Integration Using U-Substitution involving Trig Functions and Identities

In summary, we discussed the best approach for solving the problem and determined that using the substitution u = cos(x) would be most effective. We then solved for the integral and confirmed our answer by differentiating it. It was also noted that paying attention to proper notation and writing things precisely can be helpful in understanding the material.
  • #1
seanoe25
7
0
1.) ∫[(7 sin (x))/[1+cos^2(x)]] * dx



2.) I'm looking at the trig identity sin^2 x+cos^2 x=1, and am wondering if I could use that in solving the problem. Or should I use u=sin x, then du= cos x, then plug those in?



3.) so I thought maybe it would be easier to separate the two integrals. I came up with
7∫sin (x)*dx + ∫[1+cos^2(x)]^-1*dx.

which equals
7 cos (x)dx+∫1/[1+cos^2(x)]

u=sin(x)
du=cos(x)*dx

= ∫dx/[1+du^2]

And now I'm stuck. Please help
 
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  • #2
seanoe25 said:
1.) ∫[(7 sin (x))/[1+cos^2(x)]] * dx



2.) I'm looking at the trig identity sin^2 x+cos^2 x=1, and am wondering if I could use that in solving the problem. Or should I use u=sin x, then du= cos x, then plug those in?
If you use u = sin(x), do you see a du=cos(x)dx in your problem? What if you try u = cos(x)?
3.) so I thought maybe it would be easier to separate the two integrals. I came up with
7∫sin (x)*dx + ∫[1+cos^2(x)]^-1*dx.
No hope of getting a correct answer with that kind of algebra.

which equals
7 cos (x)dx+∫1/[1+cos^2(x)]

Yellow is unreadable.
 
  • #3
I'm sorry, but first of all: What makes you think that

[itex]\int \frac{7 \sin{x}}{1+cos^{2}{x}} dx = \int 7 \sin{x} dx + \int \frac{1}{1+cos^{2}{x}} dx[/itex]
 
  • #4
Don't be sorry, you all are helping me


I took the integral identity ∫f(x)+g(x)=∫f(x)+∫g(x), and somehow came up with some bogus new rule.

I used u =cos (x)
du= - sin (x)*dx
dx= du/-sin (X)

∫ 7 sin (x)*dx/[1+u^2]
∫ [7 sin (x)* [du/-sin (x)]/ 1+u^2
=
∫-7du/[1+u^2]
=
7∫1/[1+u^2] * du. Is that correct?
 
  • #5
seanoe25 said:
Don't be sorry, you all are helping me


I took the integral identity ∫f(x)+g(x)=∫f(x)+∫g(x), and somehow came up with some bogus new rule.

I used u =cos (x)
du= - sin (x)*dx
dx= du/-sin (X)

∫ 7 sin (x)*dx/[1+u^2]
∫ [7 sin (x)* [du/-sin (x)]/ 1+u^2
=
∫-7du/[1+u^2]
=
7∫1/[1+u^2] * du. Is that correct?

Yes, but why did you drop the minus sign? And, of course, you have to finish...
 
  • #6
I just forgot to type it in.

so now I have

-7 ∫1/(1+u^2), but this is where it gets hazy for me.

would it be

-7* 1/(-2+1) *u^(-2+1)?

which would equal

-7*-1*u^-1, which gives me 7/cos(x), but that doesn't seem right.
 
  • #7
actually I got it. I forgot arctan (u) = 1/1+u^2. So it becomes

-7 ∫arctan (u)*du
=
-7 arctan (cos x) +C

Thanks for the push. And putting up with my crappy algebra
 
  • #8
seanoe25 said:
actually I got it. I forgot arctan (u) = 1/1+u^2. So it becomes

-7 ∫arctan (u)*du
=
-7 arctan (cos x) +C

Thanks for the push. And putting up with my crappy algebra

Yes. And you can verify it is correct by differentiating it.
 
  • #9
seanoe25 said:
so now I have

-7 ∫1/(1+u^2), but this is where it gets hazy for me.

would it be

-7* 1/(-2+1) *u^(-2+1)?
The rule you tried to use,
$$\int x^n\,dx = \frac{x^{n+1}}{n+1}+C,$$ doesn't apply here because ##\frac{1}{1+u^2}## isn't of the form ##u^n##. I know you were just making a wild guess here, but you should have known it was wrong.

seanoe25 said:
actually I got it. I forgot arctan (u) = 1/1+u^2.
What you've written isn't correct. You left out the parentheses around the denominator, and arctan u isn't equal to 1/(1+u2). What you meant was that
$$\frac{d}{du}\arctan u = \frac{1}{1+u^2}.$$ You can't just leave out the d/du part.

You should really make an effort to understand the notation and get in the habit of writing things down precisely and correctly, if only so you don't annoy whoever is grading your exams. (Trust me, they find it annoying when you're sloppy with the notation.) You'll also probably find that making this effort will help in your understanding of the material.

So it becomes

-7 ∫arctan (u)*du
=
-7 arctan (cos x) +C
Same thing here. You're not integrating arctan u; you're integrating 1/(1+u2). What you meant was
$$-7\int \frac{1}{1+u^2}\,du = -7\arctan u + C = -7\arctan(\cos x)+C$$
 
  • #10
vela said:
Same thing here. You're not integrating arctan u; you're integrating 1/(1+u2). What you meant was
$$-7\int \frac{1}{1+u^2}\,du = -7\arctan u + C = -7\arctan(\cos x)+C$$

Yes, and I should have read what he actually wrote more carefully instead of just observing he had the final answer correct. :frown:
 

1. What is U-substitution in calculus?

U-substitution is a technique used in calculus to simplify the integration of a function by replacing a variable with a new variable, u, that is a function of the original variable. This allows for easier integration by using known integration rules.

2. How do I know when to use U-substitution?

U-substitution is typically used when the integrand contains a function and its derivative, or when it involves a composition of functions. You can also look for patterns such as polynomial or trigonometric functions that can be simplified using U-substitution.

3. Can U-substitution be used for trigonometric functions?

Yes, U-substitution can be used for trigonometric functions. In fact, it is often used to simplify integrals involving trigonometric functions and identities. It can also be used for inverse trigonometric functions.

4. How do I choose the right u for U-substitution involving trigonometric functions?

The choice of u for U-substitution is not always obvious, but a good rule of thumb is to choose u as the inner function in a composition of functions. For trigonometric functions, u can be chosen as the inside function in a trigonometric identity or as the argument of a trigonometric function.

5. Are there any special cases when using U-substitution for trigonometric functions?

Yes, there are some special cases to consider when using U-substitution for trigonometric functions. This includes using half-angle or double-angle identities, completing the square, or using the Pythagorean identities to simplify the integrand before applying U-substitution.

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