- #1
Prove It
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What is the indefinite integral (with respect to t) of $\displaystyle \begin{align*} 50\,t\cos{ \left( 5\,t^2 \right) } \end{align*}$?
$\displaystyle \begin{align*} \int{ 50\,t\cos{\left( 5\,t^2 \right) } \,\mathrm{d}t } &= 5\int{ 10\,t\cos{ \left( 5\,t^2 \right) }\,\mathrm{d}t } \end{align*}$
Let $\displaystyle \begin{align*} u = 5\,t^2 \implies \mathrm{d}u = 10\,t\,\mathrm{d}t \end{align*}$ and the integral becomes
$\displaystyle \begin{align*} 5\int{ 10\,t\cos{ \left( 5\,t^2 \right) } \,\mathrm{d}t } &= 5\int{ \cos{(u)}\,\mathrm{d}u } \\ &= 5\sin{(u)} + C \\ &= 5\sin{ \left( 5\,t^2 \right) } + C \end{align*}$
$\displaystyle \begin{align*} \int{ 50\,t\cos{\left( 5\,t^2 \right) } \,\mathrm{d}t } &= 5\int{ 10\,t\cos{ \left( 5\,t^2 \right) }\,\mathrm{d}t } \end{align*}$
Let $\displaystyle \begin{align*} u = 5\,t^2 \implies \mathrm{d}u = 10\,t\,\mathrm{d}t \end{align*}$ and the integral becomes
$\displaystyle \begin{align*} 5\int{ 10\,t\cos{ \left( 5\,t^2 \right) } \,\mathrm{d}t } &= 5\int{ \cos{(u)}\,\mathrm{d}u } \\ &= 5\sin{(u)} + C \\ &= 5\sin{ \left( 5\,t^2 \right) } + C \end{align*}$