Prove that $\lim_{t \to \infty} \int_1^t \sin(x)\sin(x^2)\,dx$ converges

You can write $$ \sin(x) \sin(x^2) = \frac{1}{2} \left[\cos(x^2-x) - \cos(x^2+x)\right]$$ and let $u = x^2-x$, $v=x^2+x$ to obtain \begin{align} \int \limits_1^t \sin(x) \sin(x^2) \, \mathrm{d} x &= \frac{1}{2} \int \limits_1^t \left[\cos(x^2-x) - \cos(x^2+x)\right] \, \mathrm{d} x \\ &= \frac{1}{4} \left[\int \limits_0^{t^2-t} \frac{\cos(u)}{\sqrt{u + \frac{1}{4}}} \, \mathrm{d}u - \int \limits_2^{t^2+t} \frac{\cos(v)}{\sqrt{v + \frac{1}{4}}} \, \mathrm{d}v\right] . \end{align} The convergence of this expression as $t \to \infty$ is ensured by Dirichlet's test for integrals or integration by parts.

This is an old Putnam problem [2000, A4]: show that $\displaystyle{\lim_{B\to \infty}\int _0^B \sin(x)\sin(x^2)\,dx}$ exists.

Since we are interested in the limit, we can assume $B>1$. To simplify matters, we introduce a factor of 4. $$ \lim_{B\to \infty}2\int _0^B \sin(x)\cdot 2\sin(x^2)\,dx $$$\sin(x^2)$ doesn't have an elementary antiderivative, but $2x\sin(x^2)$ does. We multiply and divide by $x$ to introduce this factor $$ \lim_{B\to \infty}2\int _0^B\frac{ \sin(x)}{x}\cdot2 x\sin(x^2)\,dx $$Let's use IBP to trade one integral for another. Let $u=\frac{\sin(x)}{x}$, $dv=2x\sin(x^2)dx$: $$ 2\int _0^B\frac{ \sin(x)}{x}\cdot2 x\sin(x^2)\,dx = \left.-2\cos(x^2)\frac{\sin(x)}{x}\right|_0^B + \int _0^B 2\cos(x^2)\cdot \frac{x \cos (x)-\sin (x)}{x^2}\,dx $$The boundary term evaluates to 2 as $B\to\infty$. Now we use that $B>1$ to split up the new integral: $$ =2 + 2\int _0^1 \cos(x^2)\cdot \frac{x \cos (x)-\sin (x)}{x^2}\,dx+ \int _1^B 2\cos(x^2)\cdot \frac{x \cos (x)-\sin (x)}{x^2}\,dx $$For the first integral, note that cosine is positive and continuous on $[0,1]$. Then by the Mean Value Theorem for Integrals, for some $\xi\in(0,1)$ we have $$ \int _0^1 2\cos(x^2)\cdot \frac{x \cos (x)-\sin (x)}{x^2}\,dx = 2\cos(\xi^2)\int _0^1 \frac{x \cos (x)-\sin (x)}{x^2}\,dx = 2\cos(\xi^2) \left.\frac{\sin(x)}{x}\right|_0^1 $$ $$ = 2\cos(\xi^2)\left(\frac{\sin(1)}{1}-1\right); $$this is a finite number. The presence of $x^2$ in the denominator of the second integral is a good sign: we know that $x^{-2}$ is improperly integrable on $[1,\infty).$ We have to tweak things a little but the goal is to get a function comparable with $x^{-2}$ on $[1,B]$. Now we reintroduce a factor of $x/x$ as earlier and use IBP again: $$ \int _1^B 2x\cos(x^2)\cdot \frac{x \cos (x)-\sin (x)}{x^3}\,dx $$ $$ = \left.\sin(x^2)\cdot \frac{x \cos (x)-\sin (x)}{x^3} \right|_1^B - \int_1^B \sin(x^2)\cdot \frac{-x^2\sin (x)+3 \sin (x)-3 x \cos (x)}{x^4}\,dx $$The boundary terms are finite in the limit. This new integral has the form we want: it is roughly of the form $x^{-2}$. We now use several basic comparisons to show it converges as $B\to \infty$. First, take absolute values: $$ \left|\int_1^B \sin(x^2)\cdot \frac{-x^2\sin (x)+3 \sin (x)-3 x \cos (x)}{x^4}\,dx\right| $$ $$ \leq \int_1^B\left| \sin(x^2)\cdot \frac{-x^2\sin (x)+3 \sin (x)-3 x \cos (x)}{x^4}\right|\,dx $$ $|\sin(\theta)|\leq1$ for any real $\theta$, and dividing the numerator and denominator by $x^2$ gives $$ \leq \int_1^B1\cdot\left| \frac{-x^2\sin (x)+3 \sin (x)-3 x \cos (x)}{x^4}\right|\,dx $$ $$ = \int_1^B\frac{\left|-\sin (x)+3 \sin (x)/x^2-3 \cos (x)/x\right|}{x^2}\,dx $$ $$ \leq \int_1^B\frac{1+3+3}{x^2}\,dx=\int_1^B\frac{7}{x^2}\,dx $$This last integral is finite. The original integral is bounded by a number of finite pieces, hence is finite.