Recent content by footmath

Hello , please guide me . How can I transformed the equation x^{5}-x^{4}-x^{3}-x^{2}-x-1 =0 to y^{5}+2y^{2}+47y+122 ? I studied a lecture that the writer had written :<< by using y=x^{2}-3x we can transformed x^{5}-x^{4}-x^{3}-x^{2}-x-1 =0 to y^{5}+2y^{2}+47y+122 But how ? if in y=x^2-3x...

What is your opinion about$ x^5 + 20x^3 + 20x^2 + 30x + 10 =0 $ Are you accept that the one root of this equation is $\sqrt[5]{2}-\sqrt[5]{2^2}+\sqrt[5]{2^3}-\sqrt[5]{2^4} $ ? if you substitute this answer in the equation you will see very, very close to 0 but NOT 0. $ \sqrt[5]{5.40985+0i}$ is...

When we want to calculate the root of a equation for example x^3+3x+1=0. The root of x^3+3x+1=0 is $-0.3221853546 $ but if you substitute $-0.3221853546 $ in the equation x^3+3x+1=0 you will see (-0.3221853546)^3+3(-0.3221853546)+1=0.00000000000846 . we cannot obtain a rational root but...

Hello. Galois theory tell us $x^5-6x+3$ is not solvable by radical but every equation lower than fifth can solve by radical. If $G$ is solvable and $H$ is solvable too $G*H$ are solvable. For $x^5-6x+3$ we can use Newton’s method and find one root of this equation. We obtain $x=1.4$ and...

Is it correct ? $ \sqrt{(-1)^2} $ = $ \sqrt{(-1)}\sqrt{(-1)} $= $ i*i $=$ i^2 $ =$ -1 $ or $ \sqrt{(-1)^2} $ = $ \sqrt{(1)} $=1

no sorry.I transformed$A=\int\sqrt{1+\sin^{2}x}\,dx$to$ \int\sqrt{\frac{t+1}{t^{2}-t}}\,dt$ but how can I solve it ?

$ \int\sqrt{\frac{1}{t^{2}-t}}dt =\int\frac{dt}{\sqrt{(t-\frac{1}{2})^{2}-\frac{1}{4}}}=ln\mid{t-\frac{1}{2}+\sqrt{t^{2}-t}}\mid+c $

I work this integral during one mouth . I am in grade two high school and my teacher gave me this integral : $ \int\sqrt{sin(u)}du $ and I transformed to : (sinx)^1/2=t => sinx=t^2 => x= arc sint^2 dx=2t/(1-t^4)^1/2 so $ \int\sqrt{sin(u)}du $=int (2t^2/(1-t^4)^1/2) if t=cosx we have ...

would you do it?

you forget radical ? [Integral]Sqrt[1/(u - 1) - 1/u] du

sqrt{1-\sin^{2}x} equal cosx not sqrt{1+\sin^{2}x}

yes . this integral at first was : $ A=\int\sqrt{1+\sin^{2}x}\,dx $

please solve this integral : \[Integral]Sqrt[1/(u - 1) - 1/u] du

Thank you but I can not solve this integral:$ A=\int\sqrt{1+\sin^{2}x}\,dx $ please explain about solution .

would you please explain the solution of elliptic integral