What is the integral of tan^3(x)sec^{2n+1}(x) ?

The integral of tan^3(x)sec^{2n+1}(x) is 1/(-2n-1)cos^{-2n-1}(x)-1/(-2n-3)cos^{-2n-3}(x)+C

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integral of tan^3(x)sec^{2n+1}(x)

\int tan^{3} (x) sec^{2 n + 1} (x) d x

\frac{1}{- 2 n - 1} cos^{- 2 n - 1} (x) - \frac{1}{- 2 n - 3} cos^{- 2 n - 3} (x) + C

Solution steps

\int tan^{3} (x) sec^{2 n + 1} (x) d x

Apply u-substitution

: \int - \frac{- u ^{2} + 1}{u ^{2 n + 4}} d u

= \int - \frac{- u ^{2} + 1}{u ^{2 n + 4}} d u

Take the constant out:

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

= - \int \frac{- u ^{2} + 1}{u ^{2 n + 4}} d u

Expand

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

= - \int - u^{- 2 n - 2} + \frac{1}{u ^{2 n + 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 u^{- 2 n - 2} d u + \int \frac{1}{u ^{2 n + 4}} d u)

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

\int \frac{1}{u ^{2 n + 4}} d u = \frac{u ^{- 2 n - 3}}{- 2 n - 3}

= - (- \frac{u ^{- 2 n - 1}}{- 2 n - 1} + \frac{u ^{- 2 n - 3}}{- 2 n - 3})

Substitute back

u = cos (x)

= - (- \frac{cos ^{- 2 n - 1} ( x )}{- 2 n - 1} + \frac{cos ^{- 2 n - 3} ( x )}{- 2 n - 3})

Simplify

- (- \frac{cos ^{- 2 n - 1} ( x )}{- 2 n - 1} + \frac{cos ^{- 2 n - 3} ( x )}{- 2 n - 3}) : \frac{1}{- 2 n - 1} cos^{- 2 n - 1} (x) - \frac{1}{- 2 n - 3} cos^{- 2 n - 3} (x)

= \frac{1}{- 2 n - 1} cos^{- 2 n - 1} (x) - \frac{1}{- 2 n - 3} cos^{- 2 n - 3} (x)

Add a constant to the solution

= \frac{1}{- 2 n - 1} cos^{- 2 n - 1} (x) - \frac{1}{- 2 n - 3} cos^{- 2 n - 3} (x) + C

\int x^{3} - 1 d x

\int (x + 3) (x^{2} + 6 x + 7)^{6} d x

f (x) = 2 x^{4} - 16 x^{2} + 6

\int (3 cos (x) - e^{x} + \frac{5}{x}) d x

\frac{d y}{d x} = 0. 4^{\frac{x}{3}}

What is the integral of tan^3(x)sec^{2n+1}(x) ?
The integral of tan^3(x)sec^{2n+1}(x) is 1/(-2n-1)cos^{-2n-1}(x)-1/(-2n-3)cos^{-2n-3}(x)+C