What is the integral of 5tan^5(x)sec^3(x) ?

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

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integral of 5tan^5(x)sec^3(x)

\int 5 tan^{5} (x) sec^{3} (x) d x

\frac{5}{7} sec^{7} (x) - 2 sec^{5} (x) + \frac{5}{3} sec^{3} (x) + C

Solution steps

\int 5 tan^{5} (x) sec^{3} (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 \cdot \int tan^{5} (x) sec^{3} (x) d x

Rewrite using trig identities

= 5 \cdot \int (- 1 + sec^{2} (x))^{2} tan (x) sec^{3} (x) d x

Apply u-substitution

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

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

Expand

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

= 5 \cdot \int u^{6} - 2 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

= 5 (\int u^{6} d u - \int 2 u^{4} d u + \int u^{2} d u)

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

\int 2 u^{4} d u = \frac{2 u ^{5}}{5}

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

= 5 (\frac{u ^{7}}{7} - \frac{2 u ^{5}}{5} + \frac{u ^{3}}{3})

Substitute back

u = sec (x)

= 5 (\frac{sec ^{7} ( x )}{7} - \frac{2 sec ^{5} ( x )}{5} + \frac{sec ^{3} ( x )}{3})

Simplify

5 (\frac{sec ^{7} ( x )}{7} - \frac{2 sec ^{5} ( x )}{5} + \frac{sec ^{3} ( x )}{3}) : \frac{5}{7} sec^{7} (x) - 2 sec^{5} (x) + \frac{5}{3} sec^{3} (x)

= \frac{5}{7} sec^{7} (x) - 2 sec^{5} (x) + \frac{5}{3} sec^{3} (x)

Add a constant to the solution

= \frac{5}{7} sec^{7} (x) - 2 sec^{5} (x) + \frac{5}{3} sec^{3} (x) + C

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