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

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

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

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

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

Solution steps

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

Rewrite using trig identities

= \int (1 + tan^{2} (x)) sec^{2} (x) tan^{6} (x) d x

Apply u-substitution

: \int u^{6} (1 + u^{2}) d u

= \int u^{6} (1 + u^{2}) d u

Expand

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

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

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

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

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

Substitute back

u = tan (x)

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

Add a constant to the solution

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

t^{2} (\frac{d y}{d t}) + y = 0

\frac{\partial}{\partial y} (y e^{z})

\frac{d y}{d x} + x y = x y^{- 3}

x \to - 4 π lim (1 + csc (x))

x^{2} - 6 x, - x^{2} + 2 x

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