MHB Integration of trig. functions II

[paulmdrdo1]

How about these?

∫tan^3xdx

∫(sin^4x cos^2x)dx

 

[MarkFL]

Re: integration of trig func.

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

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

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

I would write the integrand as:

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

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...

 

[Prove It]

How about these?

∫tan^3xdx

∫(sin^4x cos^2x)dx

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

[math]\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*}[/math]

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

[math]\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*}[/math]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)...

[math]\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*}[/math]

 

[MarkFL]

...I would try to use more double angle identities to avoid the power reduction formulae (which I can never remember)...

We are actually doing the same thing. The power reduction identity I had in mind is:

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

 

[Prove It]

We are actually doing the same thing. The power reduction identity I had in mind is:

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

I thought you were referring to the power reduction formula to integrate [math]\displaystyle \begin{align*} \sin^{n}{(x)} \end{align*}[/math], not the double angle identity for cosine...

 

[MarkFL]

I thought you were referring to the power reduction formula to integrate [math]\displaystyle \begin{align*} \sin^{n}{(x)} \end{align*}[/math], not the double angle identity for cosine...

This is my "cheat sheet" and why I use the term "power reduction:"

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.

 

[paulmdrdo1]

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

[math]\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*}[/math]

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

[math]\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*}[/math]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)...

[math]\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*}[/math]

where did you get the sin4x/4 part?

 

[paulmdrdo1]

i understood the first 8 lines of your solution but after that I'm stucked! this is what i do..

∫ 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!

 

[MarkFL]

Let's look at:

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

Now, if we let:

$$u=4x\,\therefore\,du=4\,dx$$

then we have:

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

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

 

[Petrus]

i understood the first 8 lines of your solution but after that I'm stucked! this is what i do..

∫ 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!

Hello Paul,
indeed that $$\cos(4x)\neq \frac{\sin(4x)}{4}$$ BUT what prove it did Was that he integrate it! Let's check if it is correct! So if we derivate $$\frac{\sin(4x)}{4}$$ we shall get $$\cos(4x)$$ so if we use chain rule to derivate that we get $$\frac{4\cos(4x)}{4}$$ and if we simplifie that we get $$\cos(4x)$$ which we wanted to get:)
Edit:Mark Was faster:(
Regards,
$$|\pi\rangle$$

 

[paulmdrdo1]

Let's look at:

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

Now, if we let:

$$u=4x\,\therefore\,du=4\,dx$$

then we have:

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

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

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?

 

[MarkFL]

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?

You asked (twice):

where did you get the sin4x/4 part?

...how can the cos4x be sin4x/4? please help me!

So this is what I was addressing. To evaluate:

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

I would use the substitution:

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

and so we have:

$$\frac{1}{2}\int u^2\,du$$

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

 

[paulmdrdo1]

You asked (twice):So this is what I was addressing. To evaluate:

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

I would use the substitution:

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

and so we have:

$$\frac{1}{2}\int u^2\,du$$

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

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.

 

[Petrus]

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.

Hello Paul,
Just a fast respond as I have to think more when I am home but that Dont work, Cause you Will have $$du$$ on one side... You need on both side:)
Regards,
$$|\pi\rangle$$

 

[Petrus]

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.

Hello Paul,
I think now this should work
subsitute $$u=2x <=> du=2$$ and integrate them separate so you got
$$\frac{1}{16}\int\sin^2(u) du-\frac{1}{16}\int\cos(u)\sin^2(u)du$$
does this make it simpler?

Regards
$$|\pi\rangle$$