Where Does the $\frac{1}{2}$ in Step 3 Come From?

In summary, the conversation discusses the process of evaluating an integral using the substitution method. It involves substituting a function inside the composition and using the appropriate derivative to simplify the integral. The final answer can be expressed in terms of the original variable.
  • #1
karush
Gold Member
MHB
3,269
5
https://www.physicsforums.com/attachments/5246
I'm trying to follow this example but where does the $\frac{1} {2 } $ come from in step 3
 
Physics news on Phys.org
  • #2
They are letting:

\(\displaystyle u=\sin(2x)\implies \frac{1}{2}du=\cos(2x)\,dx\)

Does that make sense?
 
  • #3
Sorta, So to get the 2 they multiplied by $\frac{1}{2}$
Letters me try one

$$\int\sin^3\left({x}\right) dx $$

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

$$u=\sin\left({x}\right)\ \ \ du =\cos\left({x}\right)dx$$

So far?
 
Last edited:
  • #4
karush said:
Sorta, So to get the 2 they multiplied by $\frac{1}{2}$
Letters me try one

$$\int\sin^3\left({x}\right) dx $$

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

$$u=\sin\left({x}\right)\ \ \ du =\cos\left({x}\right)dx$$

So far?

You substituted the wrong function. You want to substitute whatever is INSIDE the composition, not what is outside, so you would want u = cos(x) and du = -sin(x) dx.

Also [sin(x)]^2 = 1 - [cos(x)]^2, not [cos(x)]^2 - 1...
 
  • #5
$$\int\sin^3 \left({x}\right)dx
\implies \int\sin^2\left({x}\right)\sin\left({x}\right)dx
\implies\int\left(1-\cos^2 \left({x}\right)\right)\sin\left({x}\right)dx $$

$$u=\cos\left({x}\right)\ \ \ du=-\sin\left({x}\right)dx $$

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

I continued but didn't get this TI-Nspire answer

$$\left(\frac{-\sin^2 \left({x}\right)}{3}-\frac{2}{3}\right)\cos\left({x}\right)$$
 
  • #6
karush said:
$$\int\sin^3 \left({x}\right)dx
\implies \int\sin^2\left({x}\right)\sin\left({x}\right)dx
\implies\int\left(1-\cos^2 \left({x}\right)\right)\sin\left({x}\right)dx $$

$$u=\cos\left({x}\right)\ \ \ du=-\sin\left({x}\right)dx $$

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

I continued but didn't get this TI-Nspire answer

$$\left(\frac{-\sin^2 \left({x}\right)}{3}-\frac{2}{3}\right)\cos\left({x}\right)$$

What you have done is correct, so keep going.
 
  • #7
$$- \int 1 du + \int {u^{2 }} du $$

$$-u+\frac{{u}^{3}}{3}+C$$
 
  • #8
So now put it back as a function of x. You should be able to see why the two answers are equivalent.
 
  • #9
$$-\cos\left({x}\right)-\frac{\cos^3\left({x}\right)}{3}
\implies\left[-1-\frac{\cos^2 \left({x}\right)}{3}\right]\cos\left({x}\right)
\implies\left[\frac{-3+1+\sin^2 \left({x}\right)}{3}\right]\cos\left({x}\right)$$

Thus

$$\left[\frac{-\sin^2 \left({x}\right)}{3}-\frac{2}{3}\right]\cos\left({x}\right)$$
 
Last edited:
  • #10
\(\displaystyle -\int\cos^3{2x}\,dx=-\frac{1}{4}\int\cos{6x}\,dx-\frac{3}{4}\int\cos{2x}\,dx=-\frac{1}{4}(\frac{1}{6})\sin{6x}-\frac{3}{4}(\frac{1}{2})\sin{2x}+C\)
\(\displaystyle \therefore-\int\cos^3{2x}\,dx=-\frac{1}{24}\sin{6x}-\frac{3}{8}\sin{2x}+C\)
This, of course, requires the reader the ability to prove \(\displaystyle \cos{6x}=4\cos^3{2x}-3\cos{2x}\).
 

FAQ: Where Does the $\frac{1}{2}$ in Step 3 Come From?

1. What is an integral involving trig example?

An integral involving trig example is a mathematical expression that involves both trigonometric functions (such as sine, cosine, tangent, etc.) and integration. It is used to find the area under a curve or the value of a function at a specific point.

2. How do I solve an integral involving trig example?

To solve an integral involving trig example, you can use various integration techniques such as substitution, integration by parts, or trigonometric identities. It is important to have a good understanding of trigonometric functions and integration rules to solve these types of integrals.

3. What are some common trigonometric identities used in integrals?

Some common trigonometric identities used in integrals include the Pythagorean identities (sin^2x + cos^2x = 1), the double-angle identities (sin2x = 2sinxcosx), and the power-reducing identities (sin^2x = 1/2(1-cos2x)). These identities help simplify the integrals and make them easier to solve.

4. Are there any special cases in integrals involving trig functions?

Yes, there are some special cases in integrals involving trig functions. For example, when the integrand contains a product of sine and cosine functions, you can use the double-angle identity (sin2x) to simplify the expression. Also, when integrating trigonometric functions raised to an odd power, you can use the power-reducing identities.

5. How can I check if my solution to an integral involving trig example is correct?

You can check the solution to an integral involving trig example by differentiating the result and seeing if it matches the original integrand. You can also use online calculators or graphing software to graph the original function and the result of the integral and check if they are the same.

Similar threads

Replies
8
Views
578
Replies
4
Views
2K
Replies
6
Views
545
Replies
8
Views
1K
Replies
2
Views
2K
Replies
1
Views
1K
Back
Top