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

The integral of sec^4(u)tan(u) is (sec^4(u))/4+C

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integral of sec^4(u)tan(u)

\int sec^{4} (u) tan (u) d u

\frac{sec ^{4} ( u )}{4} + C

Solution steps

\int sec^{4} (u) tan (u) d u

Apply u-substitution

: \int v^{3} d v

= \int v^{3} d v

Apply the Power Rule

: \frac{v ^{4}}{4}

= \frac{v ^{4}}{4}

Substitute back

v = sec (u)

= \frac{sec ^{4} ( u )}{4}

Add a constant to the solution

= \frac{sec ^{4} ( u )}{4} + C

2 x (y + 1) d x - y d y = 0

x \to 0 lim (arccos (x))

x \to \infty lim (6 x)

\int 5 sin^{3} (x) cos^{2} (x) d x

f (x) = arcsin (6 x + 1)

What is the integral of sec^4(u)tan(u) ?
The integral of sec^4(u)tan(u) is (sec^4(u))/4+C