What is the integral of tan^5(θ)sec^3(θ) ?

The integral of tan^5(θ)sec^3(θ) is (sec^7(θ))/7-(2sec^5(θ))/5+(sec^3(θ))/3+C

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integral of tan^5(θ)sec^3(θ)

\int tan^{5} (θ) sec^{3} (θ) d θ

\frac{sec ^{7} ( θ )}{7} - \frac{2 sec ^{5} ( θ )}{5} + \frac{sec ^{3} ( θ )}{3} + C

Solution steps

\int tan^{5} (θ) sec^{3} (θ) d θ

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

u^{2} (u^{2} - 1)^{2} : u^{6} - 2 u^{4} + u^{2}

= \int u^{6} - 2 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^{6} d u - \int 2 u^{4} d u + \int u^{2} d u

\int u^{6} d u = \frac{u ^{7}}{7}

\int 2 u^{4} d u = \frac{2 u ^{5}}{5}

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

= \frac{u ^{7}}{7} - \frac{2 u ^{5}}{5} + \frac{u ^{3}}{3}

Substitute back

u = sec (θ)

= \frac{sec ^{7} ( θ )}{7} - \frac{2 sec ^{5} ( θ )}{5} + \frac{sec ^{3} ( θ )}{3}

Add a constant to the solution

= \frac{sec ^{7} ( θ )}{7} - \frac{2 sec ^{5} ( θ )}{5} + \frac{sec ^{3} ( θ )}{3} + C

\frac{\partial}{\partial x} (x^{3} z + z^{3} x - 2 yz)

f (x) = x^{2} x^{3} + 1, at x = 1

x \to 1 lim (x^{2} + 3 x - 2)

\int_{0}^{1} x \cdot e^{- x} d x

\int_{0}^{1} (e^{- x}) d x

What is the integral of tan^5(θ)sec^3(θ) ?
The integral of tan^5(θ)sec^3(θ) is (sec^7(θ))/7-(2sec^5(θ))/5+(sec^3(θ))/3+C