What is the integral of tan^4(x)sec^3(x)tan(x)sec(x) ?

The integral of tan^4(x)sec^3(x)tan(x)sec(x) is (tan^6(x))/6+(tan^8(x))/8+C

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integral of tan^4(x)sec^3(x)tan(x)sec(x)

\int tan^{4} (x) sec^{3} (x) tan (x) sec (x) d x

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

Solution steps

\int tan^{4} (x) sec^{3} (x) tan (x) sec (x) d x

Simplify

tan^{4} (x) sec^{3} (x) tan (x) sec (x) : tan^{5} (x) sec^{4} (x)

= \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

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What is the integral of tan^4(x)sec^3(x)tan(x)sec(x) ?
The integral of tan^4(x)sec^3(x)tan(x)sec(x) is (tan^6(x))/6+(tan^8(x))/8+C