What is the integral of 18tan^{17}(x)sec^2(x) ?

The integral of 18tan^{17}(x)sec^2(x) is tan^{18}(x)+C

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integral of 18tan^{17}(x)sec^2(x)

\int 18 tan^{17} (x) sec^{2} (x) d x

tan^{18} (x) + C

Solution steps

\int 18 tan^{17} (x) sec^{2} (x) d x

Take the constant out:

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

= 18 \cdot \int tan^{17} (x) sec^{2} (x) d x

Rewrite using trig identities

= 18 \cdot \int (1 + tan^{2} (x)) tan^{17} (x) d x

Apply u-substitution

: \int u^{17} d u

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

Apply the Power Rule

: \frac{u ^{18}}{18}

= 18 \cdot \frac{u ^{18}}{18}

Substitute back

u = tan (x)

= 18 \cdot \frac{tan ^{18} ( x )}{18}

Simplify

18 \cdot \frac{tan ^{18} ( x )}{18} : tan^{18} (x)

= tan^{18} (x)

Add a constant to the solution

= tan^{18} (x) + C

x \to 9 lim (\frac{3 - x}{9 x - x ^{2}})

\int \frac{1}{u ^{2} + 3} d u

\frac{d x}{d y} = \frac{y ^{3} - 3 x}{y}

n = 1 \sum \infty \frac{4}{2 n}

What is the integral of 18tan^{17}(x)sec^2(x) ?
The integral of 18tan^{17}(x)sec^2(x) is tan^{18}(x)+C