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

The integral of 9sec^2(x)tan^2(x) is 3tan^3(x)+C

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

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

3 tan^{3} (x) + C

Solution steps

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

Take the constant out:

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

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

Rewrite using trig identities

= 9 \cdot \int (1 + tan^{2} (x)) tan^{2} (x) d x

Apply u-substitution

: \int u^{2} d u

= 9 \cdot \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= 9 \cdot \frac{u ^{3}}{3}

Substitute back

u = tan (x)

= 9 \cdot \frac{tan ^{3} ( x )}{3}

Simplify

9 \cdot \frac{tan ^{3} ( x )}{3} : 3 tan^{3} (x)

= 3 tan^{3} (x)

Add a constant to the solution

= 3 tan^{3} (x) + C

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What is the integral of 9sec^2(x)tan^2(x) ?
The integral of 9sec^2(x)tan^2(x) is 3tan^3(x)+C