# Evaluating integrals by trigonometric substitution

• Integralien
In summary, if you're having trouble with integrating by trigonometric substitution, you may want to try practicing with worked out examples, or see if someone else can help you with the problem.
Integralien
I have a few quick problems concering evaluating integrals by trigonometric substitution. I guess I will just post five that way if anyone can help with any, would be greatly appreciated. Also: if anyone could inform me on how to input the actual equations onto this forum as I have seen in some posts, I would also greatly appreciate that.

1) The integral of (3x² + x - 1) onto the square root of (1 - x²)dx

2) The integral of 1 all over the square root of (16 + 25x²) dx

3) The integral of 1 all over the square root of (36x² - 25) dx

4) The integral of 1 all over the square root of (3 - 2x - x²) dx

5) The integral of the square root of (10 - 4x + 4x²)dx

Again I apologize if that is hard to understand if anyone could tell me quickly how to offer an easier way to display problems, I would be very thankful.

Brian

use LATEX!

The best way to learn how to do these kind of integrals is to practice with some worked out examples.

If you google for 'trig sub', you'll find plenty of tutorials on how to do it.

1.) $$\int \frac{3x^2 + x - 1}{\sqrt{1 - x^2}}dx$$

2.) $$\int \frac{dx}{\sqrt{16 + 25x^2}}$$

3.) $$\int \frac{dx}{\sqrt{36x^2 - 25}}$$

4.) $$\int \frac{dx}{\sqrt{3 - 2x - x^2}}$$

5.) $$\int \sqrt{10-4x+x^2}dx$$

For the last two you must complete the square to get the radicands in the form:
$$\int\frac{dx}{\sqrt{a^2 \pm (x+c)^2}}=\int\frac{du}{\sqrt{a^2 \pm u^2}}$$

Notice that in all cases you have sums or differences of squares inside a radical.
The method of trigonometric substitution resolves these using one of the forms of the Pythagorean identity:

I. $$\cos^2(\theta) + \sin^2(\theta) = 1 \quad \Rightarrow \cos^2(\theta) = 1-\sin^2(\theta)$$
which helps if you substitute $$x = \frac{a}{b}\sin(\theta)$$ when you must deal with the form $$a^2 - b^2 x^2$$.

II. $$1 + \tan^2(\theta) = \sec^2(\theta) \Leftrightarrow \tan^2(\theta) = \sec^2(\theta) - 1$$
which helps if you substitute $$x = \frac{a}{b}\tan(\theta)$$ when you must deal with the form $$a^2 + b^2 x^2$$
and helps if you substitute $$x = \frac{a}{b}\sec(\theta)$$ when you must deal with the form $$b^2 x^2 + a^2$$.

Making these substitutions should give you an integral which is a product of various trigonometric functions each of which you must deal with on a case by case basis.

This should all be outlined in your class notes and text. I'm not sure what else to tell you unless/until you state what exactly is tripping you up on a particular problem.

I'll demonstrate another example to help get you started:

$$\int \sqrt{4x^2 + 4x - 35}dx$$
First you must complete the square:
$$4x^2 + 4x - 35$$
you want to rewrite $$4x^2+4x$$ as: $$(2x+b)^2 - b^2=4x^2+4bx$$ so $$b= 1$$.
$$4x^2 + 4x - 35 = 4x^2+4x+1 - 1 - 35 = (2x+1)^2 -36$$
and you've completed the square:

Let's call $$u=(2x+1)$$ and note that $$du=2dx$$
then the integral becomes:
$$\int \sqrt{2x^2 + 12x + 9}dx=\frac{1}{2}\int\sqrt{u^2 - 36}du$$

Now we are ready to execute a trigonometric substitution. We will use the identity
$$\sec^2(\theta) - 1=\tan^2(\theta)$$
so choosing $$u = 6\sec{\theta}$$ we get $$\sqrt{u^2 - 36}=\sqrt{36sec^2(\theta) - 36} = \sqrt{36}\sqrt{\sec^2(\theta)-1}=6\sqrt{\tan^2(\theta)}= 6 \tan(\theta)$$

Note also that $$du = 6 \sec(\theta)\tan(\theta)d\theta$$
So the integral becomes:
$$\frac{1}{2}\int\sqrt{u^2 - 36}du=\frac{1}{2}\int 6\tan(\theta)du=\frac{1}{2}\int 6\tan(\theta)\cdot 6\sec(\theta)\tan(\theta)d\theta$$

$$= 18\int \tan^2(\theta)\sec(\theta)d\theta$$
We've finished the trigonometric substitution and now must evaluate the integral.
This example is do-able but not easy. I'll leave it for now. The substitution is what I wanted to demonstrate.

## 1. What is trigonometric substitution in integral calculus?

Trigonometric substitution is a technique used to evaluate integrals involving expressions that contain algebraic and trigonometric functions. This method involves substituting trigonometric identities for certain variables in the integral, which simplifies the expression and makes it easier to integrate.

## 2. When should I use trigonometric substitution to evaluate an integral?

Trigonometric substitution is typically used when the integral contains expressions of the form √(a²-x²), √(x²-a²), or √(a²+x²), where a is a constant. These expressions can be simplified using trigonometric identities, making the integral easier to solve.

## 3. What are the common trigonometric substitutions used in integrals?

The most common trigonometric substitutions used in integrals are:
- √(a²-x²) is substituted with x = a sin θ
- √(x²-a²) is substituted with x = a sec θ
- √(a²+x²) is substituted with x = a tan θ

## 4. How do I know which trigonometric substitution to use?

To determine which trigonometric substitution to use, look at the expression inside the square root in the integral. If it is of the form √(a²-x²), then use the substitution x = a sin θ. If it is of the form √(x²-a²), use x = a sec θ. And if it is of the form √(a²+x²), use x = a tan θ.

## 5. Are there any special cases when using trigonometric substitution?

Yes, there are a few special cases when using trigonometric substitution. If the integral contains expressions of the form √(x²+a²) or √(a²-x²) with odd powers, or if the expression inside the square root is a difference of squares, then a different substitution may be needed. Additionally, if the integral contains trigonometric functions raised to even powers, a different approach may be necessary.

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