What is the integral of sec^4(x ?

The integral of sec^4(x is (x^2sec^4(d))/2+C

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integral of sec^4(x

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

\frac{x ^{2} sec ^{4} ( d )}{2} + C

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

: \frac{x ^{2}}{2}

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Add a constant to the solution

= \frac{x ^{2} sec ^{4} ( d )}{2} + C

\frac{d h}{d t} = 4 - 16 t^{2}

g (x) = x^{2} + 3 x^{3}

\frac{d y}{d x} (x y) = 7

f (x) = \frac{x ^{2} - 7}{16 - x ^{2}}

\int lo g_{a} (x) d x