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 sin^{3} (x) cos^{- 4} (x) d x

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

Solution steps

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

Simplify

sin^{3} (x) : sin^{2} (x) sin (x)

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

Simplify

sin^{2} (x) sin (x) cos^{- 4} (x) : \frac{sin ^{3} ( x )}{cos ^{4} ( x )}

= \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

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3 y^{^{''}} + 4 y^{^{'}} = 0

\int_{1}^{e} x^{4} ln (x) d x

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