What is the integral of sec^4(t) ?

The integral of sec^4(t) is tan(t)+(tan^3(t))/3+C

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integral of sec^4(t)

\int sec^{4} (t) d t

tan (t) + \frac{tan ^{3} ( t )}{3} + C

Solution steps

\int sec^{4} (t) d t

Rewrite using trig identities

= \int (1 + tan^{2} (t)) sec^{2} (t) d t

Apply u-substitution

: \int 1 + u^{2} d u

= \int 1 + u^{2} d u

Apply the Sum Rule:

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

= \int 1 d u + \int u^{2} d u

\int 1 d u = u

\int u^{2} d u = \frac{u ^{3}}{3}

= u + \frac{u ^{3}}{3}

Substitute back

u = tan (t)

= tan (t) + \frac{tan ^{3} ( t )}{3}

Add a constant to the solution

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

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