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

The integral of sec^3(x)tan(x) is 1/3 sec^3(x)+C

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integral of sec^3(x)tan(x)

\int sec^{3} (x) tan (x) d x

\frac{1}{3} sec^{3} (x) + C

Solution steps

\int sec^{3} (x) tan (x) d x

Apply u-substitution

: \int \frac{1}{3} d u

= \int \frac{1}{3} d u

Integral of a constant:

\int a d x = a x

= \frac{1}{3} u

Substitute back

u = sec^{3} (x)

= \frac{1}{3} sec^{3} (x)

Add a constant to the solution

= \frac{1}{3} sec^{3} (x) + C

x \to (- \frac{π}{2}) - lim (sec (x))

\frac{\partial}{\partial x} (6 x^{2} + 3 y^{2} + x y)

7 x + 1 (9.8)

f^{^{'}} (x) = e^{x^{2}}

\int \frac{1}{3 x + 4} d x

What is the integral of sec^3(x)tan(x) ?
The integral of sec^3(x)tan(x) is 1/3 sec^3(x)+C