What is the integral of (tan^2(x))/(sec^5(x)) ?

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

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integral of (tan^2(x))/(sec^5(x))

\int \frac{tan ^{2} ( x )}{sec ^{5} ( x )} d x

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

Solution steps

\int \frac{tan ^{2} ( x )}{sec ^{5} ( x )} d x

Rewrite using trig identities

= \int (1 - sin^{2} (x)) cos (x) sin^{2} (x) d x

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 = sin (x)

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

Add a constant to the solution

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

\int \frac{w}{16 - w ^{2}} d w

\frac{d}{d x} (\frac{x}{x ^{2} + 7})

x \to 0 lim (7 cot^{4} (3 x))

\int \frac{cos ( x )}{x} d x

f (x) = 5 (7 + x)

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