As we can see in this integral, there is no reverse chain-rule u-substitution possible. The logical step is to use a trigonometric substitution. If one recalls that trig substitutions of the type \(\displaystyle x^2-a^2\) could be solved with the substitution \(\displaystyle \frac{2}{5}sec\theta\), then the answer is easily seen. However, we can also use a right triangle:                                                       

And thus we have: 

\(\displaystyle sec \theta = \frac{5x}{2}\)

or:

 \(\displaystyle x = \frac{2}{5}sec\theta\)

At angle \(\displaystyle 0\) the graph as a radius of \(\displaystyle 1\). As it approaches \(\displaystyle \frac{\pi }{2}\), the radius approaches \(\displaystyle 0\).

As the graph approaches \(\displaystyle \pi\), the radius approaches \(\displaystyle -1\).

Because this is a negative radius, the curve is drawn in the opposite quadrant between \(\displaystyle \frac{3\pi }{2}\) and \(\displaystyle 2\pi\).

Between \(\displaystyle \pi\) and \(\displaystyle \frac{3\pi }{2}\), the radius approaches \(\displaystyle 0\) from \(\displaystyle -1\) and redraws the curve in the first quadrant.

Between \(\displaystyle \frac{3\pi }{2}\) and \(\displaystyle 2\pi\), the graph redraws the curve in the fourth quadrant as the radius approaches \(\displaystyle 1\) from \(\displaystyle 0\).    

Between \(\displaystyle 0\) and \(\displaystyle \frac{\pi }{2}\), the radius approaches \(\displaystyle 1\) from \(\displaystyle 0\).

From \(\displaystyle \frac{\pi }{2}\) to \(\displaystyle \pi\) the radius goes from \(\displaystyle 1\) to \(\displaystyle 0\).

Between \(\displaystyle \pi\) and \(\displaystyle \frac{3\pi }{2}\), the curve is redrawn in the opposite quadrant, the first quadrant as the radius approaches \(\displaystyle -1\).

From \(\displaystyle \frac{3\pi }{2}\) and \(\displaystyle 2\pi\), the curve is redrawn in the second quadrant as the radius approaches \(\displaystyle 0\) from \(\displaystyle -1\).   

Because this function has a period of \(\displaystyle \pi\), the x-intercepts of the graph \(\displaystyle y=cos(2x)\)  happen at a reference angle of \(\displaystyle \frac{\pi }{4}\) (angles halfway between the angles of the axes).  

Between \(\displaystyle 0\) and \(\displaystyle \frac{\pi }{4}\) the radius approaches \(\displaystyle 0\) from \(\displaystyle 1\).

Between \(\displaystyle \frac{\pi }{4}\) and \(\displaystyle \frac{\pi }{2}\), the radius approaches \(\displaystyle -1\) from \(\displaystyle 0\) and is drawn in the opposite quadrant, the third quadrant because it has a negative radius.

From \(\displaystyle \frac{\pi }{2}\) to \(\displaystyle \frac{3\pi }{4}\) the radius approaches \(\displaystyle 0\) from \(\displaystyle -1\) , and is drawn in the fourth quadrant, the opposite quadrant. 

Between \(\displaystyle \frac{3\pi }{4}\) and \(\displaystyle \pi\), the radius approaches \(\displaystyle 1\) from \(\displaystyle 0\).

From \(\displaystyle \pi\) and \(\displaystyle \frac{5\pi }{4}\), the radius approaches \(\displaystyle 0\) from \(\displaystyle 1\).

Between \(\displaystyle \frac{5\pi }{4}\) and \(\displaystyle \frac{3\pi }{2}\), the radius approaches \(\displaystyle -1\) from \(\displaystyle 0\). Because it is a negative radius, it is drawn in the opposite quadrant, the first quadrant.

Then between \(\displaystyle \frac{3\pi }{2}\) and \(\displaystyle \frac{7\pi }{4}\) the radius approaches \(\displaystyle 0\) from \(\displaystyle -1\) and is draw in the second quadrant.

Finally between \(\displaystyle \frac{7\pi }{4}\) and \(\displaystyle 2\pi\), the radius approaches \(\displaystyle 1\) from \(\displaystyle 0\).                  

Because this function has a period of \(\displaystyle \pi\), the amplitude of the graph \(\displaystyle y=sin(2x)\)  appear at a reference angle of \(\displaystyle \frac{\pi }{4}\) (angles halfway between the angles of the axes).  

Between \(\displaystyle 0\) and \(\displaystyle \frac{\pi }{4}\) the radius approaches 1 from 0.

Between \(\displaystyle \frac{\pi }{4}\) and \(\displaystyle \frac{\pi }{2}\), the radius approaches 0 from 1.

From \(\displaystyle \frac{\pi }{2}\) to \(\displaystyle \frac{3\pi }{4}\) the radius approaches -1 from 0 and is drawn in the opposite quadrant, the fourth quadrant because it has a negative radius.

