Indefinite integral $\int t \cdot \cos^3(t^2)dt$

Question

I am having trouble integrating $$\int t \cdot \cos^3(t^2)dt$$

Progress

I have made $u=t^2$ which makes the problem $1/2 \int \cos^3(u) du$.

After writing that out I subsituted $v=\sin(u)$ and got $1/2 \int (1-v^2)dv $ ... then I integrated and got $\frac12[v-v^3/3]$, after that I plugged in $v$ which would make it $\frac12 [\sin(u)-\sin^3(u)/3]$.

Is this correct so far, and what else should be done?

Answer 1

$$\int t\cos^3(t^2)\,dt= \frac{1}{2}\int\cos^3(u)\,du= \frac{1}{2}\int\cos^2(u)\,d\big(\sin(u)\big)= \frac{1}{2}\int\big(1-\sin^2(u)\big)\,d\big(\sin(u)\big)= \frac{1}{2}\Big(\int\,d\big(\sin(u)\big)-\int\sin^2(u)\,d\big(\sin(u)\big)\Big)= \frac{1}{2}\Big(\sin(u)-\frac{1}{3}\sin^3(u)\Big)+c$$

subsitute $u=t^2$ to get $$\frac{1}{2}\Big(\sin(t^2)-\frac{1}{3}\sin^3(t^2)\Big)+c$$