e^{0}\ne 0$. To obtain a contradiction assume that $cos(1)$ is ...

5

I think the other answers indicating that this is an error made by WolframAlpha are perfectly reasonable and I upvoted them all. I also think it's a good idea to push the feedback button as @AlexR suggests. However, we can place the query in a broader context that makes WolframAlpha's response reasonable and shows us how to fix it at the same time. I want ...

1

Why not??? $\sec \theta = \tan \theta \leftrightarrow \dfrac{1}{cos \theta}=\dfrac{\sin \theta}{\cos \theta}$. We should conver this into $\cos \theta$. $\dfrac{1}{cos \theta}=\dfrac{\pm \sqrt{1-cos^2 \theta}}{\cos \theta}$. Now, we can cut $\cos \theta$ from both sides when $\cos \theta\ne0$. But at last we gets $\cos \theta = 0$.. so, it is a ...

3

Note that the solution W|A claims is $\frac12 (4\pi n + \pi) = \frac\pi2 + 2\pi n$ but $\tan\theta$ is never defined for $\theta = \frac\pi2 + 2\pi n$, thus you have found an error in W|A, wich you should report to them using their feedback form.

5

Suppose that $$\sec\theta = \tan\theta$$ then it follows, from the Pythagorean identity that $$\sec^2\theta=\tan^2+1$$ $$\tan^2\theta=\tan^2+1$$ $$0=1$$ which is a contradiction, therefore my original supposition (that $\sec\theta=\tan\theta$) was false.

3

$$\frac1{\cos\theta}=\frac{\sin\theta}{\cos\theta}\iff\begin{cases}\sin\theta=1\\{}\\and\\{}\\\cos\theta\neq0\end{cases}$$ and since the lower case always happens when the upper one happens, there is no solution. If WA tells otherwise then it is, again, wrong.

2

Recall that if $y = \cot^{-1} x$, this means that $y$ is an angle whose cotangent is $x$; i.e., $$\cot y = x.$$ Due to the periodicity of the circular trigonometric functions, it is clear that for a given value of $x$, there are infinitely many values of $y$ satisfying the above relationship; namely, $$\cot(y + k\pi) = x$$ for any integer $k \in \mathbb Z$. ...

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