# How To Integrate a Trigonometric Substitution

By Kathleen Knowles, 03 Jul 2020

After looking at the title, you must be wondering, “What is integration by trigonometric substitution?” Even just saying it out loud makes it seem like a particularly complicated topic.

However, the best way to understand complicated things is to break it down from the beginning. To answer this question properly, you'll first need to have an understanding of what integration is.

## What is Integration?

Integration is a part of calculus, a branch in mathematics. Calculus has two major branches: differential calculus and integral calculus.

Differential calculus is all about finding the instant rate of change of anything. A good, applicable example of this would be calculating the infection rate of COVID-19 per hour. We will *not* be focusing on this branch of calculus. We will instead be focusing on Integral Calculus. This is all about the accumulation of quantities and finding the area under or between a curve.

The method used to find the area under the curve or between the curve is called integration. The symbol used for integration is “⨛.”

### Understanding the Graph

Now let us assume a curve as shown below:

This is a plot of a function of x, which we denote as *f(x) = x2*.

The domain of this function, that is, the numbers that are valid for this function, is any real number. For any value of x, we get a value of y, which helps us plot a graph for x2.

So if x = 1, y = 12 = 1x1, which is 1. If x = 2, y = 22 = 2x2, which is 4, and so on. That is how this graph is generated.

### Finding the Area Between the Curve

If I want to find the area between a specific part of this curve, say, from x = -2 to x = 2, how do I do that? It's simple!

- Mark the plot above between x = -2 to x = 2.
- Now, divide the graph into four parts and make rectangles.
- After that, calculate the area of the rectangle. To find the area of a rectangle, multiply the length of the rectangle base by its height (
*l x h)*. - You will get an answer, but it will be incorrect. This is because you must have noticed that when you made the rectangles, they either did not cover all the area under the plot or they covered an extra area outside the plot.
- Repeat the same step with eight rectangles. We will find that we now have an answer that’s more precise than the one we got with four rectangles.
- Now imaging creating an infinite number of rectangles and finding the area of all those rectangles. That will look like the plot below:

In order to calculate the area of the plot between x = -2 and x = 2, we add the areas of all the rectangles. This is called **integration**. It is physically impossible to add the area of infinite rectangles, so we use integration to do it for us.

### Performing Integration

We use the integrate function with respect to the limits given to us. That is, we integrate *f(x)* between -2 and 2 to find the area under the curve from *x = -2* to *x = 2.* This is the following calculation:

*-22x2 dx*

Now, this symbol with the -2 and 2 at the bottom and at the top of the integral symbol is called the definite integral symbol. This means we are integrating with knowledge about the bounds. The previous symbol is used for indefinite integral, which is integrating a function without knowing the bounds.

In this type, we always add *c* after the integration. This is called the integral constant. The integration of x2 requires us to make the infinite rectangles. After calculating, we find that the integration of x2 is actually x33. So, on calculating further, we get:

*-22x2dx = [x33]2-2*

* = 83 - [-83] = 163*

Therefore, we can say the area under the curve of x2 from x = -2 to x = 2 is 163. After adding the areas of the infinite rectangles, you should obtain the same answer.

## Incorporating the Trigonometric Substitution

Now that we know what integration is, we now move towards integration using trigonometric substitution.

Let's say we are finding the area of an ellipse from x = 0 to x = 2. The equation of the ellipse is *(xa)2 + (yb)2 = 1*. We get a value *y = ba(a2 - x2)*.

Now solving a square root in integration is quite difficult, and you usually reach a point where you want to pull your hair out! Trigonometric substitution makes it really simple.

Let's say we are evaluating the integral from x = 0 to x = a. The plot of an ellipse is shown below:

Integrate y from x = 0 to x = a. Since the area of this will be only for the first quadrant of the plot, we will need to multiply the area by 4 in order to get the area of the whole ellipse.

To evaluate this integral we substitute *x = asin(u)*. Then *dx = acos(u)du*.

We now change the limit.

So, x = 0, sin(u) = 0, which means u = 0, and when x = a, sin(u) = 1, which means u = π/2.

So we are integrating *4ba(a2 - x2)*, where *x = asin(u)*. On solving, we get *4ab0π/2(cos(u))2du*.

Then, on further solving, we get *πab*.

Therefore, we have proved that the area of an ellipse with semi-axes a and b is *πab.* For* a = b = r*, we get the very famous area of a circle, πr2.

In conclusion, trigonometric substitution is a technique used to make an integrand easy to solve.

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