What is the integral of tan^7(x)sec^3(x) ?

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

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integral of tan^7(x)sec^3(x)

\int tan^{7} (x) sec^{3} (x) d x

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

Solution steps

\int tan^{7} (x) sec^{3} (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

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

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

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

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

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

Substitute back

u = sec (x)

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{sin ( x )}{x ^{2} - 1})

y = 14 x sin (x), (\frac{π}{2}, 7 π)

x \to \infty lim (xx)

\int 3 cos^{3} (6 x) d x

\int cos^{10} (2 y) sin^{3} (2 y) d y

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