What is the integral of 14tan^{13}(x)sec^2(x) ?

The integral of 14tan^{13}(x)sec^2(x) is tan^{14}(x)+C

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integral of 14tan^{13}(x)sec^2(x)

\int 14 tan^{13} (x) sec^{2} (x) d x

tan^{14} (x) + C

Solution steps

\int 14 tan^{13} (x) sec^{2} (x) d x

Take the constant out:

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

= 14 \cdot \int tan^{13} (x) sec^{2} (x) d x

Rewrite using trig identities

= 14 \cdot \int (1 + tan^{2} (x)) tan^{13} (x) d x

Apply u-substitution

: \int u^{13} d u

= 14 \cdot \int u^{13} d u

Apply the Power Rule

: \frac{u ^{14}}{14}

= 14 \cdot \frac{u ^{14}}{14}

Substitute back

u = tan (x)

= 14 \cdot \frac{tan ^{14} ( x )}{14}

Simplify

14 \cdot \frac{tan ^{14} ( x )}{14} : tan^{14} (x)

= tan^{14} (x)

Add a constant to the solution

= tan^{14} (x) + C

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What is the integral of 14tan^{13}(x)sec^2(x) ?
The integral of 14tan^{13}(x)sec^2(x) is tan^{14}(x)+C