What is the integral of tan^2(θ)sec^4(θ) ?

The integral of tan^2(θ)sec^4(θ) is (tan^3(θ))/3+(tan^5(θ))/5+C

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integral of tan^2(θ)sec^4(θ)

\int tan^{2} (θ) sec^{4} (θ) d θ

\frac{tan ^{3} ( θ )}{3} + \frac{tan ^{5} ( θ )}{5} + C

Solution steps

\int tan^{2} (θ) sec^{4} (θ) d θ

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

u^{2} (1 + u^{2}) : u^{2} + u^{4}

= \int u^{2} + u^{4} 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 u^{2} d u + \int u^{4} d u

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

\int u^{4} d u = \frac{u ^{5}}{5}

= \frac{u ^{3}}{3} + \frac{u ^{5}}{5}

Substitute back

u = tan (θ)

= \frac{tan ^{3} ( θ )}{3} + \frac{tan ^{5} ( θ )}{5}

Add a constant to the solution

= \frac{tan ^{3} ( θ )}{3} + \frac{tan ^{5} ( θ )}{5} + C

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What is the integral of tan^2(θ)sec^4(θ) ?
The integral of tan^2(θ)sec^4(θ) is (tan^3(θ))/3+(tan^5(θ))/5+C