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

The integral of tan^4(x)sec^8(x) is (tan^5(x))/5+(3tan^7(x))/7+(tan^9(x))/3+(tan^{11}(x))/(11)+C

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

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

\frac{tan ^{5} ( x )}{5} + \frac{3 tan ^{7} ( x )}{7} + \frac{tan ^{9} ( x )}{3} + \frac{tan ^{11} ( x )}{11} + C

Solution steps

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

Rewrite using trig identities

= \int sec^{2} (x) (1 + tan^{2} (x))^{3} tan^{4} (x) d x

Apply u-substitution

: \int u^{4} (1 + u^{2})^{3} d u

= \int u^{4} (1 + u^{2})^{3} d u

Expand

u^{4} (1 + u^{2})^{3} : u^{4} + 3 u^{6} + 3 u^{8} + u^{10}

= \int u^{4} + 3 u^{6} + 3 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^{4} d u + \int 3 u^{6} d u + \int 3 u^{8} d u + \int u^{10} d u

\int u^{4} d u = \frac{u ^{5}}{5}

\int 3 u^{6} d u = \frac{3 u ^{7}}{7}

\int 3 u^{8} d u = \frac{u ^{9}}{3}

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

= \frac{u ^{5}}{5} + \frac{3 u ^{7}}{7} + \frac{u ^{9}}{3} + \frac{u ^{11}}{11}

Substitute back

u = tan (x)

= \frac{tan ^{5} ( x )}{5} + \frac{3 tan ^{7} ( x )}{7} + \frac{tan ^{9} ( x )}{3} + \frac{tan ^{11} ( x )}{11}

Add a constant to the solution

= \frac{tan ^{5} ( x )}{5} + \frac{3 tan ^{7} ( x )}{7} + \frac{tan ^{9} ( x )}{3} + \frac{tan ^{11} ( x )}{11} + C

\int \frac{x - 4}{x ^{2} - 5 x + 6} d x

x \to - 2 lim (\frac{5}{x ^{2}})

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

x \to 1 - lim (\frac{x + 1}{x ^{4} - 1})

x^{2} y^{^{''}} - 2 x y^{^{'}} - 4 y = 0, y (1) = 0, y^{^{'}} (1) = - 5

What is the integral of tan^4(x)sec^8(x) ?
The integral of tan^4(x)sec^8(x) is (tan^5(x))/5+(3tan^7(x))/7+(tan^9(x))/3+(tan^{11}(x))/(11)+C