What is the integral of tan^4(2x)sec(2x) ?

The integral of tan^4(2x)sec(2x) is

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integral of tan^4(2x)sec(2x)

\int tan^{4} (2 x) sec (2 x) d x

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

Solution steps

\int tan^{4} (2 x) sec (2 x) d x

Apply u-substitution

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

Take the constant out:

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

= \frac{1}{2} \cdot \int tan^{4} (u) sec (u) d u

Rewrite using trig identities

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

Apply u-substitution

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

Take the constant out:

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

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

Take the partial fraction of

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Substitute back

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

Simplify

\frac{1}{2} \cdot 32 (\frac{3}{256} ln tan (\frac{2 x}{2}) + 1 - \frac{3}{256 ( tan ( \frac{2 x}{2} ) + 1 )} - (- \frac{1}{256 ( tan ( \frac{2 x}{2} ) + 1 ) ^{2}}) - (- \frac{1}{64 ( tan ( \frac{2 x}{2} ) + 1 ) ^{3}}) - \frac{1}{128 ( tan ( \frac{2 x}{2} ) + 1 ) ^{4}} - \frac{3}{256} ln tan (\frac{2 x}{2}) - 1 - \frac{3}{256 ( tan ( \frac{2 x}{2} ) - 1 )} - \frac{1}{256 ( tan ( \frac{2 x}{2} ) - 1 ) ^{2}} - (- \frac{1}{64 ( tan ( \frac{2 x}{2} ) - 1 ) ^{3}}) - (- \frac{1}{128 ( tan ( \frac{2 x}{2} ) - 1 ) ^{4}})) : 16 (\frac{3}{256} ln ∣ tan (x) + 1 ∣ - \frac{3}{256 ( tan ( x ) + 1 )} + \frac{1}{256 ( tan ( x ) + 1 ) ^{2}} + \frac{1}{64 ( tan ( x ) + 1 ) ^{3}} - \frac{1}{128 ( tan ( x ) + 1 ) ^{4}} - \frac{3}{256} ln ∣ tan (x) - 1 ∣ - \frac{3}{256 ( tan ( x ) - 1 )} - \frac{1}{256 ( tan ( x ) - 1 ) ^{2}} + \frac{1}{64 ( tan ( x ) - 1 ) ^{3}} + \frac{1}{128 ( tan ( x ) - 1 ) ^{4}})

= 16 (\frac{3}{256} ln ∣ tan (x) + 1 ∣ - \frac{3}{256 ( tan ( x ) + 1 )} + \frac{1}{256 ( tan ( x ) + 1 ) ^{2}} + \frac{1}{64 ( tan ( x ) + 1 ) ^{3}} - \frac{1}{128 ( tan ( x ) + 1 ) ^{4}} - \frac{3}{256} ln ∣ tan (x) - 1 ∣ - \frac{3}{256 ( tan ( x ) - 1 )} - \frac{1}{256 ( tan ( x ) - 1 ) ^{2}} + \frac{1}{64 ( tan ( x ) - 1 ) ^{3}} + \frac{1}{128 ( tan ( x ) - 1 ) ^{4}})

Add a constant to the solution

= 16 (\frac{3}{256} ln ∣ tan (x) + 1 ∣ - \frac{3}{256 ( tan ( x ) + 1 )} + \frac{1}{256 ( tan ( x ) + 1 ) ^{2}} + \frac{1}{64 ( tan ( x ) + 1 ) ^{3}} - \frac{1}{128 ( tan ( x ) + 1 ) ^{4}} - \frac{3}{256} ln ∣ tan (x) - 1 ∣ - \frac{3}{256 ( tan ( x ) - 1 )} - \frac{1}{256 ( tan ( x ) - 1 ) ^{2}} + \frac{1}{64 ( tan ( x ) - 1 ) ^{3}} + \frac{1}{128 ( tan ( x ) - 1 ) ^{4}}) + C