What is the integral of sec^{-3}(x)tan(x) ?

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

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integral of sec^{-3}(x)tan(x)

\int sec^{- 3} (x) tan (x) d x

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

Solution steps

\int sec^{- 3} (x) tan (x) d x

Simplify

sec^{- 3} (x) tan (x) : cos^{2} (x) sin (x)

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

Apply u-substitution

: \int - u^{2} d u

= \int - u^{2} d u

Take the constant out:

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

= - \int u^{2} d u

Apply the Power Rule

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

= - \frac{u ^{3}}{3}

Substitute back

u = cos (x)

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

Add a constant to the solution

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

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