What is the integral of 9tan(x)sec^3(x) ?

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

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integral of 9tan(x)sec^3(x)

\int 9 tan (x) sec^{3} (x) d x

3 sec^{3} (x) + C

Solution steps

\int 9 tan (x) sec^{3} (x) d x

Take the constant out:

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

= 9 \cdot \int tan (x) sec^{3} (x) d x

Apply u-substitution

: \int \frac{1}{3} d u

= 9 \cdot \int \frac{1}{3} d u

Integral of a constant:

\int a d x = a x

= 9 \cdot \frac{1}{3} u

Substitute back

u = sec^{3} (x)

= 9 \cdot \frac{1}{3} sec^{3} (x)

Simplify

9 \cdot \frac{1}{3} sec^{3} (x) : 3 sec^{3} (x)

= 3 sec^{3} (x)

Add a constant to the solution

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

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