What is the integral of 3sec^4(x) ?

The integral of 3sec^4(x) is 3tan(x)+tan^3(x)+C

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integral of 3sec^4(x)

\int 3 sec^{4} (x) d x

3 tan (x) + tan^{3} (x) + C

Solution steps

\int 3 sec^{4} (x) d x

Take the constant out:

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

= 3 \cdot \int sec^{4} (x) d x

Rewrite using trig identities

= 3 \cdot \int (1 + tan^{2} (x)) sec^{2} (x) d x

Apply u-substitution

: \int 1 + u^{2} d u

= 3 \cdot \int 1 + u^{2} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 (\int 1 d u + \int u^{2} d u)

\int 1 d u = u

\int u^{2} d u = \frac{u ^{3}}{3}

= 3 (u + \frac{u ^{3}}{3})

Substitute back

u = tan (x)

= 3 (tan (x) + \frac{tan ^{3} ( x )}{3})

Simplify

3 (tan (x) + \frac{tan ^{3} ( x )}{3}) : 3 tan (x) + tan^{3} (x)

= 3 tan (x) + tan^{3} (x)

Add a constant to the solution

= 3 tan (x) + tan^{3} (x) + C

x \to \frac{3 π}{4} lim (- 2 sec (x))

\frac{d x}{d t} = 6 \cdot 0.3 - 5 \cdot \frac{x}{100 + t}, x (0) = 0

f (x) = 2 \cdot 7^{2 x^{2} + 3}

\int \frac{( x ^{3} )}{x ^{2} + 25} d x

\frac{\partial}{\partial t} (s cos (t))

What is the integral of 3sec^4(x) ?
The integral of 3sec^4(x) is 3tan(x)+tan^3(x)+C