What is the integral of tan^8(7x)sec^4(7x) ?

The integral of tan^8(7x)sec^4(7x) is 1/7 ((tan^9(7x))/9+(tan^{11}(7x))/(11))+C

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integral of tan^8(7x)sec^4(7x)

\int tan^{8} (7 x) sec^{4} (7 x) d x

\frac{1}{7} (\frac{tan ^{9} ( 7 x )}{9} + \frac{tan ^{11} ( 7 x )}{11}) + C

Solution steps

\int tan^{8} (7 x) sec^{4} (7 x) d x

Apply u-substitution

: \int tan^{8} (u) sec^{4} (u) \frac{1}{7} d u

= \int tan^{8} (u) sec^{4} (u) \frac{1}{7} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{7} \cdot \int tan^{8} (u) sec^{4} (u) d u

Rewrite using trig identities

= \frac{1}{7} \cdot \int (1 + tan^{2} (u)) sec^{2} (u) tan^{8} (u) d u

Apply u-substitution

: \int v^{8} (1 + v^{2}) d v

= \frac{1}{7} \cdot \int v^{8} (1 + v^{2}) d v

Expand

v^{8} (1 + v^{2}) : v^{8} + v^{10}

= \frac{1}{7} \cdot \int v^{8} + v^{10} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{7} (\int v^{8} d v + \int v^{10} d v)

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

\int v^{10} d v = \frac{v ^{11}}{11}

= \frac{1}{7} (\frac{v ^{9}}{9} + \frac{v ^{11}}{11})

Substitute back

= \frac{1}{7} (\frac{tan ^{9} ( 7 x )}{9} + \frac{tan ^{11} ( 7 x )}{11})

Add a constant to the solution

= \frac{1}{7} (\frac{tan ^{9} ( 7 x )}{9} + \frac{tan ^{11} ( 7 x )}{11}) + C

f (x) = x^{2} - 1, at x = - 3

\frac{\partial}{\partial x} (3 y x^{2})

\frac{d}{d x} (5 - 8 x)

\frac{\partial}{\partial y} (- \frac{y ^{3}}{3} - 8 x)

n \to \infty lim (\frac{n}{10 + n})

What is the integral of tan^8(7x)sec^4(7x) ?
The integral of tan^8(7x)sec^4(7x) is 1/7 ((tan^9(7x))/9+(tan^{11}(7x))/(11))+C