What is the integral of 6sec^2(x)-9x^3 ?

The integral of 6sec^2(x)-9x^3 is 6tan(x)-(9x^4)/4+C

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integral of 6sec^2(x)-9x^3

\int 6 sec^{2} (x) - 9 x^{3} d x

6 tan (x) - \frac{9 x ^{4}}{4} + C

Solution steps

\int 6 sec^{2} (x) - 9 x^{3} d x

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 sec^{2} (x) d x - \int 9 x^{3} d x

\int 6 sec^{2} (x) d x = 6 tan (x)

\int 9 x^{3} d x = \frac{9 x ^{4}}{4}

= 6 tan (x) - \frac{9 x ^{4}}{4}

Add a constant to the solution

= 6 tan (x) - \frac{9 x ^{4}}{4} + C

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What is the integral of 6sec^2(x)-9x^3 ?
The integral of 6sec^2(x)-9x^3 is 6tan(x)-(9x^4)/4+C