Help to Integral Calculus Solutions & Examples

Proving the double differential of z = -z implies ${\displaystyle z=\sin x}$
${\displaystyle \frac{d^{2}z}{{dx}^2}=-z}$
implies z is of the form ${\displaystyle a\sin x+b\cos x}$. Is there a proof for the same. I was trying to arrive at the desired function but couldn't understand how to get these trigonometric functions in the equations by integration. Does it require the use of taylor polynomial expansion of ${\displaystyle \sin x\text{or}\cos x}$?

I am approximating a solution to a first order LODE using Euler's method. I made two tables, one using a step size of .01 and another using .05 ( I had to start at x=0 and end at x=1). I am not understanding the directions for the second part of my assignment:

It states that the order of numerical methods (like Euler's) is based upon the bound for the cummulative error; i.e. for the cummulative error at, say x=2, is bounded by $C{h}^n$, where $C$ is a generally unknown constant and $n$ is the order. For Euler's method, plot the points:
$(0.1,\mathrm{\varphi <!--\; \varphi \; -->}(1)-<!--\; -\; -->y_{10}),$
$(.05,\mathrm{\varphi <!--\; \varphi \; -->}(1)-<!--\; -\; -->y_{20}),$
$(.025,\mathrm{\varphi <!--\; \varphi \; -->}(1)-<!--\; -\; -->y_{40}),$
$(.0125,\mathrm{\varphi <!--\; \varphi \; -->}(1)-<!--\; -\; -->y_{80}),$
$(.00625,\mathrm{\varphi <!--\; \varphi \; -->}(1)-<!--\; -\; -->y_{160})$
And then fit a line to the above data of the form $Ch$. I don't understand, am I supposed to plot these using a step size of .1 or .05? Or am I supposed to use another step size?
Any clarification is appreciated.
Thanks
Edit:
The LODE I am given is $y^{\prime }=x+2y,y(0)=1$ and the exact solution I found was $\mathrm{\varphi <!--\; \varphi \; -->}(x)=\frac{1}{4}(-<!--\; -\; -->2x+5e^{2x}-<!--\; -\; -->1)$.

I am trying to evaluate the integral
${\displaystyle {\int }_0^{\frac{\pi }{2}}\frac{\log (e^{ix}+e^{-ix})}{-}\log \left\{2\right\}\}\{e^{2ix}-1\}{e}^{ix}dx}$
I am wondering if a complex variable substitution the likes of ${\displaystyle iu=e^{ix}}$, ${\displaystyle du=e^{ix}dx}$ is justifiable? In my mind. when x=0,u=0 and when ${\displaystyle x=\frac{\pi }{2},u=1}$ which would make the integral
${\displaystyle -{\int }_0^1\frac{\log (iu-\frac{i}{u})-\log \left(2\right)}{1+u^2},du}$

Integrating ${\displaystyle {\int }_0^{\frac{\pi }{4}}\frac{\sin x+\cos x}{{\sin }^{4}x+{\cos }^{2}x}dx}$

Integrate ${\displaystyle \int \frac{dx}{\sqrt{{\sin }^{3}x\sin (x+\alpha )}}}$
${\displaystyle \int \frac{1}{\sqrt{{\sin }^{3}x\sin (x+\alpha )}}dx=\int \frac{1}{\sin x\sqrt{\sin x\sin (x+\alpha )}}dx}$
${\displaystyle =\frac{1}{\sin \alpha }\int \frac{\sin (x+\alpha -x)}{\sin x\sqrt{\sin x\sin (x+\alpha )}}dx=\frac{1}{\sin \alpha }\int \frac{\sin (x+\alpha )\cos x-\cos (x+\alpha )\sin \left\{x\right\}}{\sin x\sqrt{\sin x\sin (x+\alpha )}}dx}$
${\displaystyle =\frac{1}{\sin \alpha }\int \frac{\sin (x+\alpha )\cos x}{\sin x\sqrt{\sin x\sin (x+\alpha )}}dx-\frac{1}{\sin \alpha }\int \frac{\cos (x+\alpha )\sin \left\{x\right\}}{\sin x\sqrt{\sin x\sin (x+\alpha )}}dx}$
Is it possible to proceed further and complete the integration or what is the right substitution to find the solution ?

Integrate ${\displaystyle \int x\cos \left(x\right)\sin \left(x\right)dx}$
I've been trying to integrate by parts, but I can't! I know there's a trigonometric function about ${\displaystyle \cos \left(2x\right)}$ but I don't know how to integrate that function wit ${\displaystyle \sin x}$

Name or Adjective for Ellipse with very different (or very similar) scales
I am looking for an adjective or word to describe an ellipse (or ellipsoid, in more dims) where the length of the principal axes are of roughly the same order of magnitude ${\displaystyle O\left(a\right)=O\left(b\right)}$ (including equal, i.e. a circle).
For instance:
- ${\displaystyle a=2,b=5}$ would be something like an "isometric" ellipse (or ellipsoid)
- ${\displaystyle a=2,b=50}$ would be something like an "asometric" ellipse (or ellipsoid) since they are of different orders of magnitude

integral ${\displaystyle \int \frac{1-a\cos x}{1+a^2-2a\cos x}dx}$
My trouble: Using software, i found that if ${\displaystyle \left|a\right|>1}$, the integral is equal to
${\displaystyle \arctan \left(\frac{x}{2}\right)+\frac{1-a^2}{|a-1||a+1|}\arctan \left(\frac{|a+1|}{|a-1|}\tan \left(\frac{x}{2}\right)\right)+C}$
If ${\displaystyle \left|a\right|<1}$, then the antiderivate is
${\displaystyle x-\arctan \left(\frac{x}{2}\right)+\frac{1-a^2}{|a-1||a+1|}\arctan \left(\frac{|a+1|}{|a-1|}\tan \left(\frac{x}{2}\right)\right)+C}$

Integrate ${\displaystyle \int \frac{\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx}$

Integral of ${\displaystyle \int \left\{\frac{{\cos }^{4}x+{\sin }^{4}x}{\sqrt{1+\cos 4x}}dx\right\}}$

Integral n for ${\displaystyle {\sin }^n\left(x\right)+{\cos }^n\left(x\right)=1-\frac{n}{2}{\sin }^2\left(x\right){\cos }^2\left(x\right)}$ is n=4,6.
I have proved for n=4,6 it is true, for other integer, I tried to check range of functions on left and right side. This is not as effective, and I failed. Please suggest a method in this.

If x and y are real numbers and ${\displaystyle 4x^2+2xy+9y^2=100}$ then what are all possible values of ${\displaystyle x^2+2xy+3y^2}$.