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?

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}
