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Calculus Level 3

If $\cos x = \large{\frac{\sqrt{16 - \pi^2}}{4}}$ and $\sin x = a$ , when $a > 0$ , find the value of: $4\int{at^e}\, dt,$

where $e$ is a constant.

Clarification : $C$ denotes the arbitrary constant of integration .

\frac{\pi t^{e+1}}{e+1} + C

\frac{t^{e+1}}{e(\pi+1)} + C

\frac{{t^{e+1}}\sqrt{16 - \pi^2}}{4(e+1)} + C

\pi t^e + C

-\cos x \frac{\pi t^{e+1}}{e+1} + C

1 solution

Guilherme Niedu
Aug 5, 2016

$cos(x) = \frac{\sqrt{16 - \pi^2}}{4}$ then $sin(x) = \frac{\pi}{4} = a$

The integration will be equal to $\frac{4at^{e+1}}{e+1} + C$

Which, substituting a is:

$\frac{\pi t^{e+1}}{e+1} + C$