What is the integral of tan^3(2-x)sec(2-x) ?

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

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integral of tan^3(2-x)sec(2-x)

\int tan^{3} (2 - x) sec (2 - x) d x

sec (2 - x) - \frac{1}{3} sec^{3} (2 - x) + C

Solution steps

\int tan^{3} (2 - x) sec (2 - x) d x

Apply u-substitution

: \int - tan^{3} (u) sec (u) d u

= \int - tan^{3} (u) sec (u) d u

Take the constant out:

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

= - \int tan^{3} (u) sec (u) d u

Rewrite using trig identities

= - \int (- 1 + sec^{2} (u)) tan (u) sec (u) d u

Apply u-substitution

: \int - 1 + v^{2} d v

= - \int - 1 + v^{2} d v

Apply the Sum Rule:

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

= - (- \int 1 d v + \int v^{2} d v)

\int 1 d v = v

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

= - (- v + \frac{v ^{3}}{3})

Substitute back

= - (- sec (2 - x) + \frac{sec ^{3} ( 2 - x )}{3})

Simplify

- (- sec (2 - x) + \frac{sec ^{3} ( 2 - x )}{3}) : sec (2 - x) - \frac{1}{3} sec^{3} (2 - x)

= sec (2 - x) - \frac{1}{3} sec^{3} (2 - x)

Add a constant to the solution

= sec (2 - x) - \frac{1}{3} sec^{3} (2 - x) + C

\int \frac{x ^{3}}{9 x ^{2} + 16} d x

\int x^{6} (x^{7} + 5) d x

\int e^{3 x} d x

x^{- 1}

\int \frac{1}{x} \frac{1 - x}{1 + x} d x

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