What is the integral of 27(sec^2(x)-1)tan(x)sec(x) ?

The integral of 27(sec^2(x)-1)tan(x)sec(x) is 27((sec^3(x))/3-sec(x))+C

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integral of 27(sec^2(x)-1)tan(x)sec(x)

\int 27 (sec^{2} (x) - 1) tan (x) sec (x) d x

27 (\frac{sec ^{3} ( x )}{3} - sec (x)) + C

Solution steps

\int 27 (sec^{2} (x) - 1) tan (x) sec (x) d x

Take the constant out:

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

= 27 \cdot \int (sec^{2} (x) - 1) tan (x) sec (x) d x

Apply u-substitution

: \int u^{2} - 1 d u

= 27 \cdot \int u^{2} - 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 27 (\int u^{2} d u - \int 1 d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int 1 d u = u

= 27 (\frac{u ^{3}}{3} - u)

Substitute back

u = sec (x)

= 27 (\frac{sec ^{3} ( x )}{3} - sec (x))

Add a constant to the solution

= 27 (\frac{sec ^{3} ( x )}{3} - sec (x)) + C

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What is the integral of 27(sec^2(x)-1)tan(x)sec(x) ?
The integral of 27(sec^2(x)-1)tan(x)sec(x) is 27((sec^3(x))/3-sec(x))+C