What is the integral of tan^5(x)sec^4(x) ?

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

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integral of tan^5(x)sec^4(x)

\int tan^{5} (x) sec^{4} (x) d x

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

Solution steps

\int tan^{5} (x) sec^{4} (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

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

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

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

Substitute back

u = tan (x)

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

Add a constant to the solution

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

\int \frac{t}{t + 1} d t

d y + e^{3 y} d x = 0

\frac{d}{d x} (4 sec (x) - 3 x)

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

\frac{d}{d x} (\frac{x ^{6} - 1}{x - 1})

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