Between \(\displaystyle \frac{3\pi }{4}\) and \(\displaystyle \pi\), the radius approaches 0 from -1, and is also drawn in the fourth quadrant.

From \(\displaystyle \pi\) and \(\displaystyle \frac{5\pi }{4}\), the radius approaches 1 from 0. Between \(\displaystyle \frac{5\pi }{4}\) and \(\displaystyle \frac{3\pi }{2}\), the radius approaches 0 from 1.

Then between \(\displaystyle \frac{3\pi }{2}\) and \(\displaystyle \frac{7\pi }{4}\) the radius approaches -1 from 0. Because it is a negative radius, it is drawn in the opposite quadrant, the second quadrant. Likewise, as the radius approaches 0 from -1. Between \(\displaystyle \frac{7\pi }{4}\) and \(\displaystyle 2\pi\), the curve is drawn in the second quadrant.                  

Taking the graph of \(\displaystyle y=cos(2x)\), we only want the areas in the positive first quadrant because the radius is squared and cannot be negative.

This leaves us with the areas from \(\displaystyle 0\) to \(\displaystyle \frac{\pi }{4}\), \(\displaystyle \frac{3\pi }{4}\) to \(\displaystyle \frac{5\pi }{4}\), and \(\displaystyle \frac{7\pi }{4}\) to \(\displaystyle 2\pi\).

Then, when we take the square root of the radius, we get both a positive and negative answer with a maximum and minimum radius of \(\displaystyle \pm 1\).

To draw the graph, the radius is 1 at \(\displaystyle 0\) and traces to 0 at \(\displaystyle \frac{\pi }{4}\). As well, the negative part of the radius starts at -1 and traces to zero in the opposite quadrant, the third quadrant.

From \(\displaystyle \frac{3\pi }{4}\) to \(\displaystyle \pi\), the curves are traced from 0 to 1 and 0 to -1 in the fourth quadrant. Following this pattern, the graph is redrawn again from the areas included in \(\displaystyle \pi\) to \(\displaystyle 2\pi\).    

Taking the graph of \(\displaystyle y=sin(2x)\), we only want the areas in the positive first quadrant because the radius is squared and cannot be negative.

This leaves us with the areas from \(\displaystyle 0\) to \(\displaystyle \frac{\pi }{2}\) and \(\displaystyle \pi\) to \(\displaystyle \frac{3\pi }{2}\). 

Then, when we take the square root of the radius, we get both a positive and negative answer with a maximum and minimum radius of \(\displaystyle \pm 1\).

To draw the graph, the radius is 0 at \(\displaystyle 0\) and traces to 1 at \(\displaystyle \frac{\pi }{4}\). As well, the negative part of the radius starts at 0 and traces to-1 in the opposite quadrant, the third quadrant.

From \(\displaystyle \frac{\pi }4{}\) to \(\displaystyle \frac{\pi }{2}\), the curves are traced from 1 to 0 and -1 to 0 in the third quadrant.

Following this pattern, the graph is redrawn again from the areas included in \(\displaystyle \pi\) to \(\displaystyle 2\pi\).    

A point in polar coordinates is described by a position vector to a point A and an angle between horizontal axis and said vector. A convention for a positive angle is counter-clockwise.

Note, that in polar coordinates, position vector may also be of negative value, meaning pointing in the opposite direction.

Graphing polar equations is different that plotting cartesian equations. Instead of plotting an \(\displaystyle (x,y)\) coordinate, polar graphs consist of an \(\displaystyle (r,\theta)\) coordinate where \(\displaystyle r\) is the radial distance of a point from the origin and \(\displaystyle \theta\) is the angle above the x-axis.

When the graph of an equation in the form \(\displaystyle \theta=\alpha\), where \(\displaystyle \alpha\) is an angle, the angle of the graph is constant and independent of the radius. This creates a straight line \(\displaystyle \alpha\) radians above the x-axis passing through the origin.

In this problem,  \(\displaystyle \theta=\pi/4\) is a straight line \(\displaystyle \pi/4\) radians or  \(\displaystyle 45^{\circ}\) about the x-axis passing through the origin.

Graphing polar equations is different that plotting cartesian equations.  Instead of plotting an \(\displaystyle (x,y)\) coordinate, polar graphs consist of an \(\displaystyle (r,\theta)\) coordinate where \(\displaystyle r\) is the radial distance of a point from the origin and \(\displaystyle \theta\) is the angle above the x-axis.

When the graph of an equation in the form \(\displaystyle r=\lambda\), where \(\displaystyle \lambda\) is a constant, the graph is a circle centered around the origin with a radius of \(\displaystyle \lambda\).

In this problem, \(\displaystyle r=4\) is a circle centered around the origin with a radius of \(\displaystyle 4\).