I tried to solve the integral below using integration by parts $$\int_0^t\cos(x)\cos(t-x)dx=\frac{1}{2}(\sin(t)+t\cos(t))$$
It seemed solvable through doing integration by parts twice, but it hasn't worked for me yet... tcos(t) doesn't come up!
I know how it can be solved using properties of trig function, why can't it be solved by integration by parts?
On
Hint: use the formula
$$\cos(x)\cos(y)=\frac{1}{2}(\cos(x-y)+\cos(x+y))$$ for your Control: $$1/2\,\sin \left( t \right) +1/2\,\cos \left( t \right) t$$
On
I’m going to guess you haven’t learned about Laplace transforms or convolutions, as you didn’t recognize it in this case, but the form of the integral is such that it is
$$\cos(t)*\cos(t)$$
Where $*$ is the convolution operator
In such case the answer is obtained by
$$\mathcal{L}^{-1}\left(\left({\frac{s}{s^2+1}}\right)^2\right)$$
$$=\frac{1}{2}(\sin(t)+t\cos(t))$$
On
First you can observe that
$ \cos(x)\cos(t-x)-\sin(x)\sin(t-x)=$
$=\cos(x+(t-x))=\cos(t)$
so
$\cos(x)\cos(t-x)=\cos(t)+\sin(x)\sin(t-x)$
Then
$\int_0^t \cos(x)\cos(t-x)dx =$
$=\int_0^t \cos(t)dx +\int_0^t \sin(x)\sin(t-x)dx=$
and using integration by parts you have that
$=t \cos(t)+[\sin(x)\cos(t-x)]_0^t-\int_0^t \cos(x)\cos(t-x)dx=$
$=t \cos(t)+\sin(t)-\int_0^t \cos(x)\cos(t-x)dx$
so
$2\int_0^t \cos(x)\cos(t-x)dx =t \cos(t)+\sin(t)$
and
$\int_0^t \cos(x)\cos(t-x)dx =\frac{1}{2}(t \cos(t)+\sin(t))$
On
$$ \begin{align} \int_o^t{\cos(x)\cos(t-x)dx} &= \displaystyle{ 1 \over 2}\int_o^t[\cos(2x-t)+\cos(t)]dx \\ &= \displaystyle{ 1 \over 2}t\cos(t) + \displaystyle{ 1 \over 2} \int_o^t{\cos(2x-t)}dx \\ &= \displaystyle{ 1 \over 2}t\cos(t) + \displaystyle{ 1 \over 4}\sin(2x-t)\Big|_{0}^{t} \\ & =\displaystyle{ 1 \over 2}(\sin(t)+t\cos(t)) \end{align} $$
You have the last equality since $\sin(-t)=-\sin(t)$
You can do it by integration by parts, but you have to start off in a slightly funny direction to stop the final integral cancelling with the original: $$ I = \int_0^t 1 \cdot \cos{x}\cos{(t-x)} \, dx = [x\cos{x}\cos{(t-x)}]_0^t - \int_0^t x(-\sin{x}\cos{(t-x)}+\cos{x}\sin{(t-x)}) \, dx \\ = t\cos{t} + \int_0^t x(\sin{x}\cos{(t-x)}-\cos{x}\sin{(t-x)}) \, dx, $$ giving something that looks like one of the terms. Now integrate the first term in the integral by parts, $$ \int_0^t x\sin{x}\cdot \cos{(t-x)} \, dx = [-x\sin{x}\sin{(t-x)}]_0^t + \int_0^t (x\cos{x}-\sin{x}) \sin{(t-x)} \, dx. $$ The first term in the remaining integral cancels with the second term in the integral from the first integration by parts, so $$ \int_0^t \cos{x}\cos{(t-x)} \, dx = t\cos{t} - \int_0^t \sin{x}\sin{(t-x)} \, dx. $$ To deal with the remaining integral we integrate by parts again: $$ - \int_0^t \sin{x}\sin{(t-x)} \, dx = [\cos{x}\sin{(t-x)}]_0^t-\int_0^t \cos{x}\cos{(t-x)} \, dx \\ = \sin{t}-I, $$ so $$ I = t\cos{t}+\sin{t}-I, $$ or $$ I = \frac{1}{2}(t\cos{t}+\sin{t}), $$ as expected.
It was not at all obvious that this would work. In general, $$ \int fg' \, dx = xfg' - \int x(f'g'+fg'') \, dx \\ = x(fg' - f'g) + \int [(xf''+f')g-xfg''] \, dx \\ = x(fg'-f'g) + fg-\int fg' \, dx + \int x(f''g-fg'') \, dx, $$ so $$ \int fg' \, dx = \frac{1}{2}(x(fg'-f'g)+fg)+\frac{1}{2}\int x(f''g-fg'') \, dx, $$ and it so happens that in this case the integral on the right vanishes (as it will for any two solutions to a differential equation of the form $y''+q(x)y=0$, since $f''g-fg''$ is the derivative of the Wronskian, and this vanishes since the coefficient of $y'$ is zero).
A way to do this without a strange integration by parts is to look at $$ \lim_{a \to 1} \int_0^t \cos{ax}\cos{(t-x)} \, dx: $$ the integral and the limit can be interchanged because everything is continuous and the limit function is continuous, but $\int_0^t \cos{ax}\cos{(t-x)} \, dx$ can be done by parts in the usual way: we find $$ \int_0^t \cos{ax}\cos{(t-x)} \, dx = \dotsb = \sin{t}-a\sin{t} + \int_0^t \cos{ax}\cos{(t-x)} \, dx, $$ so $$ \int_0^t \cos{ax}\cos{(t-x)} \, dx = \frac{\sin{t}+a\sin{at}}{1-a^2}, $$ and taking the limit gives the result (although will require some trig identities).