What is the integral of sec^4(θ) ?

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

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integral of sec^4(θ)

\int sec^{4} (θ) d θ

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

Solution steps

\int sec^{4} (θ) d θ

Rewrite using trig identities

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

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 (θ)

= tan (θ) + \frac{tan ^{3} ( θ )}{3}

Add a constant to the solution

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

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