What is the integral of 18sec^2(9x)tan(9x) ?

The integral of 18sec^2(9x)tan(9x) is tan^2(9x)+C

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integral of 18sec^2(9x)tan(9x)

\int 18 sec^{2} (9 x) tan (9 x) d x

tan^{2} (9 x) + C

Solution steps

\int 18 sec^{2} (9 x) tan (9 x) d x

Take the constant out:

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

= 18 \cdot \int sec^{2} (9 x) tan (9 x) d x

Apply u-substitution

: \int \frac{u}{9} d u

= 18 \cdot \int \frac{u}{9} d u

Take the constant out:

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

= 18 \cdot \frac{1}{9} \cdot \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= 18 \cdot \frac{1}{9} \cdot \frac{u ^{2}}{2}

Substitute back

u = tan (9 x)

= 18 \cdot \frac{1}{9} \cdot \frac{tan ^{2} ( 9 x )}{2}

Simplify

18 \cdot \frac{1}{9} \cdot \frac{tan ^{2} ( 9 x )}{2} : tan^{2} (9 x)

= tan^{2} (9 x)

Add a constant to the solution

= tan^{2} (9 x) + C

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What is the integral of 18sec^2(9x)tan(9x) ?
The integral of 18sec^2(9x)tan(9x) is tan^2(9x)+C