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

The integral of (sin^5(x))/(cos^4(x)) is 1/3 sec^3(x)-2sec(x)-cos(x)+C

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

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

\frac{1}{3} sec^{3} (x) - 2 sec (x) - cos (x) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{( 1 - cos ^{2} ( x ) ) ^{2} sin ( x )}{cos ^{4} ( x )} d x

Apply u-substitution

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

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

Expand

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

= \int - \frac{1}{u ^{4}} + \frac{2}{u ^{2}} - 1 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}{u ^{4}} d u + \int \frac{2}{u ^{2}} d u - \int 1 d u

\int \frac{1}{u ^{4}} d u = - \frac{1}{3 u ^{3}}

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

\int 1 d u = u

= - (- \frac{1}{3 u ^{3}}) - \frac{2}{u} - u

Substitute back

u = cos (x)

= - (- \frac{1}{3 cos ^{3} ( x )}) - \frac{2}{cos ( x )} - cos (x)

Simplify

- (- \frac{1}{3 cos ^{3} ( x )}) - \frac{2}{cos ( x )} - cos (x) : \frac{1}{3} sec^{3} (x) - 2 sec (x) - cos (x)

= \frac{1}{3} sec^{3} (x) - 2 sec (x) - cos (x)

Add a constant to the solution

= \frac{1}{3} sec^{3} (x) - 2 sec (x) - cos (x) + C

t \to 1 + lim (t^{\frac{1}{1 - t}})

e^{(t - 2)}

\int (3) d x

\frac{d}{d x} (\frac{1}{x ^{2} + x})

\int \frac{x}{sin ^{2} ( 2 x ^{2} )} d x

What is the integral of (sin^5(x))/(cos^4(x)) ?
The integral of (sin^5(x))/(cos^4(x)) is 1/3 sec^3(x)-2sec(x)-cos(x)+C