What is the integral of 3xsec^2(4x^2) ?

The integral of 3xsec^2(4x^2) is 3/8 tan(4x^2)+C

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integral of 3xsec^2(4x^2)

\int 3 x sec^{2} (4 x^{2}) d x

\frac{3}{8} tan (4 x^{2}) + C

Solution steps

\int 3 x sec^{2} (4 x^{2}) d x

Take the constant out:

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

= 3 \cdot \int x sec^{2} (4 x^{2}) d x

Apply u-substitution

: \int \frac{sec ^{2} ( u )}{8} d u

= 3 \cdot \int \frac{sec ^{2} ( u )}{8} d u

Take the constant out:

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

= 3 \cdot \frac{1}{8} \cdot \int sec^{2} (u) d u

Use the common integral:

\int sec^{2} (u) d u = tan (u)

= 3 \cdot \frac{1}{8} tan (u)

Substitute back

u = 4 x^{2}

= 3 \cdot \frac{1}{8} tan (4 x^{2})

Simplify

3 \cdot \frac{1}{8} tan (4 x^{2}) : \frac{3}{8} tan (4 x^{2})

= \frac{3}{8} tan (4 x^{2})

Add a constant to the solution

= \frac{3}{8} tan (4 x^{2}) + C

\int e^{x} + x^{2} + x^{3} d x

\int_{0}^{1} t e^{- s t} d t

\frac{d}{d x} (60 x^{4} + 180 x^{3} - 240 x^{2})

f (x) = (- 2 x^{2} - 4)^{3}

\frac{1}{1 + s}

What is the integral of 3xsec^2(4x^2) ?
The integral of 3xsec^2(4x^2) is 3/8 tan(4x^2)+C