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  could be solved with the substitution , then the answer is easily seen. However, we can also use a right triangle:                                                       

And thus we have: 

or:

At angle  the graph as a radius of . As it approaches , the radius approaches .

As the graph approaches , the radius approaches .

Because this is a negative radius, the curve is drawn in the opposite quadrant between  and .

Between  and , the radius approaches  from  and redraws the curve in the first quadrant.

Between  and , the graph redraws the curve in the fourth quadrant as the radius approaches  from .    

Between  and , the radius approaches  from .

From  to  the radius goes from  to .

Between  and , the curve is redrawn in the opposite quadrant, the first quadrant as the radius approaches .

From  and , the curve is redrawn in the second quadrant as the radius approaches  from .   

Because this function has a period of , the x-intercepts of the graph   happen at a reference angle of  (angles halfway between the angles of the axes).  

Between  and  the radius approaches  from .

Between  and , the radius approaches  from  and is drawn in the opposite quadrant, the third quadrant because it has a negative radius.

From  to  the radius approaches  from  , and is drawn in the fourth quadrant, the opposite quadrant. 

Between  and , the radius approaches  from .

From  and , the radius approaches  from .

Between  and , the radius approaches  from . Because it is a negative radius, it is drawn in the opposite quadrant, the first quadrant.

Then between  and  the radius approaches  from  and is draw in the second quadrant.

Finally between  and , the radius approaches  from .                  

Because this function has a period of , the amplitude of the graph   appear at a reference angle of  (angles halfway between the angles of the axes).  

Between  and  the radius approaches 1 from 0.

Between  and , the radius approaches 0 from 1.

From  to  the radius approaches -1 from 0 and is drawn in the opposite quadrant, the fourth quadrant because it has a negative radius.

Between  and , the radius approaches 0 from -1, and is also drawn in the fourth quadrant.

From  and , the radius approaches 1 from 0. Between  and , the radius approaches 0 from 1.

Then between  and  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  and , the curve is drawn in the second quadrant.                  

Taking the graph of , 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  to ,  to , and  to .

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 .

To draw the graph, the radius is 1 at  and traces to 0 at . As well, the negative part of the radius starts at -1 and traces to zero in the opposite quadrant, the third quadrant.

From  to , 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  to .    

Taking the graph of , 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  to  and  to . 

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 .

To draw the graph, the radius is 0 at  and traces to 1 at . As well, the negative part of the radius starts at 0 and traces to-1 in the opposite quadrant, the third quadrant.

From  to , 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  to .    

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  coordinate, polar graphs consist of an  coordinate where  is the radial distance of a point from the origin and  is the angle above the x-axis.

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

In this problem,   is a straight line  radians or   about the x-axis passing through the origin.

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

When the graph of an equation in the form , where  is a constant, the graph is a circle centered around the origin with a radius of .

In this problem,  is a circle centered around the origin with a radius of .