Solve $\int_{0}^{t}\cos x\sin (t-x)dx$ I want to solve the following integral $$\int_{0}^{t}\cos x\sin (t-x)dx$$
What I tried to do is integration by parts twice, but it just brought me straight back to $$\int_{0}^{t}\cos x\sin (t-x)dx=\int_{0}^{t}\cos x\sin (t-x)dx$$ Not very informative indeed.
Is there any way to do this without resorting to trigonometric identities?
 A: Let $x \mapsto t-x$ then 
$\displaystyle I = \int_0^t \cos{x}\sin(t-x) \;{dx} = \int_0^t \sin{x}\cos(t-x) \;{dx} $  
So $ \displaystyle  2I = \int_0^t [\cos{x}\sin(t-x)+\sin{x}\cos(t-x)]\;{dx}$
But $ \displaystyle \cos{x}\sin(t-x)+\sin{x}\cos(t-x)  = \sin(x+t-x) = \sin(t)$.
So $I \displaystyle = \frac{1}{2}\int_0^t \sin{x}\;{dx} = \frac{1}{2}(t\sin{t}).$
A: Hint: $$\sin(t - x) \cos x = \frac{1}{2}\Big(\sin t + \sin (t - 2x)\Big)$$
A: There is a nice symmetry trick. We have:
$$ I = \int_{-t/2}^{t/2}\cos(x+t/2)\sin(t/2-x)\,dx\\ = \int_{0}^{t/2}\left(\cos(t/2+x)\sin(t/2-x)+\cos(t/2-x)\sin(t/2+x)\right)\,dx\\ =\int_{0}^{t/2}\sin(t/2+x+t/2-x)\,dx\\=\int_{0}^{t/2}\sin t\,dx\\=\color{red}{\frac{t\sin t}{2}}.$$
A: I think the best solution is to use the trigonometric identities. Even though, if you really want to avoid those, here is a way :
Note that : $\displaystyle \sin(t-x) = \mathrm{Im} \Big( e^{i(t-x)} \Big) = \mathrm{Im} \Big( e^{it} e^{-ix} \Big)$. Therefore,
$$
\begin{align*}
\int_{0}^{t} \cos(x) \sin(t-x) \, dx &= {} \mathrm{Im} \Bigg( \int_{0}^{t} \cos(x) e^{it} e^{-ix} \, dx \Bigg) \\[3mm]
 &= \mathrm{Im} \Bigg( e^{it} \int_{0}^{t} \cos(x) e^{-ix} \, dx \Bigg) \\
\end{align*}$$
But, for all $x \in \mathbb{R}$, $\displaystyle \cos(x) = \frac{e^{ix}+e^{-ix}}{2}$. As a consequence :
$$
\begin{align*}
e^{it} \int_{0}^{t} \cos(x) e^{-ix} \, dx &= {} e^{it} \int_{0}^{t} \Big( \frac{e^{ix}+e^{-ix}}{2} \Big) e^{-ix} \, dx \\[2mm] 
 &= e^{it} \int_{0}^{t} \frac{1}{2} \, dx + e^{it} \int_{0}^{t} \frac{e^{-2ix}}{2} \, dx \\[2mm]
 &= \frac{te^{it}}{2} + \frac{1}{i} e^{it} \big( 1 - e^{-2it} \big) \\[2mm]
 &= \frac{te^{it}}{2} - ie^{it} + ie^{-it}. 
\end{align*}
$$
It follows that :
$$ \int_{0}^{t} \cos(x) \sin(t-x) \, dx = \frac{t\sin(t)}{2}. $$
A: If you don't wish to use trig, you could try solving the integral via Laplace transforms. In particular, the Laplace transform of your integral will look like the derivative of the transform of $\sin t$. Of course, this itself requires knowing Laplace identities... it's probably easier to just remember the basic trig identities.
A: We have
\begin{align}
\int_0^t\cos\left(x\right)\sin\left(t-x\right)\:dx,\tag{1}
\end{align}
and since $\sin\left(t-x\right)=\sin\left(t\right)\cos\left(x\right)-\sin\left(x\right)\cos\left(t\right)$,we may re-write this integral as
\begin{align}
&\int_0^t \cos\left(x\right)\left[\sin\left(t\right)\cos\left(x\right)-\sin\left(x\right)\cos\left(t\right)\right]\:dx\tag{2}\\
&=\sin\left(t\right)\int_0^t \cos^2\left(x\right)\:dx-\cos\left(t\right)\int_0^t\sin\left(x\right)\cos\left(x\right)\:dx,\tag{3}
\end{align}
and because of the reduction formula
\begin{align}
\int\cos^{n}\left(x\right)\:dx=\frac{\cos^{n-1}\left(x\right)\sin\left(x\right)}{n}+\frac{n-1}{n}\int\cos^{n-2}\left(x\right)\:dx,
\end{align}
we have
\begin{align}
&=\sin\left(t\right)\left[\frac{\cos\left(x\right)\sin\left(x\right)}{2}\bigg|_0^t+\frac{1}{2}\int_0^t\:dx\right]-\cos\left(t\right)\left[\frac{1}{2}\sin^2\left(x\right)\right]_0^t\tag{4}\\
&=\frac{\cos\left(t\right)\sin^2\left(t\right)}{2}+\frac{\sin\left(t\right)t}{2}-\frac{\cos\left(t\right)\sin^2\left(t\right)}{2}\tag{5}\\
&=\boxed{\frac{t\sin\left(t\right)}{2}.}\tag{6}
\end{align}

Another way to look at this problem is via convolution such that
\begin{align}
\left(f\ast g\right)\left(t\right)=\int_0^t f\left(t-x\right)g\left(x\right)\:dx,\tag{7}
\end{align}
and in this case our $f\left(t-x\right)=\sin\left(t-x\right)$ and $g\left(x\right)=\cos\left(x\right)$. Therefore, this is simply
\begin{align}
\sin\left(x\right)\ast\cos\left(x\right)=\cos\left(x\right)\ast\sin\left(x\right)=\int_0^t\sin\left(t-x\right)\cos\left(x\right)\:dx\\=\int_0^t \cos\left(t-x\right)\sin\left(x\right)\:dx.\tag{8}
\end{align}
