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#### paulmdrdo

##### Active member

- May 13, 2013

- 386

How about these?

∫tan^3xdx

∫(sin^4x cos^2x)dx

∫tan^3xdx

∫(sin^4x cos^2x)dx

- Thread starter paulmdrdo
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- Thread starter
- #1

- May 13, 2013

- 386

How about these?

∫tan^3xdx

∫(sin^4x cos^2x)dx

∫tan^3xdx

∫(sin^4x cos^2x)dx

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- #2

1.) \(\displaystyle \int\tan^3(x)\,dx\)

Rewrite the integrand as \(\displaystyle \tan(x)\tan^2(x)\), then use the Pythagorean identity \(\displaystyle \tan^2(\theta)=\sec^2(\theta)-1\).

2.) \(\displaystyle \int\sin^4(x)\cos^2(x)\,dx\)

I would write the integrand as:

\(\displaystyle \sin^2(x)(\sin(x)\cos(x))^2\)

Now, try applying the double-angle identity for sine on the second factor and the power reduction identity for sine on the first. There will be further work after that, but this should get you started...

I usually prefer converting to sines and cosines first...How about these?

∫tan^3xdx

∫(sin^4x cos^2x)dx

\(\displaystyle \displaystyle \begin{align*} \int{\tan^3{(x)}\,dx} &= \int{\frac{\sin^3{(x)}}{\cos^3{(x)}}\,dx} \\ &= \int{ \frac{\sin{(x)}\sin^2{(x)}}{\cos^3{(x)}}\,dx} \\ &= \int{ \frac{\sin{(x)} \left[ 1 - \cos^2{(x)} \right] }{\cos^3{(x)}} \, dx} \\ &= \int{ \frac{-\sin{(x)}\left[ \cos^2{(x)} - 1 \right] }{\cos^3{(x)}}\,dx} \end{align*}\)

Now let \(\displaystyle \displaystyle \begin{align*} u = \cos{(x)} \implies du = -\sin{(x)}\,dx \end{align*}\) and the integral becomes

\(\displaystyle \displaystyle \begin{align*} \int{\frac{ -\sin{(x)}\left[ \cos^2{(x)} - 1 \right] }{\cos^3{(x)}}\,dx} &= \int{ \frac{u^2 - 1}{u^3}\,du} \\ &= \int{ \frac{1}{u} - u^{-3}\,du} \\ &= \ln{ |u|} + \frac{1}{2}u^{-2} + C \\ &= \ln{ \left| \cos{(x)} \right| } + \frac{1}{2\cos^2{(x)}} + C \end{align*}\)

As for the second, following Mark's initial suggestion, I would try to use more double angle identities to avoid the power reduction formulae (which I can never remember)...

\(\displaystyle \displaystyle \begin{align*} \int{\sin^4{(x)}\cos^2{(x)}\,dx} &= \int{\sin^2{(x)}\sin^2{(x)}\cos^2{(x)}\,dx} \\ &= \int{ \sin^2{(x)} \left[ \sin{(x)}\cos{(x)} \right] ^2 \, dx} \\ &= \int{\sin^2{(x)} \left[ \frac{1}{2}\sin{(2x)} \right] ^2 \, dx} \\ &= \frac{1}{4} \int{ \sin^2{(x)} \sin^2{(2x)} \, dx} \\ &= \frac{1}{4} \int{ \frac{1}{2}\left[ 1 - \cos{(2x)} \right] \sin^2{(2x)}\, dx} \\ &= \frac{1}{8} \int{ \sin^2{(2x)} - \cos{(2x)}\sin^2{(2x)} \, dx} \\ &= \frac{1}{8} \int{ \frac{1}{2} \left[ 1 - \cos{(4x)} \right] - \cos{(2x)} \sin^2{(2x)} \, dx} \\ &= \frac{1}{16} \int{ 1\, dx} - \frac{1}{16} \int{ \cos{(4x)}\,dx} - \frac{1}{16} \int{ 2\cos{(2x)} \sin^2{(2x)} \, dx} \\ &= \frac{1}{16}x - \frac{1}{16} \left[ \frac{\sin{(4x)}}{4} \right] - \frac{1}{16} \left[ \frac{\sin^3{(2x)}}{3} \right] + C \\ &= \frac{x}{16} - \frac{\sin{(4x)}}{64} - \frac{\sin^3{(2x)}}{48} + C \end{align*}\)

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- #4

We are actually doing the same thing. The power reduction identity I had in mind is:...I would try to use more double angle identities to avoid the power reduction formulae (which I can never remember)...

\(\displaystyle \sin^2(\theta)=\frac{1-\cos(2\theta)}{2}\)

I thought you were referring to the power reduction formula to integrate \(\displaystyle \displaystyle \begin{align*} \sin^{n}{(x)} \end{align*}\), not the double angle identity for cosine...We are actually doing the same thing. The power reduction identity I had in mind is:

\(\displaystyle \sin^2(\theta)=\frac{1-\cos(2\theta)}{2}\)

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- #6

This is my "cheat sheet" and why I use the term "power reduction:"I thought you were referring to the power reduction formula to integrate \(\displaystyle \displaystyle \begin{align*} \sin^{n}{(x)} \end{align*}\), not the double angle identity for cosine...

