What is the integral of (sin^3(x))/(cos^6(x)) ?

The integral of (sin^3(x))/(cos^6(x)) is 1/5 sec^5(x)-1/3 sec^3(x)+C

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integral of (sin^3(x))/(cos^6(x))

\int \frac{sin ^{3} ( x )}{cos ^{6} ( x )} d x

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

Solution steps

\int \frac{sin ^{3} ( x )}{cos ^{6} ( x )} d x

Rewrite using trig identities

= \int \frac{( 1 - cos ^{2} ( x ) ) sin ( x )}{cos ^{6} ( x )} d x

Apply u-substitution

: \int - \frac{1 - u ^{2}}{u ^{6}} d u

= \int - \frac{1 - u ^{2}}{u ^{6}} d u

Expand

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

= \int - \frac{1}{u ^{6}} + \frac{1}{u ^{4}} 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 ^{6}} d u + \int \frac{1}{u ^{4}} d u

\int \frac{1}{u ^{6}} d u = - \frac{1}{5 u ^{5}}

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

= - (- \frac{1}{5 u ^{5}}) - \frac{1}{3 u ^{3}}

Substitute back

u = cos (x)

= - (- \frac{1}{5 cos ^{5} ( x )}) - \frac{1}{3 cos ^{3} ( x )}

Simplify

- (- \frac{1}{5 cos ^{5} ( x )}) - \frac{1}{3 cos ^{3} ( x )} : \frac{1}{5} sec^{5} (x) - \frac{1}{3} sec^{3} (x)

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

Add a constant to the solution

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

y = 3 + 4 x^{2} - 2 x^{3}, at x = a

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

y^{^{''}} + 5 t y^{^{'}} - 5 y = 0

\frac{d}{d x} (ln (cosh (x)))

\frac{\partial}{\partial y} (x cos (y) + y e^{x})

What is the integral of (sin^3(x))/(cos^6(x)) ?
The integral of (sin^3(x))/(cos^6(x)) is 1/5 sec^5(x)-1/3 sec^3(x)+C