What is the integral of sec^{4x}(x ?

The integral of sec^{4x}(x is 1/(16ln^2(sec(d)))(sec^{4x}(d)ln(sec^{4x}(d))-sec^{4x}(d))+C

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integral of sec^{4x}(x

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

\frac{1}{16 ln ^{2} ( sec ( d ) )} (sec^{4 x} (d) ln (sec^{4 x} (d)) - sec^{4 x} (d)) + C

Solution steps

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

Apply u-substitution

: \int \frac{ln ( u )}{16 ln ^{2} ( sec ( d ) )} d u

= \int \frac{ln ( u )}{16 ln ^{2} ( sec ( d ) )} d u

Take the constant out:

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

= \frac{1}{16 ln ^{2} ( sec ( d ) )} \cdot \int ln (u) d u

Apply Integration By Parts

: u ln (u) - \int 1 d u

= \frac{1}{16 ln ^{2} ( sec ( d ) )} (u ln (u) - \int 1 d u)

\int 1 d u = u

= \frac{1}{16 ln ^{2} ( sec ( d ) )} (u ln (u) - u)

Substitute back

u = sec^{4 x} (d)

= \frac{1}{16 ln ^{2} ( sec ( d ) )} (sec^{4 x} (d) ln (sec^{4 x} (d)) - sec^{4 x} (d))

Add a constant to the solution

= \frac{1}{16 ln ^{2} ( sec ( d ) )} (sec^{4 x} (d) ln (sec^{4 x} (d)) - sec^{4 x} (d)) + C

\frac{d}{d x} (\frac{x ^{2} - 64}{x + 8})

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(3 x + 2) + (3 y - 3) y^{^{'}} = 0

\int y (y + 1) e^{- 2 y} d y

\int e + e^{3 x} d x

What is the integral of sec^{4x}(x ?
The integral of sec^{4x}(x is 1/(16ln^2(sec(d)))(sec^{4x}(d)ln(sec^{4x}(d))-sec^{4x}(d))+C