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

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

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integral of tan^3(θ)sec^3(θ)

\int tan^{3} (θ) sec^{3} (θ) d θ

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

Solution steps

\int tan^{3} (θ) sec^{3} (θ) d θ

Rewrite using trig identities

= \int (- 1 + sec^{2} (θ)) tan (θ) sec^{3} (θ) d θ

Apply u-substitution

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

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

Expand

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

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

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

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

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

Substitute back

u = sec (θ)

= \frac{sec ^{5} ( θ )}{5} - \frac{sec ^{3} ( θ )}{3}

Add a constant to the solution

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

y^{^{'}} = - \frac{2 x}{( 1 + x ^{2} )} y + \frac{1}{x ( 1 + x ^{2} )}

\int_{- 1}^{1} x^{2} - x - 1 d x

(\frac{4}{3})^{^{'}}

\frac{d}{d x} (ln (5 x^{2} + 3 y^{2}))

(1 + e^{2 x})^{^{'}}

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