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

The integral of sec^3(5x)tan^5(5x) is (sec^7(5x))/(35)-(2sec^5(5x))/(25)+(sec^3(5x))/(15)+C

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

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

\frac{sec ^{7} ( 5 x )}{35} - \frac{2 sec ^{5} ( 5 x )}{25} + \frac{sec ^{3} ( 5 x )}{15} + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

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

\int \frac{2 u ^{4}}{5} d u = \frac{2 u ^{5}}{25}

\int \frac{u ^{2}}{5} d u = \frac{u ^{3}}{15}

= \frac{u ^{7}}{35} - \frac{2 u ^{5}}{25} + \frac{u ^{3}}{15}

Substitute back

u = sec (5 x)

= \frac{sec ^{7} ( 5 x )}{35} - \frac{2 sec ^{5} ( 5 x )}{25} + \frac{sec ^{3} ( 5 x )}{15}

Add a constant to the solution

= \frac{sec ^{7} ( 5 x )}{35} - \frac{2 sec ^{5} ( 5 x )}{25} + \frac{sec ^{3} ( 5 x )}{15} + C

f (x) = 16 x, at x = 4

\int_{3}^{2} (5 x^{2} + 4 x + 9) d x

y^{^{''}} + 4 y = 5 sec (2 t)

y^{^{''}} (e^{x} + 1) + y^{^{'}} = 0

\int x^{3} \cdot 4 - x^{2} d x

What is the integral of sec^3(5x)tan^5(5x) ?
The integral of sec^3(5x)tan^5(5x) is (sec^7(5x))/(35)-(2sec^5(5x))/(25)+(sec^3(5x))/(15)+C