What is the integral of tan^3(4x)sec^4(4x) ?

The integral of tan^3(4x)sec^4(4x) is (sec^6(4x))/(24)-(sec^4(4x))/(16)+C

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integral of tan^3(4x)sec^4(4x)

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

\frac{sec ^{6} ( 4 x )}{24} - \frac{sec ^{4} ( 4 x )}{16} + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

: \int \frac{u ^{3} ( u ^{2} - 1 )}{4} d u

= \int \frac{u ^{3} ( u ^{2} - 1 )}{4} d u

Expand

\frac{u ^{3} ( u ^{2} - 1 )}{4} : \frac{u ^{5}}{4} - \frac{u ^{3}}{4}

= \int \frac{u ^{5}}{4} - \frac{u ^{3}}{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 ^{5}}{4} d u - \int \frac{u ^{3}}{4} d u

\int \frac{u ^{5}}{4} d u = \frac{u ^{6}}{24}

\int \frac{u ^{3}}{4} d u = \frac{u ^{4}}{16}

= \frac{u ^{6}}{24} - \frac{u ^{4}}{16}

Substitute back

u = sec (4 x)

= \frac{sec ^{6} ( 4 x )}{24} - \frac{sec ^{4} ( 4 x )}{16}

Add a constant to the solution

= \frac{sec ^{6} ( 4 x )}{24} - \frac{sec ^{4} ( 4 x )}{16} + C

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What is the integral of tan^3(4x)sec^4(4x) ?
The integral of tan^3(4x)sec^4(4x) is (sec^6(4x))/(24)-(sec^4(4x))/(16)+C