What is the integral of tan^5(2x)sec^7(2x) ?

The integral of tan^5(2x)sec^7(2x) is (sec^{11}(2x)}{22}-(sec^9(2x))/9+\frac{sec^7(2x))/(14)+C

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integral of tan^5(2x)sec^7(2x)

\int tan^{5} (2 x) sec^{7} (2 x) d x

\frac{sec ^{11} ( 2 x )}{22} - \frac{sec ^{9} ( 2 x )}{9} + \frac{sec ^{7} ( 2 x )}{14} + C

Solution steps

\int tan^{5} (2 x) sec^{7} (2 x) d x

Rewrite using trig identities

= \int (- 1 + sec^{2} (2 x))^{2} tan (2 x) sec^{7} (2 x) d x

Apply u-substitution

: \int \frac{u ^{6} ( u ^{2} - 1 ) ^{2}}{2} d u

= \int \frac{u ^{6} ( u ^{2} - 1 ) ^{2}}{2} d u

Expand

\frac{u ^{6} ( u ^{2} - 1 ) ^{2}}{2} : \frac{u ^{10}}{2} - u^{8} + \frac{u ^{6}}{2}

= \int \frac{u ^{10}}{2} - u^{8} + \frac{u ^{6}}{2} 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{u ^{10}}{2} d u - \int u^{8} d u + \int \frac{u ^{6}}{2} d u

\int \frac{u ^{10}}{2} d u = \frac{u ^{11}}{22}

\int u^{8} d u = \frac{u ^{9}}{9}

\int \frac{u ^{6}}{2} d u = \frac{u ^{7}}{14}

= \frac{u ^{11}}{22} - \frac{u ^{9}}{9} + \frac{u ^{7}}{14}

Substitute back

u = sec (2 x)

= \frac{sec ^{11} ( 2 x )}{22} - \frac{sec ^{9} ( 2 x )}{9} + \frac{sec ^{7} ( 2 x )}{14}

Add a constant to the solution

= \frac{sec ^{11} ( 2 x )}{22} - \frac{sec ^{9} ( 2 x )}{9} + \frac{sec ^{7} ( 2 x )}{14} + C

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What is the integral of tan^5(2x)sec^7(2x) ?
The integral of tan^5(2x)sec^7(2x) is (sec^{11}(2x)}{22}-(sec^9(2x))/9+\frac{sec^7(2x))/(14)+C