What is the integral of tan^{15}(x)sec^4(x) ?

The integral of tan^{15}(x)sec^4(x) is (tan^{16}(x))/(16)+(tan^{18}(x))/(18)+C

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integral of tan^{15}(x)sec^4(x)

\int tan^{15} (x) sec^{4} (x) d x

\frac{tan ^{16} ( x )}{16} + \frac{tan ^{18} ( x )}{18} + C

Solution steps

\int tan^{15} (x) sec^{4} (x) d x

Rewrite using trig identities

= \int (1 + tan^{2} (x)) sec^{2} (x) tan^{15} (x) d x

Apply u-substitution

: \int u^{15} (1 + u^{2}) d u

= \int u^{15} (1 + u^{2}) d u

Expand

u^{15} (1 + u^{2}) : u^{15} + u^{17}

= \int u^{15} + u^{17} 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^{15} d u + \int u^{17} d u

\int u^{15} d u = \frac{u ^{16}}{16}

\int u^{17} d u = \frac{u ^{18}}{18}

= \frac{u ^{16}}{16} + \frac{u ^{18}}{18}

Substitute back

u = tan (x)

= \frac{tan ^{16} ( x )}{16} + \frac{tan ^{18} ( x )}{18}

Add a constant to the solution

= \frac{tan ^{16} ( x )}{16} + \frac{tan ^{18} ( x )}{18} + C

\int \frac{1}{x ^{2} 4 x + 1} d x

\frac{d}{d x} (- e^{- x} + 1)

\int_{- 2}^{- 1} (x - \frac{1}{x ^{2}}) d x

\int \frac{1}{x ^{2} + 3 x + 9} d x

f (x) = \frac{1}{x + 2}

What is the integral of tan^{15}(x)sec^4(x) ?
The integral of tan^{15}(x)sec^4(x) is (tan^{16}(x))/(16)+(tan^{18}(x))/(18)+C