What is the integral of tan^3(4x)sec^{9/2}(4x) ?

The integral of tan^3(4x)sec^{9/2}(4x) is 1/26 sec^{13/2}(4x)-1/18 sec^{9/2}(4x)+C

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integral of tan^3(4x)sec^{9/2}(4x)

\int tan^{3} (4 x) sec^{\frac{9}{2}} (4 x) d x

\frac{1}{26} sec^{\frac{13}{2}} (4 x) - \frac{1}{18} sec^{\frac{9}{2}} (4 x) + C

Solution steps

\int tan^{3} (4 x) sec^{\frac{9}{2}} (4 x) d x

Rewrite using trig identities

= \int (- 1 + sec^{2} (4 x)) tan (4 x) sec^{\frac{9}{2}} (4 x) d x

Apply u-substitution

: \int \frac{u ^{\frac{7}{2}} ( u ^{2} - 1 )}{4} d u

= \int \frac{u ^{\frac{7}{2}} ( u ^{2} - 1 )}{4} d u

Expand

\frac{u ^{\frac{7}{2}} ( u ^{2} - 1 )}{4} : \frac{u ^{\frac{11}{2}}}{4} - \frac{u ^{\frac{7}{2}}}{4}

= \int \frac{u ^{\frac{11}{2}}}{4} - \frac{u ^{\frac{7}{2}}}{4} 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 ^{\frac{11}{2}}}{4} d u - \int \frac{u ^{\frac{7}{2}}}{4} d u

\int \frac{u ^{\frac{11}{2}}}{4} d u = \frac{1}{26} u^{\frac{13}{2}}

\int \frac{u ^{\frac{7}{2}}}{4} d u = \frac{1}{18} u^{\frac{9}{2}}

= \frac{1}{26} u^{\frac{13}{2}} - \frac{1}{18} u^{\frac{9}{2}}

Substitute back

u = sec (4 x)

= \frac{1}{26} sec^{\frac{13}{2}} (4 x) - \frac{1}{18} sec^{\frac{9}{2}} (4 x)

Add a constant to the solution

= \frac{1}{26} sec^{\frac{13}{2}} (4 x) - \frac{1}{18} sec^{\frac{9}{2}} (4 x) + C

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What is the integral of tan^3(4x)sec^{9/2}(4x) ?
The integral of tan^3(4x)sec^{9/2}(4x) is 1/26 sec^{13/2}(4x)-1/18 sec^{9/2}(4x)+C