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17h
accepted Why is $\cos\left(\frac{3\pi}{2}-t+2k\pi\right) = -\sin(t)$
17h
comment Why is $\cos\left(\frac{3\pi}{2}-t+2k\pi\right) = -\sin(t)$
Thanks. Love it.
17h
comment Why is $\cos\left(\frac{3\pi}{2}-t+2k\pi\right) = -\sin(t)$
Ok, I think even/odd doesn't matter, you're just using the phase shift definition to do that last step.
17h
comment Why is $\cos\left(\frac{3\pi}{2}-t+2k\pi\right) = -\sin(t)$
I also know that cos(x) is even, and sin(x) is odd. So cos(-x) = cos(x) and sin(-x) = -sin(x). Does that come into play?
17h
comment Why is $\cos\left(\frac{3\pi}{2}-t+2k\pi\right) = -\sin(t)$
Not seeing what you did on the last step. I know $cos(x) = sin(x+\frac{\pi}{2})$ and $sin(x)=cos(x-\frac{\pi}{2})$
18h
asked Why is $\cos\left(\frac{3\pi}{2}-t+2k\pi\right) = -\sin(t)$
Feb
1
accepted Setting up this integral?
Feb
1
asked Setting up this integral?
Jan
22
comment Is this a $u$-substitution? Weird integral.
Can you demonstrate your method with a basic simple example?
Jan
22
comment Is this a $u$-substitution? Weird integral.
What is the same? Can you do your method on the example I gave above, or better yet, a simpler one?
Jan
22
comment Is this a $u$-substitution? Weird integral.
I've never seen x used as a substitution when x is already in the equation. The simplest traditional u-sub I can think of would be: $y=\int\frac{1}{1+x}$ You would let $u=1+x$ and $du=dx$. Then you would have $\int\frac{1}{u}du$ I have never seen a problem go from $y=\frac{1}{1+x}$ to something like $x=......$ the most unnatural thing I've encountered in Calculus, so far.
Jan
22
comment Is this a $u$-substitution? Weird integral.
I've never seen x used as a substitution when x is already in the equation. The simplest traditional u-sub I can think of would be: $y=\int\frac{1}{1+x}$ You would let $u=1+x$ and $du=dx$. Then you would have $\int\frac{1}{u}du$ I have never seen a problem go from $y=\frac{1}{1+x}$ to something like $x=......$ the most unnatural thing I've encountered in Calculus, so far.
Jan
22
comment Is this a $u$-substitution? Weird integral.
I've never seen x used as a substitution when x is already in the equation. The simplest traditional u-sub I can think of would be: $y=\int\frac{1}{1+x}$ You would let $u=1+x$ and $du=dx$. Then you would have $\int\frac{1}{u}du$ I have never seen a problem go from $y=\frac{1}{1+x}$ to something like $x=......$ the most unnatural thing I've encountered in Calculus, so far.
Jan
22
comment Is this a $u$-substitution? Weird integral.
What does "right to left" mean?
Jan
22
comment Is this a $u$-substitution? Weird integral.
What is this method called? I want to look up some tutorials and sample examples that are not as complex.
Jan
22
comment Is this a $u$-substitution? Weird integral.
What is this method called? I want to look up some tutorials and sample examples that are not as complex.
Jan
22
comment Is this a $u$-substitution? Weird integral.
What is this method called? I want to look up some tutorials and sample examples that are not as complex.
Jan
22
comment Is this a $u$-substitution? Weird integral.
I have no idea what you wrote. Are you doing some sort of reverse chain rule? What is this method called? I want to look up some tutorials and sample examples that are not as complex.
Jan
22
comment Is this a $u$-substitution? Weird integral.
The only composition of functions I can see is $f(x)=\frac{1}{x^2} ,,,and,,, g(x)=1+x^2$
Jan
22
comment Is this a $u$-substitution? Weird integral.
There is absolutely nothing natural or intuitive about whatever they did. U-sub is as natural as breathing, but this thing, I have zero clue what I am looking at.