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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

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

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

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

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

Substitute back

u = cos (x)

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

Simplify

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

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

Add a constant to the solution

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

n = 1 \sum \infty \frac{1}{n ( n + 6 )}

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

\int 24 x^{2} + 2 x + 10 d x

n = 1 \sum \infty (- 2)^{n}

\frac{( s + 2 )}{s ^{2} + 7 s + 12}

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