What is the integral of sec^6(x)tan^5(x) ?

The integral of sec^6(x)tan^5(x) is (tan^6(x))/6+(tan^8(x))/4+(tan^{10}(x))/(10)+C

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integral of sec^6(x)tan^5(x)

\int sec^{6} (x) tan^{5} (x) d x

\frac{tan ^{6} ( x )}{6} + \frac{tan ^{8} ( x )}{4} + \frac{tan ^{10} ( x )}{10} + C

Solution steps

\int sec^{6} (x) tan^{5} (x) d x

Rewrite using trig identities

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

Apply u-substitution

: \int u^{5} (1 + u^{2})^{2} d u

= \int u^{5} (1 + u^{2})^{2} d u

Expand

u^{5} (1 + u^{2})^{2} : u^{5} + 2 u^{7} + u^{9}

= \int u^{5} + 2 u^{7} + u^{9} 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^{5} d u + \int 2 u^{7} d u + \int u^{9} d u

\int u^{5} d u = \frac{u ^{6}}{6}

\int 2 u^{7} d u = \frac{u ^{8}}{4}

\int u^{9} d u = \frac{u ^{10}}{10}

= \frac{u ^{6}}{6} + \frac{u ^{8}}{4} + \frac{u ^{10}}{10}

Substitute back

u = tan (x)

= \frac{tan ^{6} ( x )}{6} + \frac{tan ^{8} ( x )}{4} + \frac{tan ^{10} ( x )}{10}

Add a constant to the solution

= \frac{tan ^{6} ( x )}{6} + \frac{tan ^{8} ( x )}{4} + \frac{tan ^{10} ( x )}{10} + C

x \to 0 lim (\frac{x ^{2}}{2})

\frac{d}{d x} (\frac{( x ^{2} + 3 x - 4 )}{x + 2})

\frac{d}{d x} (- cos (x) - sin (x))

\frac{1}{y} y^{^{'}} = - \frac{1}{x}

y^{^{''}} + 3 y^{^{'}} + 2 y = sin^{2} (t)

What is the integral of sec^6(x)tan^5(x) ?
The integral of sec^6(x)tan^5(x) is (tan^6(x))/6+(tan^8(x))/4+(tan^{10}(x))/(10)+C