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

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

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

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

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

Solution steps

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

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} (t)

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

Add a constant to the solution

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

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