What is the integral of sec^6(x)tan^4(x) ?

The integral of sec^6(x)tan^4(x) is (tan^5(x))/5+(2tan^7(x))/7+(tan^9(x))/9+C

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integral of sec^6(x)tan^4(x)

\int sec^{6} (x) tan^{4} (x) d x

\frac{tan ^{5} ( x )}{5} + \frac{2 tan ^{7} ( x )}{7} + \frac{tan ^{9} ( x )}{9} + C

Solution steps

\int sec^{6} (x) tan^{4} (x) d x

Rewrite using trig identities

= \int sec^{2} (x) (1 + tan^{2} (x))^{2} tan^{4} (x) d x

Apply u-substitution

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

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

Expand

u^{4} (1 + u^{2})^{2} : u^{4} + 2 u^{6} + u^{8}

= \int u^{4} + 2 u^{6} + u^{8} 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 2 u^{6} d u + \int u^{8} d u

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

\int 2 u^{6} d u = \frac{2 u ^{7}}{7}

\int u^{8} d u = \frac{u ^{9}}{9}

= \frac{u ^{5}}{5} + \frac{2 u ^{7}}{7} + \frac{u ^{9}}{9}

Substitute back

u = tan (x)

= \frac{tan ^{5} ( x )}{5} + \frac{2 tan ^{7} ( x )}{7} + \frac{tan ^{9} ( x )}{9}

Add a constant to the solution

= \frac{tan ^{5} ( x )}{5} + \frac{2 tan ^{7} ( x )}{7} + \frac{tan ^{9} ( x )}{9} + C

\int - x^{- 1} d x

\int (1 + \frac{1}{x})^{3} (\frac{1}{x ^{2}}) d x

x \to \frac{π}{6} lim (sin (2 x - π))

\int x^{2} - 9 x + 5 d x

\frac{d}{d t} (e^{t} (a cos (2 t) + b sin (2 t)))

What is the integral of sec^6(x)tan^4(x) ?
The integral of sec^6(x)tan^4(x) is (tan^5(x))/5+(2tan^7(x))/7+(tan^9(x))/9+C