What is the integral of sec^n(x)tan(x) ?

The integral of sec^n(x)tan(x) is 1/n cos^{-n}(x)+C

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integral of sec^n(x)tan(x)

\int sec^{n} (x) tan (x) d x

\frac{1}{n} cos^{- n} (x) + C

Solution steps

\int sec^{n} (x) tan (x) d x

Rewrite using trig identities

= \int \frac{sin ( x )}{cos ^{n + 1} ( x )} d x

Apply u-substitution

: \int - \frac{1}{u ^{n + 1}} d u

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

Take the constant out:

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

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

Apply the Power Rule

: - \frac{u ^{- n}}{n}

= - (- \frac{u ^{- n}}{n})

Substitute back

u = cos (x)

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

Simplify

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

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

Add a constant to the solution

= \frac{1}{n} cos^{- n} (x) + C

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