# Trigonometric Integrals

I got these two questions wrong and I'm wondering what i did wrong.

Homework Statement [/b]
1. http://i324.photobucket.com/albums/k327/ProtoGirlEXE/q3.jpg [Broken]

2. http://i324.photobucket.com/albums/k327/ProtoGirlEXE/q4.jpg [Broken]

The attempt at a solution[/b]
1. http://i324.photobucket.com/albums/k327/ProtoGirlEXE/100_0639.jpg [Broken]

2.http://i324.photobucket.com/albums/k327/ProtoGirlEXE/100_0643.jpg [Broken]
(I put t instead of x while I was doing the problem on accident)

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jgens
Gold Member
Your work is really difficult to read from those pages, but from what I can gather, here are some of the mistakes that you made ...

First Problem: You forgot to distribute the 3 across when you expanded 3cos2(2x). Check your work there. Second, I have no idea how you got your expansion for cos3(2x).

Second Problem: Review your trig identities jgens
Gold Member
First Problem: cos(2x)cos(4x) =/= cos2(8x). I'm not really sure how you got that in the first place.

Second Problem: Do it again, and this time be very careful about how you distribute things so that you don't end up with extra terms.

First Problem: cos(2x)cos(4x) =/= cos2(8x). I'm not really sure how you got that in the first place.

Second Problem: Do it again, and this time be very careful about how you distribute things so that you don't end up with extra terms.

for the second one, is it just (1/3)tan^3 x +c?

for the first one: there was a -cos^3(2x), so I thought I should split it up like cos^2(2x) and cos (2x) then make cos^2(2x)---> -(1/2)+(1/2)cos(4x)(cos(2x)....

jgens
Gold Member
for the second one, is it just (1/3)tan^3 x +c?

No. Again, be careful about how you distribute things.

for the first one: there was a -cos^3(2x), so I thought I should split it up like cos^2(2x) and cos (2x) then make cos^2(2x)---> -(1/2)+(1/2)cos(4x)(cos(2x)....

Sure, you can split it up like that, and I followed your work through there. However, cos(4x)cos(2x) =/= cos2(8x). I don't know what would make you think that that was the case.

No. Again, be careful about how you distribute things.

I tried it with u substitution, instead of distributing, and got (1/3)tan^3 + x + c

Sure, you can split it up like that, and I followed your work through there. However, cos(4x)cos(2x) =/= cos2(8x). I don't know what would make you think that that was the case.

oh yeah duh, so it's just cos(8x)
then I should get (2/8)x-(3/16)sin(2x)+(3/64)cos(4x)+(1/128)cos(16x)+c, right?

jgens
Gold Member
No. What makes you think that cos(2x)cos(4x) = cos(8x)? They're not equal.

No. What makes you think that cos(2x)cos(4x) = cos(8x)? They're not equal.

they're not? so then what can I do, if they won't separate?

omg (1/3)tan^3 x + x +c is wrong too :(
this is so frustrating

jgens
Gold Member
they're not? so then what can I do, if they won't separate?

They're quite clearly not equal. My question is still what makes you think that they should be in the first place? And I would have tackled this problem using integration by parts. It's really easy to derive a reduction formula and that does pretty much all of the work for you.

jgens
Gold Member
omg (1/3)tan^3 x + x +c is wrong too :(
this is so frustrating

Why don't you show your steps? From the answers that you've been posting, it looks like you've been distributing terms incorrectly.

They're quite clearly not equal. My question is still what makes you think that they should be in the first place? And I would have tackled this problem using integration by parts. It's really easy to derive a reduction formula and that does pretty much all of the work for you.

how can you apply integration of parts to sin^6 x, its just one term?

jgens
Gold Member
how can you apply integration of parts to sin^6 x, its just one term?

You have to manipulate sin6(x) using trig identities first.

integral sec^4 x dx
integral (tan^2 x+1)sec^2 x dx
u= tanx du=sec^2 x
integral (u^2 + 1) du
(1/3)tan^3 x + x + c

u =/= x

You have to manipulate sin6(x) using trig identities first.

u =/= x

now I'm terribly lost

jgens
Gold Member

$$\int\sin{(x)}^{2n}dx=\int[1-\cos{(x)}^2]\sin{(x)}^{2(n-1)}dx=\int\sin{(x)}^{2(n-1)}dx-\int\cos{(x)}^2\sin{(x)}^{2(n-1)}dx$$

$$\int\cos{(x)}^2\sin{(x)}^{2(n-1)}dx=\frac{\cos{(x)}\sin{(x)}^{2n-1}}{2n-1}+\frac{1}{2n-1}\int\sin{(x)}^{2n}dx$$

Putting these two together, we see that

$$\int\sin{(x)}^{2n}dx=\int\sin{(x)}^{2(n-1)}dx-\frac{\cos{(x)}\sin{(x)}^{2n-1}}{2n-1}-\frac{1}{2n-1}\int\sin{(x)}^{2n}dx$$

$$\frac{2n}{2n-1}\int\sin{(x)}^{2n}dx=\int\sin{(x)}^{2(n-1)}dx-\frac{\cos{(x)}\sin{(x)}^{2n-1}}{2n-1}$$

$$\int\sin{(x)}^{2n}dx=\frac{2n-1}{2n}\left[\int\sin{(x)}^{2(n-1)}dx-\frac{\cos{(x)}\sin{(x)}^{2n-1}}{2n-1}\right]$$

Can you see how this applies to the first problem that you're working? What I posted here is a generalization, so you'll need to figure the specifics out.

jgens
Gold Member

$$\int u^2+1du=\frac{u^3}{3}+u+C$$

Since $u=\tan{x}$, it's now just a matter of replacing each $u$ in the equation above with $\tan{x}$. The last few times, you've haven't been consistent in this last regard and that's what's causing you so much grief.

jgens
Gold Member
x/8 - 3/16 sin2x + 3/16 ∫ (1 + cos 4x) dx - 1/16 ∫ (1 + cos 4x) cos 2x dx =
x/8 - 3/16 sin 2x + 3/16 x + 3/64 sin 4x - 1/16x + (1/64)sin 4x + C

What happened to the factor of cos(2x)? You can't just ignore it, so yes, your answer is wrong. The approach that I outlined in one of my previous posts will let you evaluate this integral with very little work, so why don't you try to apply that?

∫ sec^4 x dx = ∫ sec^2 x (sec^2 x dx) = ∫ (1 + tan^2 x) sec^2 x dx = tan x + tan^3 x/3 + C
u = tan x. So, ∫ (u^2 + 1) du = u^3/3 + u + C = tan^2 x/3 + tan x + C
****I got that from somewhere, but I think it's wrong because I really think it should be tan^3 not tan^2.

I'm not going to resolve this one for you since you're making such an obvious mistake. Look at what you wrote above carefully and tell me if it makes sense. It doesn't pay off to be sloppy with your work.