MHB Effie's question via email about an indefinite integral.

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The indefinite integral of 50t cos(5t²) with respect to t is calculated using substitution. By letting u = 5t², the differential du becomes 10t dt, simplifying the integral to 5 times the integral of cos(u) du. This results in the expression 5 sin(u) + C, which translates back to 5 sin(5t²) + C. Verifying the result through differentiation confirms that it matches the original function. The final answer is 5 sin(5t²) + C.
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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

$\displaystyle \begin{align*} 5\int{ 10\,t\cos{ \left( 5\,t^2 \right) } \,\mathrm{d}t } &= 5\int{ \cos{(u)}\,\mathrm{d}u } \\ &= 5\sin{(u)} + C \\ &= 5\sin{ \left( 5\,t^2 \right) } + C \end{align*}$
 
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And the easy check is to differentiate the result to see if you get back the original antiderivative.
 
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