What is the integral of (tan^5(x))/(sec(x)) ?

The integral of (tan^5(x))/(sec(x)) is -(-1/3 sec^3(x)+2sec(x)+cos(x))+C

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integral of (tan^5(x))/(sec(x))

\int \frac{tan ^{5} ( x )}{sec ( x )} d x

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

Solution steps

\int \frac{tan ^{5} ( x )}{sec ( x )} d x

Apply Trigonometric Substitution

: \int - \frac{( 1 - u ^{2} ) ^{2}}{u ^{4}} d u

= \int - \frac{( 1 - u ^{2} ) ^{2}}{u ^{4}} d u

Take the constant out:

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

= - \int \frac{( 1 - u ^{2} ) ^{2}}{u ^{4}} d u

Expand

\frac{( 1 - u ^{2} ) ^{2}}{u ^{4}} : \frac{1}{u ^{4}} - \frac{2}{u ^{2}} + 1

= - \int \frac{1}{u ^{4}} - \frac{2}{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

= - (\int \frac{1}{u ^{4}} d u - \int \frac{2}{u ^{2}} d u + \int 1 d u)

\int \frac{1}{u ^{4}} d u = - \frac{1}{3 u ^{3}}

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

\int 1 d u = u

= - (- \frac{1}{3 u ^{3}} - (- \frac{2}{u}) + u)

Substitute back

u = cos (x)

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} ((4 x - 1)^{3} (5 x + 3)^{2})

\int \frac{1}{( 1 + cos ^{2} ( x ) )} d x

e^{\frac{x}{2}}

3 x^{2} - 4

What is the integral of (tan^5(x))/(sec(x)) ?
The integral of (tan^5(x))/(sec(x)) is -(-1/3 sec^3(x)+2sec(x)+cos(x))+C