What is the integral of sin(x)cos^{-4}(x) ?

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

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integral of sin(x)cos^{-4}(x)

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

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

Substitute back

u = cos (x)

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

Simplify

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

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

Add a constant to the solution

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

\int_{1}^{6} x^{- 4} d x

\int \frac{3 x ^{3}}{49 - x ^{2}} d x

x^{2} - 3 x - 2

\frac{d}{d x} (arccos (\frac{x + 1}{3}))

y^{^{'}} = x + y, y (0) = 0

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