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

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

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

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

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

Solution steps

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

Rewrite using trig identities

= \int (1 + tan^{2} (x)) sec^{2} (x) tan^{8} (x) d x

Apply u-substitution

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

= \int u^{8} (1 + u^{2}) d u

Expand

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

= \int u^{8} + u^{10} 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^{8} d u + \int u^{10} d u

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

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

= \frac{u ^{9}}{9} + \frac{u ^{11}}{11}

Substitute back

u = tan (x)

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

Add a constant to the solution

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

y = 4 - x^{2}, (4, - 12)

\int \frac{6 x}{( 9 x ^{2} - 4 ) ^{3}} d x

\frac{\partial}{\partial x} x y \cdot e^{(\frac{( - 5 x ^{2} + 5 y ^{2} )}{2} + 1)}

\frac{d y}{d x} = y^{2} (6 - 2 x)

\frac{\partial}{\partial x} (ln (\frac{x}{k}))

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