What is the integral of 4x^2sec(5x^3) ?

The integral of 4x^2sec(5x^3) is 4/15 ln|tan(5x^3)+sec(5x^3)|+C

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integral of 4x^2sec(5x^3)

\int 4 x^{2} sec (5 x^{3}) d x

\frac{4}{15} ln tan (5 x^{3}) + sec (5 x^{3}) + C

Solution steps

\int 4 x^{2} sec (5 x^{3}) d x

Take the constant out:

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

= 4 \cdot \int x^{2} sec (5 x^{3}) d x

Apply u-substitution

: \int \frac{u sec ( 5 u ^{\frac{3}{2}} )}{2 x} d u

= 4 \cdot \int \frac{sec ( 5 u )}{3} d u

Take the constant out:

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

= 4 \cdot \frac{1}{3} \cdot \int sec (5 u) d u

Apply u-substitution

: \int sec (v) \frac{1}{5} d v

= 4 \cdot \frac{1}{3} \cdot \int sec (v) \frac{1}{5} d v

Take the constant out:

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

= 4 \cdot \frac{1}{3} \cdot \frac{1}{5} \cdot \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= 4 \cdot \frac{1}{3} \cdot \frac{1}{5} ln ∣ tan (v) + sec (v) ∣

Substitute back

= 4 \cdot \frac{1}{3} \cdot \frac{1}{5} ln tan (5 x^{3}) + sec (5 x^{3})

Simplify

4 \cdot \frac{1}{3} \cdot \frac{1}{5} ln tan (5 x^{3}) + sec (5 x^{3}) : \frac{4}{15} ln tan (5 x^{3}) + sec (5 x^{3})

= \frac{4}{15} ln tan (5 x^{3}) + sec (5 x^{3})

Add a constant to the solution

= \frac{4}{15} ln tan (5 x^{3}) + sec (5 x^{3}) + C

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What is the integral of 4x^2sec(5x^3) ?
The integral of 4x^2sec(5x^3) is 4/15 ln|tan(5x^3)+sec(5x^3)|+C