What is the integral from 0 to 1 of (x^2tan(x^3))/4 ?

The integral from 0 to 1 of (x^2tan(x^3))/4 is -1/12 ln(cos(1))

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integral from 0 to 1 of (x^2tan(x^3))/4

\int_{0}^{1} \frac{x ^{2} tan ( x ^{3} )}{4} d x

- \frac{1}{12} ln (cos (1))

Solution steps

\int_{0}^{1} \frac{x ^{2} tan ( x ^{3} )}{4} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int_{0}^{1} x^{2} tan (x^{3}) d x

Apply u-substitution

: \int_{0}^{1} \frac{tan ( u )}{3} d u

= \frac{1}{4} \cdot \int_{0}^{1} \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}{4} \cdot \frac{1}{3} \cdot \int_{0}^{1} tan (u) d u

Rewrite using trig identities

= \frac{1}{4} \cdot \frac{1}{3} \cdot \int_{0}^{1} \frac{sin ( u )}{cos ( u )} d u

Apply u-substitution

: \int_{1}^{c o s (1)} - \frac{1}{v} d v

= \frac{1}{4} \cdot \frac{1}{3} \cdot \int_{1}^{c o s (1)} - \frac{1}{v} d v

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= \frac{1}{4} \cdot \frac{1}{3} (- \int_{c o s (1)}^{1} - \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}{4} \cdot \frac{1}{3} (- (- \int_{c o s (1)}^{1} \frac{1}{v} d v))

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{4} \cdot \frac{1}{3} (- (- [ln ∣ v ∣]_{c o s (1)}^{1}))

Simplify

\frac{1}{4} \cdot \frac{1}{3} (- (- [ln ∣ v ∣]_{c o s (1)}^{1})) : \frac{1}{12} [ln ∣ v ∣]_{c o s (1)}^{1}

= \frac{1}{12} [ln ∣ v ∣]_{c o s (1)}^{1}

Compute the boundaries:

- ln (cos (1))

= \frac{1}{12} (- ln (cos (1)))

Simplify

= - \frac{1}{12} ln (cos (1))

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What is the integral from 0 to 1 of (x^2tan(x^3))/4 ?
The integral from 0 to 1 of (x^2tan(x^3))/4 is -1/12 ln(cos(1))