What is the integral of tan^2(x)sec^8(x) ?

The integral of tan^2(x)sec^8(x) is (tan^3(x))/3+(3tan^5(x))/5+(3tan^7(x))/7+(tan^9(x))/9+C

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integral of tan^2(x)sec^8(x)

\int tan^{2} (x) sec^{8} (x) d x

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

Solution steps

\int tan^{2} (x) sec^{8} (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

= \int u^{2} + 3 u^{4} + 3 u^{6} + u^{8} 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^{2} d u + \int 3 u^{4} d u + \int 3 u^{6} d u + \int u^{8} d u

\int u^{2} d u = \frac{u ^{3}}{3}

\int 3 u^{4} d u = \frac{3 u ^{5}}{5}

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

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

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

Substitute back

u = tan (x)

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

Add a constant to the solution

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

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What is the integral of tan^2(x)sec^8(x) ?
The integral of tan^2(x)sec^8(x) is (tan^3(x))/3+(3tan^5(x))/5+(3tan^7(x))/7+(tan^9(x))/9+C