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

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?

• Hint: there is quite an obvious substitution which might help – Mark Bennet Oct 4 '14 at 22:24
• I have made u=t^2 which makes the problem 1/2 integral cos^3(u) du – Zak Oct 4 '14 at 22:26
• Well if you've done that, explain it in the question, because it always helps us to know what progress you've made. – Mark Bennet Oct 4 '14 at 22:28
• Next write $\cos^{3}u=\cos^{2}u\cos u=(1-\sin^{2}u)\cos u$ and substitute again. – user84413 Oct 4 '14 at 22:33
• You just integrated, so you're done. Just substitute $u$ back in and don't forget your constant. – Andrey Kaipov Oct 4 '14 at 22:44

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