What is the integral of tan^4(x)sec(x) ?

The integral of tan^4(x)sec(x) is

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integral of tan^4(x)sec(x)

\int tan^{4} (x) sec (x) d x

32 (\frac{3}{256} ln tan (\frac{x}{2}) + 1 - \frac{3}{256 ( tan ( \frac{x}{2} ) + 1 )} + \frac{1}{256 ( tan ( \frac{x}{2} ) + 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{x}{2} ) + 1 ) ^{3}} - \frac{1}{128 ( tan ( \frac{x}{2} ) + 1 ) ^{4}} - \frac{3}{256} ln tan (\frac{x}{2}) - 1 - \frac{3}{256 ( tan ( \frac{x}{2} ) - 1 )} - \frac{1}{256 ( tan ( \frac{x}{2} ) - 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{x}{2} ) - 1 ) ^{3}} + \frac{1}{128 ( tan ( \frac{x}{2} ) - 1 ) ^{4}}) + C

Solution steps

\int tan^{4} (x) sec (x) d x

Rewrite using trig identities

= \int \frac{sin ^{4} ( x )}{cos ^{5} ( x )} d x

Apply u-substitution

: \int \frac{32 u ^{4}}{( 1 - u ^{2} ) ^{5}} d u

= \int \frac{32 u ^{4}}{( 1 - u ^{2} ) ^{5}} d u

Take the constant out:

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

= 32 \cdot \int \frac{u ^{4}}{( 1 - u ^{2} ) ^{5}} d u

Take the partial fraction of

\frac{u ^{4}}{( 1 - u ^{2} ) ^{5}} : \frac{3}{256 ( u + 1 )} + \frac{3}{256 ( u + 1 ) ^{2}} - \frac{1}{128 ( u + 1 ) ^{3}} - \frac{3}{64 ( u + 1 ) ^{4}} + \frac{1}{32 ( u + 1 ) ^{5}} - \frac{3}{256 ( u - 1 )} + \frac{3}{256 ( u - 1 ) ^{2}} + \frac{1}{128 ( u - 1 ) ^{3}} - \frac{3}{64 ( u - 1 ) ^{4}} - \frac{1}{32 ( u - 1 ) ^{5}}

= 32 \cdot \int \frac{3}{256 ( u + 1 )} + \frac{3}{256 ( u + 1 ) ^{2}} - \frac{1}{128 ( u + 1 ) ^{3}} - \frac{3}{64 ( u + 1 ) ^{4}} + \frac{1}{32 ( u + 1 ) ^{5}} - \frac{3}{256 ( u - 1 )} + \frac{3}{256 ( u - 1 ) ^{2}} + \frac{1}{128 ( u - 1 ) ^{3}} - \frac{3}{64 ( u - 1 ) ^{4}} - \frac{1}{32 ( u - 1 ) ^{5}} 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

= 32 (\int \frac{3}{256 ( u + 1 )} d u + \int \frac{3}{256 ( u + 1 ) ^{2}} d u - \int \frac{1}{128 ( u + 1 ) ^{3}} d u - \int \frac{3}{64 ( u + 1 ) ^{4}} d u + \int \frac{1}{32 ( u + 1 ) ^{5}} d u - \int \frac{3}{256 ( u - 1 )} d u + \int \frac{3}{256 ( u - 1 ) ^{2}} d u + \int \frac{1}{128 ( u - 1 ) ^{3}} d u - \int \frac{3}{64 ( u - 1 ) ^{4}} d u - \int \frac{1}{32 ( u - 1 ) ^{5}} d u)