List of trigonometric identities - Wikipedia, the free encyclopedia

You are correct though, that it is simply a rearrangement of one of the forms of the double-angle identity for cosine.

- Thread starter
- #7

- May 13, 2013

- 386

I usually prefer converting to sines and cosines first...

\(\displaystyle \displaystyle \begin{align*} \int{\tan^3{(x)}\,dx} &= \int{\frac{\sin^3{(x)}}{\cos^3{(x)}}\,dx} \\ &= \int{ \frac{\sin{(x)}\sin^2{(x)}}{\cos^3{(x)}}\,dx} \\ &= \int{ \frac{\sin{(x)} \left[ 1 - \cos^2{(x)} \right] }{\cos^3{(x)}} \, dx} \\ &= \int{ \frac{-\sin{(x)}\left[ \cos^2{(x)} - 1 \right] }{\cos^3{(x)}}\,dx} \end{align*}\)

Now let \(\displaystyle \displaystyle \begin{align*} u = \cos{(x)} \implies du = -\sin{(x)}\,dx \end{align*}\) and the integral becomes

\(\displaystyle \displaystyle \begin{align*} \int{\frac{ -\sin{(x)}\left[ \cos^2{(x)} - 1 \right] }{\cos^3{(x)}}\,dx} &= \int{ \frac{u^2 - 1}{u^3}\,du} \\ &= \int{ \frac{1}{u} - u^{-3}\,du} \\ &= \ln{ |u|} + \frac{1}{2}u^{-2} + C \\ &= \ln{ \left| \cos{(x)} \right| } + \frac{1}{2\cos^2{(x)}} + C \end{align*}\)

As for the second, following Mark's initial suggestion, I would try to use more double angle identities to avoid the power reduction formulae (which I can never remember)...

\(\displaystyle \displaystyle \begin{align*} \int{\sin^4{(x)}\cos^2{(x)}\,dx} &= \int{\sin^2{(x)}\sin^2{(x)}\cos^2{(x)}\,dx} \\ &= \int{ \sin^2{(x)} \left[ \sin{(x)}\cos{(x)} \right] ^2 \, dx} \\ &= \int{\sin^2{(x)} \left[ \frac{1}{2}\sin{(2x)} \right] ^2 \, dx} \\ &= \frac{1}{4} \int{ \sin^2{(x)} \sin^2{(2x)} \, dx} \\ &= \frac{1}{4} \int{ \frac{1}{2}\left[ 1 - \cos{(2x)} \right] \sin^2{(2x)}\, dx} \\ &= \frac{1}{8} \int{ \sin^2{(2x)} - \cos{(2x)}\sin^2{(2x)} \, dx} \\ &= \frac{1}{8} \int{ \frac{1}{2} \left[ 1 - \cos{(4x)} \right] - \cos{(2x)} \sin^2{(2x)} \, dx} \\ &= \frac{1}{16} \int{ 1\, dx} - \frac{1}{16} \int{ \cos{(4x)}\,dx} - \frac{1}{16} \int{ 2\cos{(2x)} \sin^2{(2x)} \, dx} \\ &= \frac{1}{16}x - \frac{1}{16} \left[ \frac{\sin{(4x)}}{4} \right] - \frac{1}{16} \left[ \frac{\sin^3{(2x)}}{3} \right] + C \\ &= \frac{x}{16} - \frac{\sin{(4x)}}{64} - \frac{\sin^3{(2x)}}{48} + C \end{align*}\)

where did you get the sin4x/4 part?

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- #8

- May 13, 2013

- 386

∫ sin²x (sin²x cos²x) dx

= ∫ sin²x (sinx cosx)² dx

half-angle identity:

sin²x = (1/2)[1 - cos(2x)]

the double-angle identity:

sin(2x) = 2sinx cosx → sinx cosx = (1/2) sin(2x)

the integral becomes:

∫ (1/2)[1 - cos(2x)] [(1/2) sin(2x)]² dx =

∫ (1/2)[1 - cos(2x)] (1/4) sin²(2x) dx =

pulling out the constants and expanding the integrand,

(1/8) ∫ [sin²(2x) - cos(2x) sin²(2x)] dx =

break it into:

(1/8) ∫ sin²(2x) dx - (1/8) ∫ sin²(2x) cos(2x) dx =

(1/8) ∫ (1/2){1 - cos[2(2x)]} dx - (1/8) ∫ sin²(2x) cos(2x) dx =

(1/16) ∫ dx - (1/16) ∫ cos(4x) dx - (1/8) ∫ sin²(2x) cos(2x) dx = ---> this is where i'm stuck how can the cos4x be sin4x/4? please help me!

