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

The integral of tan(x^3)x^2 is -1/3 ln|cos(x^3)|+C

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integral of tan(x^3)x^2

\int tan (x^{3}) x^{2} d x

- \frac{1}{3} ln cos (x^{3}) + C

Solution steps

\int tan (x^{3}) x^{2} d x

Apply u-substitution

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

= \int \frac{tan ( u )}{3} d u

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

: \int - \frac{1}{v} d v

= \frac{1}{3} \cdot \int - \frac{1}{v} d v

Take the constant out:

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

= \frac{1}{3} (- \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{3} (- ln ∣ v ∣)

Substitute back

= \frac{1}{3} (- ln cos (x^{3}))

Simplify

= - \frac{1}{3} ln cos (x^{3})

Add a constant to the solution

= - \frac{1}{3} ln cos (x^{3}) + C

\frac{9 x}{e ^{x}}

x \to \infty lim (x + a)

f (t) = \frac{7}{t}

y^{^{'}} + 3 x y^{3} = 0, y (0) = 1

(x + \frac{1}{x})^{5}

What is the integral of tan(x^3)x^2 ?
The integral of tan(x^3)x^2 is -1/3 ln|cos(x^3)|+C