# How Do You Integrate Using Trigonometric Substitution?

• amcavoy
In summary: In this case, with sin^2(x)cos^2(x)= \frac{1}{4}sin^2(2x) we have \int sin^5(x) dx= \int sin^4(x) sin(x) dx= \int (\frac{1}{4}sin^2(2x))^2 (sin x dx)= \int \frac{1}{16}sin^4(2x) (sin x dx). Now again, let u= cos(2x) so that \frac{1}{2}du= sin(2x) dxand that becomes \int \frac{1}{16} sin^4(2x) \frac{
amcavoy
$$\int x^3\sqrt{4-9x^2}dx$$

I tried to use $$x=\frac{2}{3}\cos{(x)}$$ but it just left me with $$\int \sin^3{(x)}\cos^2{(x)}dx$$

Any suggestions?

It's the oher way around.U missed a numerical factor (containing an essential minus)...U need to substitute

$$x=\frac{2}{3}\sin u$$

to get the powers of sine & cosine that u have there.U'd still miss the numerical factor and but at least you'd have gotten rid of the "-"...

Let's assume this one (it's your transformed integral up to a numerical factor).

$$\int \sin^{3}u \cos^{2} u \ du$$...

Use the fundamental identity of circular trigonometry to write it

$$\int \sin^{3}u \ du -\int \sin^{5}u \ du$$

The first can be written as

$$-\int (1-\cos^{2}u) \ d(\cos u)$$

The second (the integral,without the "-" preceding it).

$$-\int \left(1-\cos^{2}u\right)^{2} \ d(\cos u)$$

,which are very easy to integrate...

Daniel.

Last edited:
$$\int{\sin^3{\theta}}d\theta - \int{\sin^5{\theta}}d\theta$$

I know how to integrate the first, but the second I am having a bit more trouble with.

Is this true?: $$\int{\sin^5{\theta}}d\theta = -\int{(1-\cos{\theta})^2}d\cos{\theta}$$

Thanks again.

I told u what it is equal to.

$$\int \sin^{5}x \ dx=-\int \left(1-\cos^{2}x\right)^{2} d(\cos x)$$

Daniel.

I'm sorry, what I meant was:

What is the best way to integrate it? I haven't seen one with the square before, so I'm not sure...

If you are going to use "trigonometric substitution" you should learn to integrate trig functions first!

Any time you have a an odd power of sin or cos, factor out one of then to go with the "dx", then convert the even power using sin2x+ cos2x= 1.

For example, seeing $$\int sin^5 x dx$$, I thinkof it as $$\int sin^4 x (sin x dx)= \int (sin^2 x)^2 (sin x dx)= \int (1- cos^2 x)(sin x dx)$$. Now let u= cos(x) so that du= -sin x dx and that becomes $$\int (1-u^2)^2 du$$.

If you have only even powers of sin and cos, you can't do that- you have to use the trig identities $$cos^2(x)= \frac{1}{2}[1+ cos(2x)]$$ and $$sin^2(x)= \frac{1}{2}[1- cos(2x)]$$ to reduce the powers.

## 1. What is trigonometric substitution?

Trigonometric substitution is a technique used in calculus to solve integrals involving algebraic expressions and square roots. It involves substituting the variable in the integral with a trigonometric function in order to simplify the expression and make it easier to solve.

## 2. When should trigonometric substitution be used?

Trigonometric substitution should be used when the integral involves expressions with square roots, especially when the expression can be simplified using trigonometric identities. It is also useful when the integral involves the square of a trigonometric function or when the denominator can be expressed in terms of a trigonometric function.

## 3. How is trigonometric substitution performed?

To perform trigonometric substitution, the variable in the integral is replaced with a trigonometric function using a substitution rule. The trigonometric function is chosen based on the expression in the integral, and then trigonometric identities are used to simplify the expression and solve the integral.

## 4. What are the most commonly used trigonometric substitutions?

The most commonly used trigonometric substitutions are:
1. Substituting √(a^2-x^2) with a sinθ
2. Substituting √(a^2+x^2) with a tanθ
3. Substituting √(x^2-a^2) with a secθ

## 5. What are the benefits of using trigonometric substitution?

Using trigonometric substitution can simplify complicated integrals, making them easier to solve. It also allows for the use of trigonometric identities, which can help to solve integrals that would otherwise be difficult to evaluate. Additionally, trigonometric substitution can be used to solve integrals that cannot be solved using other techniques.

• Calculus
Replies
3
Views
1K
• Calculus
Replies
4
Views
1K
• Calculus
Replies
6
Views
2K
• Calculus
Replies
14
Views
2K
• Calculus
Replies
12
Views
2K
• Calculus
Replies
4
Views
835
• Calculus
Replies
8
Views
1K
• Calculus
Replies
5
Views
2K
• Calculus
Replies
8
Views
620
• Calculus
Replies
2
Views
1K