Help to Integral Calculus Solutions & Examples

Geometric meaning of ${\displaystyle ||z-z_1|-|z-z_2\mid \mid =a}$, where ${\displaystyle z,z_1,z_2\in \mathbb{C}}$

Can anyone suggest a simple method to solve this integral by complex variables?
${\displaystyle {\int }_0^{2\pi }\frac{d\theta }{\omega -a\sin \theta }}$
where ${\displaystyle \left|a\right|<\left|w\right|}$.

A square ABCD has all it's vertices on ${\displaystyle x^2{y}^2=1}$. The midpoints of it's sides also lie on the same curve . What is the area of this square?
It can be found very easily if we can show that diagonal of this square meet at the origin.

Can anyone help to integrat this function please?
${\displaystyle ic{\int }_{-1}^1\left(\frac{-2}{x}\frac{1}{1+{\left(\frac{tc}{x}\right)}^2}\right)dt}$
C>0, x>0 and i the imaginary part
Not really sure how to go about integrating it.

By considering,
${\displaystyle \int {\sin }^n\left(x\right){\cos }^3\left(x\right)dx}$
Prove that:
${\displaystyle \frac{{\sin }^8\left(x\right)}{8}-\frac{{\sin }^6\left(x\right)}{6}+\frac{1}{24}=\frac{{\cos }^8\left(x\right)}{8}-\frac{{\cos }^6\left(x\right)}{3}+\frac{{\cos }^4\left(x\right)}{4}}$
I have integrated the suggested integral successfully, that being:
${\displaystyle \frac{{\sin }^{n+1}\left(x\right)}{n+1}-\frac{{\sin }^{n+3}\left(x\right)}{n+3}+c}$
I see some resemblance but I fail to connect and finish the proof. Any hints on finding the relation between the powers of sin and cos through this integral? Obviously we could expand ${\displaystyle {\sin }^2\left(x\right)=(1-{\cos }^2\left(x\right))}$ but this seems far too tedious.

calculate the complex integral limit
${\displaystyle \underset{T\to \mathrm{\infty }}{lim}\frac{1}{2\pi i}{\int }_{c-iT}^{c+iT}\frac{x^s}{s^{k+1}}ds}$
where ${\displaystyle c>0}$ and ${\displaystyle k\ge 1}$ is an integer.

Angle between normal vector of ellipse and the major-axis.
I am trying to derive the angle made between the major or x-axis and the normal vector of an ellipse of general shape ${\displaystyle x=a\cos \left(t\right),y=b\sin \left(t\right)}$ with the parameter t reffering to Ellipse in polar coordinates. I need to solve it for any angle t. From standard reasoning I find the normal vector by its definition and checked it with the page on mathworld from wolfram and works well. Then since I know 2 points, namely a point ON the shape and a point on the normal vector I derive the angle of interest to be ${\displaystyle \tan \left(\varphi \right)=\frac{a}{b}\tan \left(t\right)}$ Derivation. However this is very similar to the polar angle namely its simply the term a and b flipped. But when thinking about it I keep getting confused, am I correct or do I need the polar angle? If so where did I go wrong?
I also found Normal to Ellipse and Angle at Major Axis but this page confused me a bit, one idea I had was they use the polar angle vs the angle I am in need of ${\displaystyle \left(\varphi \right)}$ then I would indeed get by combing ${\displaystyle t={\tan }^{-1}\left(\frac{a}{b}\tan \left(\theta \right)\right)}$ and ${\displaystyle \varphi ={\tan }^{-1}\left(\frac{a}{b}\tan \left(t\right)\right)}$
${\displaystyle \varphi ={\tan }^{-1}\left(\frac{a}{b}\tan \left({\tan }^{-1}big\left(\frac{a}{b}\tan \theta big\right)\right)\right)={\tan }^{-1}\left(\frac{a^2}{b^2}\tan \theta \right)}$
My excuse for my rambling, I find these angles confusing...

Algorithm to determine if a 3D ellipsoid is contained within another?
Can anyone point me to an algorithm for how to efficiently check if a 3D ellipsoid is contained within another one? We can assume their origins are collocated.
I am dealing with covariance ellipsoids constructed from matrices.

