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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?

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

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

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