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

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

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integral of sec^5(θ)tan^3(θ)

\int sec^{5} (θ) tan^{3} (θ) d θ

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

Solution steps

\int sec^{5} (θ) tan^{3} (θ) d θ

Rewrite using trig identities

= \int (- 1 + sec^{2} (θ)) tan (θ) sec^{5} (θ) d θ

Apply u-substitution

: \int u^{4} (u^{2} - 1) d u

= \int u^{4} (u^{2} - 1) d u

Expand

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

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

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

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

= \frac{u ^{7}}{7} - \frac{u ^{5}}{5}

Substitute back

u = sec (θ)

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

Add a constant to the solution

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

\int x sin (2 x^{2} - 5) d x

\int sin^{2} (x) + sec^{2} (5 x) d x

\int x 36 - x^{4} d x

\int 5 sin^{5} (x) cos^{4} (x) d x

\int x^{5} e^{5 x} d x

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