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

The integral of tan^3(5x)sec^5(5x) is (sec^7(5x))/(35)-(sec^5(5x))/(25)+C

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

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

\frac{sec ^{7} ( 5 x )}{35} - \frac{sec ^{5} ( 5 x )}{25} + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

: \int \frac{u ^{4} ( u ^{2} - 1 )}{5} d u

= \int \frac{u ^{4} ( u ^{2} - 1 )}{5} d u

Expand

\frac{u ^{4} ( u ^{2} - 1 )}{5} : \frac{u ^{6}}{5} - \frac{u ^{4}}{5}

= \int \frac{u ^{6}}{5} - \frac{u ^{4}}{5} 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 \frac{u ^{6}}{5} d u - \int \frac{u ^{4}}{5} d u

\int \frac{u ^{6}}{5} d u = \frac{u ^{7}}{35}

\int \frac{u ^{4}}{5} d u = \frac{u ^{5}}{25}

= \frac{u ^{7}}{35} - \frac{u ^{5}}{25}

Substitute back

u = sec (5 x)

= \frac{sec ^{7} ( 5 x )}{35} - \frac{sec ^{5} ( 5 x )}{25}

Add a constant to the solution

= \frac{sec ^{7} ( 5 x )}{35} - \frac{sec ^{5} ( 5 x )}{25} + C

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What is the integral of tan^3(5x)sec^5(5x) ?
The integral of tan^3(5x)sec^5(5x) is (sec^7(5x))/(35)-(sec^5(5x))/(25)+C