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

The integral of sqrt(tan(4x))(sec(4x))^2 is 1/6 tan^{3/2}(4x)+C

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

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

\frac{1}{6} tan^{\frac{3}{2}} (4 x) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{sin ( 4 x )}{cos ( 4 x )} sec^{2} (4 x) d x

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

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

Apply u-substitution

: \int \frac{u ^{2}}{2} d u

= \int \frac{u ^{2}}{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 u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= \frac{1}{2} \cdot \frac{u ^{3}}{3}

Substitute back

u = tan (4 x)

= \frac{1}{2} \cdot \frac{( tan ( 4 x ) ) ^{3}}{3}

Simplify

\frac{1}{2} \cdot \frac{( tan ( 4 x ) ) ^{3}}{3} : \frac{1}{6} tan^{\frac{3}{2}} (4 x)

= \frac{1}{6} tan^{\frac{3}{2}} (4 x)

Add a constant to the solution

= \frac{1}{6} tan^{\frac{3}{2}} (4 x) + C

\frac{( s ^{2} - 1 )}{( s ^{2} + 1 ) ^{2}}

x \to 1 - lim ((x^{2} + 1)^{2})

δ (t - 1)

\int_{0}^{π} cos (x + π) d x

\int \frac{x ^{2}}{x} d x

What is the integral of sqrt(tan(4x))(sec(4x))^2 ?
The integral of sqrt(tan(4x))(sec(4x))^2 is 1/6 tan^{3/2}(4x)+C