What is the integral of (tan^3(x))/(cos^3(x)) ?

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

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integral of (tan^3(x))/(cos^3(x))

\int \frac{tan ^{3} ( x )}{cos ^{3} ( x )} d x

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

Solution steps

\int \frac{tan ^{3} ( x )}{cos ^{3} ( x )} d x

Rewrite using trig identities

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

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 (x)

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

Add a constant to the solution

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

\int sin (cos (x)) sin (x) d x

\frac{\partial}{\partial u} (u v^{2})

\frac{d}{d x} (\frac{7}{x - 3})

- 3 cos (3 x)

f (x) = \frac{x ^{2}}{x ^{3} + 1}

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