# Integration Using U-Substitution involving Trig Functions and Identities

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]

LCKurtz
Homework Helper
Gold Member
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)]

I'm sorry, but first of all: What makes you think that

$\int \frac{7 \sin{x}}{1+cos^{2}{x}} dx = \int 7 \sin{x} dx + \int \frac{1}{1+cos^{2}{x}} dx$

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?

LCKurtz
Homework Helper
Gold Member
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...

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.

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

LCKurtz
Homework Helper
Gold Member
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.

vela
Staff Emeritus
Homework Helper
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.

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
$$-7\int \frac{1}{1+u^2}\,du = -7\arctan u + C = -7\arctan(\cos x)+C$$

LCKurtz
$$-7\int \frac{1}{1+u^2}\,du = -7\arctan u + C = -7\arctan(\cos x)+C$$
Yes, and I should have read what he actually wrote more carefully instead of just observing he had the final answer correct. 