What is the integral of 16tan^{15}(x)sec^2(x) ?

The integral of 16tan^{15}(x)sec^2(x) is tan^{16}(x)+C

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integral of 16tan^{15}(x)sec^2(x)

\int 16 tan^{15} (x) sec^{2} (x) d x

tan^{16} (x) + C

Solution steps

\int 16 tan^{15} (x) sec^{2} (x) d x

Take the constant out:

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

= 16 \cdot \int tan^{15} (x) sec^{2} (x) d x

Rewrite using trig identities

= 16 \cdot \int (1 + tan^{2} (x)) tan^{15} (x) d x

Apply u-substitution

: \int u^{15} d u

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

Apply the Power Rule

: \frac{u ^{16}}{16}

= 16 \cdot \frac{u ^{16}}{16}

Substitute back

u = tan (x)

= 16 \cdot \frac{tan ^{16} ( x )}{16}

Simplify

16 \cdot \frac{tan ^{16} ( x )}{16} : tan^{16} (x)

= tan^{16} (x)

Add a constant to the solution

= tan^{16} (x) + C

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What is the integral of 16tan^{15}(x)sec^2(x) ?
The integral of 16tan^{15}(x)sec^2(x) is tan^{16}(x)+C