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

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

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

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

sec^{4} (x) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int u^{3} d u

= 4 \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= 4 \cdot \frac{u ^{4}}{4}

Substitute back

u = sec (x)

= 4 \cdot \frac{sec ^{4} ( x )}{4}

Simplify

4 \cdot \frac{sec ^{4} ( x )}{4} : sec^{4} (x)

= sec^{4} (x)

Add a constant to the solution

= sec^{4} (x) + C

\frac{d}{d x} (\frac{x ^{5}}{5} + \frac{1}{12 x ^{3}})

x d y - (x^{2} - y^{2} + y) \cdot d x = 0

y = \frac{x + 2}{x - 2}, (3, 5)

\int \frac{8 x ^{3} + 3}{( x ^{2} + 2 ) ^{2}} d x

\frac{\partial}{\partial x} (sin^{2} (m x + n y))

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