What is the integral of sec^6(x)tan^2(x) ?

The integral of sec^6(x)tan^2(x) is (tan^3(x))/3+(2tan^5(x))/5+(tan^7(x))/7+C

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integral of sec^6(x)tan^2(x)

\int sec^{6} (x) tan^{2} (x) d x

\frac{tan ^{3} ( x )}{3} + \frac{2 tan ^{5} ( x )}{5} + \frac{tan ^{7} ( x )}{7} + C

Solution steps

\int sec^{6} (x) tan^{2} (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

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

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

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

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

Substitute back

u = tan (x)

= \frac{tan ^{3} ( x )}{3} + \frac{2 tan ^{5} ( x )}{5} + \frac{tan ^{7} ( x )}{7}

Add a constant to the solution

= \frac{tan ^{3} ( x )}{3} + \frac{2 tan ^{5} ( x )}{5} + \frac{tan ^{7} ( x )}{7} + C

x \to 2 lim (\frac{x ^{x} - 4}{x - 2})

\int \frac{1}{x \cdot ( x - 1 )} d x

\frac{\partial}{\partial x} (\frac{12 x}{y + 6})

\frac{ln ( 8 x )}{8 x}

8 x \cdot \frac{d y}{d x} = 2 x e^{x} - 8 y + 6 x^{2}

What is the integral of sec^6(x)tan^2(x) ?
The integral of sec^6(x)tan^2(x) is (tan^3(x))/3+(2tan^5(x))/5+(tan^7(x))/7+C