- Admin
- #9

- Feb 21, 2013

- 739

Hello Paul,

∫ sin²x (sin²x cos²x) dx

= ∫ sin²x (sinx cosx)² dx

half-angle identity:

sin²x = (1/2)[1 - cos(2x)]

the double-angle identity:

sin(2x) = 2sinx cosx → sinx cosx = (1/2) sin(2x)

the integral becomes:

∫ (1/2)[1 - cos(2x)] [(1/2) sin(2x)]² dx =

∫ (1/2)[1 - cos(2x)] (1/4) sin²(2x) dx =

pulling out the constants and expanding the integrand,

(1/8) ∫ [sin²(2x) - cos(2x) sin²(2x)] dx =

break it into:

(1/8) ∫ sin²(2x) dx - (1/8) ∫ sin²(2x) cos(2x) dx =

(1/8) ∫ (1/2){1 - cos[2(2x)]} dx - (1/8) ∫ sin²(2x) cos(2x) dx =

(1/16) ∫ dx - (1/16) ∫ cos(4x) dx - (1/8) ∫ sin²(2x) cos(2x) dx = ---> this is where i'm stuck how can the cos4x be sin4x/4? please help me!

indeed that \(\displaystyle \cos(4x)\neq \frac{\sin(4x)}{4}\) BUT what prove it did Was that he integrate it! Let's check if it is correct! So if we derivate \(\displaystyle \frac{\sin(4x)}{4}\) we shall get \(\displaystyle \cos(4x)\) so if we use chain rule to derivate that we get \(\displaystyle \frac{4\cos(4x)}{4}\) and if we simplifie that we get \(\displaystyle \cos(4x)\) which we wanted to get

Edit:Mark Was faster

Regards,

\(\displaystyle |\pi\rangle\)

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- #11

- May 13, 2013

- 386

no i understand that part.. my follow up question is how would i deal with the third term in the integrand ((1/8) ∫ sin²(2x) cos(2x) dx )?Let's look at:

\(\displaystyle \int\cos(4x)\,dx\)

Now, if we let:

\(\displaystyle u=4x\,\therefore\,du=4\,dx\)

then we have:

\(\displaystyle \frac{1}{4}\int\cos(u)\,du\)

Now, integrate, then back-substitute for $u$, and you have the result.

should i use the power reduction formula for sin²(2x) again?

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- #12

You asked (twice):no i understand that part.. my follow up question is how would i deal with the third term in the integrand ((1/8) ∫ sin²(2x) cos(2x) dx )?

should i use the power reduction formula for sin²(2x) again?

where did you get the sin4x/4 part?

So this is what I was addressing. To evaluate:...how can the cos4x be sin4x/4? please help me!

\(\displaystyle \int\sin^2(2x)\cos(2x)\,dx\)

I would use the substitution:

\(\displaystyle u=\sin(2x)\,\therefore\,du=2\cos(2x)\,dx\)

and so we have:

\(\displaystyle \frac{1}{2}\int u^2\,du\)

Now, integrate, then back-substitute for $u$.

- Thread starter
- #13

- May 13, 2013

- 386

hmmm..correct me if my thougth processes is wrong.You asked (twice):

So this is what I was addressing. To evaluate:

\(\displaystyle \int\sin^2(2x)\cos(2x)\,dx\)

I would use the substitution:

\(\displaystyle u=\sin(2x)\,\therefore\,du=2\cos(2x)\,dx\)

and so we have:

\(\displaystyle \frac{1}{2}\int u^2\,du\)

Now, integrate, then back-substitute for $u$.

(1/8) ∫ [sin²(2x) - cos(2x) sin²(2x)] dx = in this part of the solution can i use that substitution of u=sin2x? in that way i don't have to use the power reduction formula anymore. please bear with me.

- Feb 21, 2013

- 739

Hello Paul,hmmm..correct me if my thougth processes is wrong.

(1/8) ∫ [sin²(2x) - cos(2x) sin²(2x)] dx = in this part of the solution can i use that substitution of u=sin2x? in that way i don't have to use the power reduction formula anymore. please bear with me.

Just a fast respond as I have to think more when I am home but that Dont work, Cause you Will have \(\displaystyle du\) on one side... You need on both side

Regards,

\(\displaystyle |\pi\rangle\)

- Feb 21, 2013

- 739

Hello Paul,hmmm..correct me if my thougth processes is wrong.

(1/8) ∫ [sin²(2x) - cos(2x) sin²(2x)] dx = in this part of the solution can i use that substitution of u=sin2x? in that way i don't have to use the power reduction formula anymore. please bear with me.

I think now this should work

subsitute \(\displaystyle u=2x <=> du=2\) and integrate them separate so you got

\(\displaystyle \frac{1}{16}\int\sin^2(u) du-\frac{1}{16}\int\cos(u)\sin^2(u)du\)

does this make it simpler?

Regards

\(\displaystyle |\pi\rangle\)