What is the integral of (21x^2)/(sec(x^3-3)) ?

The integral of (21x^2)/(sec(x^3-3)) is 7sin(x^3-3)+C

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integral of (21x^2)/(sec(x^3-3))

\int \frac{21 x ^{2}}{sec ( x ^{3} - 3 )} d x

7 sin (x^{3} - 3) + C

Solution steps

\int \frac{21 x ^{2}}{sec ( x ^{3} - 3 )} d x

Take the constant out:

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

= 21 \cdot \int \frac{x ^{2}}{sec ( x ^{3} - 3 )} d x

Use the basic trigonometric identity:

\frac{1}{sec ( x )} = cos (x)

= 21 \cdot \int x^{2} cos (x^{3} - 3) d x

Apply u-substitution

: \int \frac{cos ( u )}{3} d u

= 21 \cdot \int \frac{cos ( u )}{3} d u

Take the constant out:

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

= 21 \cdot \frac{1}{3} \cdot \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= 21 \cdot \frac{1}{3} sin (u)

Substitute back

u = x^{3} - 3

= 21 \cdot \frac{1}{3} sin (x^{3} - 3)

Simplify

21 \cdot \frac{1}{3} sin (x^{3} - 3) : 7 sin (x^{3} - 3)

= 7 sin (x^{3} - 3)

Add a constant to the solution

= 7 sin (x^{3} - 3) + C

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What is the integral of (21x^2)/(sec(x^3-3)) ?
The integral of (21x^2)/(sec(x^3-3)) is 7sin(x^3-3)+C