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paulmdrdo
Active member
- May 13, 2013
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How about these?
∫tan^3xdx
∫(sin^4x cos^2x)dx
∫tan^3xdx
∫(sin^4x cos^2x)dx
I usually prefer converting to sines and cosines first...How about these?
∫tan^3xdx
∫(sin^4x cos^2x)dx
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)...
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:
\(\displaystyle \sin^2(\theta)=\frac{1-\cos(2\theta)}{2}\)
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...
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)...
\(\displaystyle \displaystyle \begin{align*} \int{\sin^4{(x)}\cos^2{(x)}\,dx} &= \int{\sin^2{(x)}\sin^2{(x)}\cos^2{(x)}\,dx} \\ &= \int{ \sin^2{(x)} \left[ \sin{(x)}\cos{(x)} \right] ^2 \, dx} \\ &= \int{\sin^2{(x)} \left[ \frac{1}{2}\sin{(2x)} \right] ^2 \, dx} \\ &= \frac{1}{4} \int{ \sin^2{(x)} \sin^2{(2x)} \, dx} \\ &= \frac{1}{4} \int{ \frac{1}{2}\left[ 1 - \cos{(2x)} \right] \sin^2{(2x)}\, dx} \\ &= \frac{1}{8} \int{ \sin^2{(2x)} - \cos{(2x)}\sin^2{(2x)} \, dx} \\ &= \frac{1}{8} \int{ \frac{1}{2} \left[ 1 - \cos{(4x)} \right] - \cos{(2x)} \sin^2{(2x)} \, dx} \\ &= \frac{1}{16} \int{ 1\, dx} - \frac{1}{16} \int{ \cos{(4x)}\,dx} - \frac{1}{16} \int{ 2\cos{(2x)} \sin^2{(2x)} \, dx} \\ &= \frac{1}{16}x - \frac{1}{16} \left[ \frac{\sin{(4x)}}{4} \right] - \frac{1}{16} \left[ \frac{\sin^3{(2x)}}{3} \right] + C \\ &= \frac{x}{16} - \frac{\sin{(4x)}}{64} - \frac{\sin^3{(2x)}}{48} + C \end{align*}\)
Hello Paul,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!
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.
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!
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 $u$.![]()
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.
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.