What is the integral of 6csc^2(t)-9e^t ?

The integral of 6csc^2(t)-9e^t is -6cot(t)-9e^t+C

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integral of 6csc^2(t)-9e^t

\int 6 csc^{2} (t) - 9 e^{t} d t

- 6 cot (t) - 9 e^{t} + C

Solution steps

\int 6 csc^{2} (t) - 9 e^{t} d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 6 csc^{2} (t) d t - \int 9 e^{t} d t

\int 6 csc^{2} (t) d t = - 6 cot (t)

\int 9 e^{t} d t = 9 e^{t}

= - 6 cot (t) - 9 e^{t}

Add a constant to the solution

= - 6 cot (t) - 9 e^{t} + C

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\int \frac{9 x ^{2}}{2} d x

\int \frac{3 x ^{2} - 7 x - 24}{x - 2} d x

What is the integral of 6csc^2(t)-9e^t ?
The integral of 6csc^2(t)-9e^t is -6cot(t)-9e^t+C