What did i do wrong in my integration?

[dnt]

Homework Statement

(dont know how to make the integral sign so bare with me please)

i will use | to signify the integral sign:

here is my problem:

|(sinx)(cos)dx

Homework Equations

n/a

The Attempt at a Solution

ok this is what i did:

first i changed it to (1/2)sin2x because sin2x=2sinxcosx

so now i have:

|(1/2)sin2xdx

then i said u = 2x and du=2dx

so now i have:

|(1/4)sinudu

which is:

(1/4)(-cos u) + C

which is:

(1/4)(-cos2x) + C

where is my error?

also, can someone give me a brief overview of the best strategy to figure out what type of integration to use? what do you look for to know when to use trig substituion, integration by parts, u-substitution, etc...

ie, how do you know? thanks.

 

[IMDerek]

let u = sin x. then du = cos x dx.

Integration is just experience. You just need to practice.

 

[cristo]

Alternatively, since you know the integral of sinx=-cosx, you should know that the integral of sin(nx)=-cos(nx)/n

 

[suspenc3]

I had the exact same problem when I started integration, there really is no "trick" to know what type of integration to use. It all comes with a lot of practice. Eventually you will just recognize things and you'll know what type of integration to use.

 

[ChaoticLlama]

also, can someone give me a brief overview of the best strategy to figure out what type of integration to use? what do you look for to know when to use trig substituion, integration by parts, u-substitution, etc...

ie, how do you know? thanks.

There are general tips you can use.

If you have a square root in the function of the form sqrt(a-x^2), sqrt(a+x^2), or sqrt(x^2-a), each has a different trig substitution.
You use substitution to create a square trig function which will undo the root.

By parts is best used if nothing is obvious (can't be more vague than that :P). Specifically if one part of the function is easily integratable (eg e^x, sin(x)).
Use byparts when you see a product of somekind (hint: |f(x)dx can be considered a product of f(x) and dx)

Partial fractions is merely an algebraic method for rewriting a function in an easier-to-integrate form. You largely only use this if you see a rational expression of one polynomial divided by another. (it may be particularly advantageous is the degree of the numerator > degree of denominator)

substitution is useful if you see a function and it's derivative in the same function (eg: |ln(x)/x dx) But most times substitution can be done by inspection.

 

[Gib Z]

What ChaoticLlama said is pretty good, but you'll learn those over practice as well. Only thing I have to add is Trig substitution is only good if you can do trigonometric integrals as well. I can always get it down to the trig integral, which is the aim, but i can't solve that one at the end..

 

[dnt]

let u = sin x. then du = cos x dx.

Integration is just experience. You just need to practice.

i should have added that i do know what the correct answer is and i knew how to get it. my question was in what step of my above work is wrong? i can't find it yet i know it comes out incorrect.

 

[d_leet]

i should have added that i do know what the correct answer is and i knew how to get it. my question was in what step of my above work is wrong? i can't find it yet i know it comes out incorrect.

I don't think it is, expand the double angle formula for cosine and I think you will find that your solution is equivalent to the other.

 

[HallsofIvy]

What makes you think you have an error?

Yes, the way most people would integrate $\int sin(x)cos(x)dx$ would be to let u= sin(x) so du= cos(x)dx and the integral becomes $\int udu= \frac{1}{2}u^2+ C= \frac{1}{2}sin^2(x)+ C$

But it is perfectly correct that sin(2x)= 2sin(x)cos(x) so that $\int sin(x)cos(x)dx= \frac{1}{2}\int sin(2x)dx= \frac{1}{4}cos(2x)+ C'$.

Have you considered the possibility that
$$\frac{1}{4}cos(2x)+ C'= \frac{1}{2}sin^2(x)+ C$$
possibly with different values for C and C'?

Since you used sin(2x)= 2sin(x)cos(x) you might consider now using
cos(2x)= cos²(x)- sin²= 1- 2sin²(x).

 

[Schrodinger's Dog]

Homework Statement

/snip

You look at the answer and go You look at your answer and theirs again, and realize it's entirely equivalent because of the double angle formulas, I've been their