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

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

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

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

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

Solution steps

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

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

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

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

Add a constant to the solution

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

\int \frac{( x ^{3} )}{x ^{2} + 1} d x

\int t sec^{2} (t) d t

\int - y cos (x y) d x

\int \frac{- cos ( x )}{sin ^{3} ( x )} d x

\int \frac{e ^{x^{2}}}{x ^{3}} d x

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