What is the integral of tan^5(x)sec^6(x) ?

The integral of tan^5(x)sec^6(x) is (tan^6(x))/6+(tan^8(x))/4+(tan^{10}(x))/(10)+C

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integral of tan^5(x)sec^6(x)

\int tan^{5} (x) sec^{6} (x) d x

\frac{tan ^{6} ( x )}{6} + \frac{tan ^{8} ( x )}{4} + \frac{tan ^{10} ( x )}{10} + C

Solution steps

\int tan^{5} (x) sec^{6} (x) d x

Rewrite using trig identities

= \int sec^{2} (x) (1 + tan^{2} (x))^{2} tan^{5} (x) d x

Apply u-substitution

: \int u^{5} (1 + u^{2})^{2} d u

= \int u^{5} (1 + u^{2})^{2} d u

Expand

u^{5} (1 + u^{2})^{2} : u^{5} + 2 u^{7} + u^{9}

= \int u^{5} + 2 u^{7} + u^{9} 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^{5} d u + \int 2 u^{7} d u + \int u^{9} d u

\int u^{5} d u = \frac{u ^{6}}{6}

\int 2 u^{7} d u = \frac{u ^{8}}{4}

\int u^{9} d u = \frac{u ^{10}}{10}

= \frac{u ^{6}}{6} + \frac{u ^{8}}{4} + \frac{u ^{10}}{10}

Substitute back

u = tan (x)

= \frac{tan ^{6} ( x )}{6} + \frac{tan ^{8} ( x )}{4} + \frac{tan ^{10} ( x )}{10}

Add a constant to the solution

= \frac{tan ^{6} ( x )}{6} + \frac{tan ^{8} ( x )}{4} + \frac{tan ^{10} ( x )}{10} + C

(2 x - 2)^{4} (x^{2} + x + 1)^{4}

7 sec^{5} (π t - 8)

\int \frac{1}{( 225 - x ^{2} ) ^{\frac{3}{2}}} d x

(y^{2} sin (x)) d x + (\frac{1}{x} - \frac{y}{x}) d y = 0

x \to 4 lim (\frac{8 - 2 x}{x - 4})

What is the integral of tan^5(x)sec^6(x) ?
The integral of tan^5(x)sec^6(x) is (tan^6(x))/6+(tan^8(x))/4+(tan^{10}(x))/(10)+C