\int \frac{3}{256 ( u + 1 )} d u = \frac{3}{256} ln ∣ u + 1 ∣

\int \frac{3}{256 ( u + 1 ) ^{2}} d u = - \frac{3}{256 ( u + 1 )}

\int \frac{1}{128 ( u + 1 ) ^{3}} d u = - \frac{1}{256 ( u + 1 ) ^{2}}

\int \frac{3}{64 ( u + 1 ) ^{4}} d u = - \frac{1}{64 ( u + 1 ) ^{3}}

\int \frac{1}{32 ( u + 1 ) ^{5}} d u = - \frac{1}{128 ( u + 1 ) ^{4}}

\int \frac{3}{256 ( u - 1 )} d u = \frac{3}{256} ln ∣ u - 1 ∣

\int \frac{3}{256 ( u - 1 ) ^{2}} d u = - \frac{3}{256 ( u - 1 )}

\int \frac{1}{128 ( u - 1 ) ^{3}} d u = - \frac{1}{256 ( u - 1 ) ^{2}}

\int \frac{3}{64 ( u - 1 ) ^{4}} d u = - \frac{1}{64 ( u - 1 ) ^{3}}

\int \frac{1}{32 ( u - 1 ) ^{5}} d u = - \frac{1}{128 ( u - 1 ) ^{4}}

= 32 (\frac{3}{256} ln ∣ u + 1 ∣ - \frac{3}{256 ( u + 1 )} - (- \frac{1}{256 ( u + 1 ) ^{2}}) - (- \frac{1}{64 ( u + 1 ) ^{3}}) - \frac{1}{128 ( u + 1 ) ^{4}} - \frac{3}{256} ln ∣ u - 1 ∣ - \frac{3}{256 ( u - 1 )} - \frac{1}{256 ( u - 1 ) ^{2}} - (- \frac{1}{64 ( u - 1 ) ^{3}}) - (- \frac{1}{128 ( u - 1 ) ^{4}}))

Substitute back

u = tan (\frac{x}{2})

= 32 (\frac{3}{256} ln tan (\frac{x}{2}) + 1 - \frac{3}{256 ( tan ( \frac{x}{2} ) + 1 )} - (- \frac{1}{256 ( tan ( \frac{x}{2} ) + 1 ) ^{2}}) - (- \frac{1}{64 ( tan ( \frac{x}{2} ) + 1 ) ^{3}}) - \frac{1}{128 ( tan ( \frac{x}{2} ) + 1 ) ^{4}} - \frac{3}{256} ln tan (\frac{x}{2}) - 1 - \frac{3}{256 ( tan ( \frac{x}{2} ) - 1 )} - \frac{1}{256 ( tan ( \frac{x}{2} ) - 1 ) ^{2}} - (- \frac{1}{64 ( tan ( \frac{x}{2} ) - 1 ) ^{3}}) - (- \frac{1}{128 ( tan ( \frac{x}{2} ) - 1 ) ^{4}}))

Simplify

= 32 (\frac{3}{256} ln tan (\frac{x}{2}) + 1 - \frac{3}{256 ( tan ( \frac{x}{2} ) + 1 )} + \frac{1}{256 ( tan ( \frac{x}{2} ) + 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{x}{2} ) + 1 ) ^{3}} - \frac{1}{128 ( tan ( \frac{x}{2} ) + 1 ) ^{4}} - \frac{3}{256} ln tan (\frac{x}{2}) - 1 - \frac{3}{256 ( tan ( \frac{x}{2} ) - 1 )} - \frac{1}{256 ( tan ( \frac{x}{2} ) - 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{x}{2} ) - 1 ) ^{3}} + \frac{1}{128 ( tan ( \frac{x}{2} ) - 1 ) ^{4}})

Add a constant to the solution

= 32 (\frac{3}{256} ln tan (\frac{x}{2}) + 1 - \frac{3}{256 ( tan ( \frac{x}{2} ) + 1 )} + \frac{1}{256 ( tan ( \frac{x}{2} ) + 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{x}{2} ) + 1 ) ^{3}} - \frac{1}{128 ( tan ( \frac{x}{2} ) + 1 ) ^{4}} - \frac{3}{256} ln tan (\frac{x}{2}) - 1 - \frac{3}{256 ( tan ( \frac{x}{2} ) - 1 )} - \frac{1}{256 ( tan ( \frac{x}{2} ) - 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{x}{2} ) - 1 ) ^{3}} + \frac{1}{128 ( tan ( \frac{x}{2} ) - 1 ) ^{4}}) + C

\int_{2}^{7} t^{2} ln (2 t) d t

\int \frac{tan ( x )}{sec ( x ) + tan ( x )} d x

\int sin^{2} (\frac{x}{5}) d x

\frac{d y}{d t} = y + 1 + 2 sin (t)

\frac{\partial}{\partial z} (x^{2} y cos (\frac{z}{t}))