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

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

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

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

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

Solution steps

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

\int (x^{3} - 2 sin (3 x))^{2} d x

x \to \infty lim (\frac{e ^{2 x} - e ^{x}}{ln ( x + 1 )})

f (x) = x^{3}, (2, 8)

\frac{d}{d x} (\frac{5}{2 x + 1})

f (x) = 3 sin (x) + 2 cos (x)

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