# Is this the correct way to solve for the integral of x^3cosx^2?

• CDevo69
In summary, the conversation discusses using the integration by parts method to solve the integral of x^3cos(x^2), with varying approaches and choices for u and dv. It also mentions the ILATE rule for selecting u and dv. The final solution is given as 1/2 (x^2sinx^2 + cosx^2) + c.
CDevo69
This has been posted before although I've come across it and got a different answer from https://www.physicsforums.com/archive/index.php/t-108378.html

x sinx cosx dx using the identity sin2x = 2sinxcosx

u = x
du = 1
dv = 1/2 sin 2x
v = -1/4 cos 2x

x sinx cosx dx = -1/4xcos2x - Int -1/4cos2x + c
= -1/4xcos2x + 1/8sin2x + c
Right?

And I need some help to work out:
Integral of x^3cosx^2

I'm always getting confused with powers on the trig for some reason.
Thankyou

I would agree with that.

Your second integral is also solvable by parts. The trick for choosing u and dv is you got to choose dv such that $\int dv$ is feasable! Based on that, what are you going to choose for u and dv?

Well let's try...

Int x^3cos(x^2)

u = cos(x^2)
du = -2x sin (x^2)
dv = x^3
v = (x^4)/4

Int x^3cos(x^2) = (x^4cos(x^2))/4 + 1/2 Int x^5 sin (x^2)

u = sin(x^2)
du = 2x cos(x^2)
dv = x^5
v = (x^6)/6

Int x^5 sin (x^2) = (x^6sin(x^2))/6 - 1/3 Int x^7 cos (x^2)

Hmmm... Doesn't seem to be working out that way... Unless I look at the pattern and find...

Int x^3cos(x^2) = (x^4cos(x^2))/4 + 1/2((x^6sin(x^2))/6) - 1/3((x^8cos(x^2))/8) + 1/4((x^10sin(x^2))/10)...

Although that's just a guess...
Damn year 12...

Some how the answer comes out to be:

1/2 (x^2sinx^2 + cosx^2) + c

and I hope I didn't make a mistake in my calculations up there...

Alright, I guess my "trick for choosing u and dv" isn't so good afterall because what you took as dv has an easily found primitive but you got nowhere with it.

What other choice of u and dv can you make? There aren't an infinity, there are only two:

- There's u=x, dv=x²cos(x²). You can try doing it with this if you know the primitive of wcos(w). I haven't tried it so I don't know if it will work.

- There's u=x², dv=xcos(x²). Can you see that this choice of dv makes it perfectly fit for an integration by change of variable?You might want to look at the ILATE rule on wiki that gives a general rule of thumb for what to take as u and dv. There is a remark at the end however that warn the reader that there are exceptions to the ILATE rule. This problem of yours is one of them.

http://en.wikipedia.org/wiki/Integration_by_parts#The_ILATE_rule

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Let $$u = x^{2}$$ and $$dv = x\cos(x^{2}) dx$$Then $$du = 2x dx$$ and $$v = \frac{1}{2} \sin(x^{2})$$

Then just apply the formula from here.

$$\int udv = uv - \int vdu$$

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Didn't know you could just move over the x. Sweet.

## 1. What is an integral?

An integral is a mathematical concept used to calculate the area under a curve or the accumulation of a quantity over a given interval.

## 2. How is an integral calculated?

An integral is calculated using a process called integration, which involves finding the anti-derivative of a function and evaluating it at the limits of integration.

## 3. What is the difference between a definite and indefinite integral?

A definite integral has specific limits of integration, while an indefinite integral does not. This means that a definite integral will give a specific numerical value, while an indefinite integral will result in a function.

## 4. What does it mean for an integral to be correct?

An integral is considered correct if it satisfies certain properties, such as the fundamental theorem of calculus and the rules of integration. It also must be evaluated correctly and give a meaningful result.

## 5. What are some common applications of integrals in science?

Integrals are used in many branches of science, including physics, engineering, and economics. Some common applications include calculating work, finding the center of mass of an object, and determining the amount of radioactive decay.

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