What is wrong in my answer? Subject: finding the integral of ${\displaystyle \cot x}$
${\displaystyle \int \cot xdx=\int \frac{\cos x\sin x}{{\sin }^{2}x}dx}$
Assume ${\displaystyle t={\sin }^{2}x}$. Then ${\displaystyle dt=2\sin x\cos xdx}$. Using the identity:
${\displaystyle \sin 2x=2\sin x\cos x}$
${\displaystyle \int \cot xdx=\int \frac{1}{2t}dt=\frac{1}{2}{\ln \left(\sin x\right)}^2+c}$

What's the best method to integrate ${\displaystyle \int \arctan \left\{\frac{1}{1-x}\right\}dx}$?

what is the integration of ${\displaystyle \int \cot \left(e^x\right)\cdot e^{x}dx}$
This is my answer is it right
${\displaystyle u=e^x}$
${\displaystyle du=e^{x}dx}$
${\displaystyle dx=\frac{1}{e^x}du}$
Then
${\displaystyle \int \cot \left(u\right)\cdot u\frac{1}{u}du=\int \cot \left(u\right)du={\csc }^2\left(u\right)}$

Which properties of integral and cos are used here?
${\displaystyle \frac{2}{l}{\int }_0^{l}x\cos \frac{k\pi x}{l},dx=\frac{2l}{{\pi }^2}{\int }_0^{\pi }x\cos kxdx}$
where x from l to l.

Why is this solution incorrect?
Prove that on the axis of any parabola there is a certain point K which has the property that, if a chord PQ of the parabola be drawn through it, then ${\displaystyle \frac{1}{P{K}^2}+\frac{1}{Q{K}^2}}$ is the same for all positions of the chord.
If it would be valid for standard parabola than it would be valid for all parabolas. Thus, proving for ${\displaystyle y^2=4ax}$.
let the point K be (c,0)
Equation of line PQ using parametric coordinates: ${\displaystyle x=c+r\cos \theta }$ (Equation 1), ${\displaystyle y=r\sin \theta }$ (Equation 2)
From equation 1 and 2: ${\displaystyle {(x-c)}^2+y^2=r^2}$ (Equation 3)
Using Equation 3 and ${\displaystyle y^2=4ax}$, we get this quadratic in r: ${\displaystyle r^2-4ax-{(x-k)}^2=0}$ (Equation 4)
Roots of this quadratic ${\displaystyle \left(r_1\text{and}{r}_2\right)}$ would be the lengths of PK and QK
From Equation 4, ${\displaystyle r_1+r_2=0}$ and ${\displaystyle r_1{r}_2=-(4ax+{(x-k)}^2)}$
We know, ${\displaystyle \frac{1}{P{K}^2}+\frac{1}{Q{K}^2}=\frac{1}{r_1^2}+\frac{1}{r_2^2}=\frac{{(r_1+r_2)}^2-2r_1{r}_2}{{\left(r_1{r}_2\right)}^2}=\frac{-2}{r_1{r}_2}}$
As value of ${\displaystyle r_1{r}_2}$ is not constant thus ${\displaystyle \frac{1}{P{K}^2}+\frac{1}{Q{K}^2}}$ does not turn out to be constant. Hence, this solution is incorrect.
I've seen the correct solution but I wanted to know why this solution is incorrect?

Why tangents to a quadratic curve never cut it again?
I have been studying conic sections lately and couldnt figure out why we can use the condition ${\displaystyle D=0}$ for the quadratic equation having only one root in many of the questions involving finding tangents, which I have read in certain places is called the envelope method. That got me to thinking why the tangent to the conic passes through one and only one point on the conic. A tangent is defined only as a line which just touches the curve at one point. However, it lays no restrictions on whether it could intersect the curve again or not.
So why exactly does a tangent to a circle/ ellipse/ parabola/ hyperbola never cut it again?

Verify that y1(x) = x^3 is a solution of x^2y''-5xy'+9y= 0. Hence find the general solution.

Linear transformation properties
A linear function ${\displaystyle T:{\mathbb{R}}^n\to {\mathbb{R}}^n}$ is given by ${\displaystyle T\left(x\right)=Px,\text{where}{P}^T=P^{-1}}$.
Prove that for any two vectors ${\displaystyle x,y\in {\mathbb{R}}^n}$ we have ${\displaystyle \left|T\left(x\right)\right|=\left|x\right|\text{and}T\left(x\right).T\left(y\right)=x.y}$. Also prove that T is continuous.