What is the integral of (sin^3(5x))/(cos^4(5x)) ?

The integral of (sin^3(5x))/(cos^4(5x)) is -1/5 sec(5x)+(sec^3(5x))/(15)+C

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integral of (sin^3(5x))/(cos^4(5x))

\int \frac{sin ^{3} ( 5 x )}{cos ^{4} ( 5 x )} d x

- \frac{1}{5} sec (5 x) + \frac{sec ^{3} ( 5 x )}{15} + C

Solution steps

\int \frac{sin ^{3} ( 5 x )}{cos ^{4} ( 5 x )} d x

Rewrite using trig identities

= \int \frac{tan ^{3} ( 5 x )}{cos ( 5 x )} d x

Use the basic trigonometric identity:

\frac{1}{cos ( x )} = sec (x)

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

\frac{- 1 + u ^{2}}{5} : - \frac{1}{5} + \frac{u ^{2}}{5}

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

\int \frac{1}{5} d u = \frac{1}{5} u

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

= - \frac{1}{5} u + \frac{u ^{3}}{15}

Substitute back

u = sec (5 x)

= - \frac{1}{5} sec (5 x) + \frac{sec ^{3} ( 5 x )}{15}

Add a constant to the solution

= - \frac{1}{5} sec (5 x) + \frac{sec ^{3} ( 5 x )}{15} + C

x \to 0 - lim (\frac{1}{x ^{\frac{2}{3}}})

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

\int e^{x} 4 + e^{x} d x

\frac{\partial}{\partial x} (2 y e^{x})

6 + 5 x^{2} - 2 x^{3}

What is the integral of (sin^3(5x))/(cos^4(5x)) ?
The integral of (sin^3(5x))/(cos^4(5x)) is -1/5 sec(5x)+(sec^3(5x))/(15)+C