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

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

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

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

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

Solution steps

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

Rewrite using trig identities

= \int (- 1 + sec^{2} (x)) tan (x) sec^{7} (x) d x

Apply u-substitution

: \int u^{6} (u^{2} - 1) d u

= \int u^{6} (u^{2} - 1) d u

Expand

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

= \int u^{8} - u^{6} 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 u^{6} d u

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

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

= \frac{u ^{9}}{9} - \frac{u ^{7}}{7}

Substitute back

u = sec (x)

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

Add a constant to the solution

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

\int 7 x^{3} x d x

n = 0 \sum \infty (5 x - 3)^{n}

y = ln (3 - 2 x), at x = - 3

\int ln (x + 3) d x

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

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