What is the integral of (4x^2)/(sin(x^3-5)) ?

The integral of (4x^2)/(sin(x^3-5)) is 4/3 ln|tan(1/2 (x^3-5))|+C

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integral of (4x^2)/(sin(x^3-5))

\int \frac{4 x ^{2}}{sin ( x ^{3} - 5 )} d x

\frac{4}{3} ln tan (\frac{1}{2} (x^{3} - 5)) + C

Solution steps

\int \frac{4 x ^{2}}{sin ( x ^{3} - 5 )} d x

Take the constant out:

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

= 4 \cdot \int \frac{x ^{2}}{sin ( x ^{3} - 5 )} d x

Apply u-substitution

: \int \frac{1}{3 sin ( u )} d u

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

Apply u-substitution

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

= 4 \cdot \frac{1}{3} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 4 \cdot \frac{1}{3} ln ∣ v ∣

Substitute back

= 4 \cdot \frac{1}{3} ln tan (\frac{x ^{3} - 5}{2})

Simplify

4 \cdot \frac{1}{3} ln tan (\frac{x ^{3} - 5}{2}) : \frac{4}{3} ln tan (\frac{1}{2} (x^{3} - 5))

= \frac{4}{3} ln tan (\frac{1}{2} (x^{3} - 5))

Add a constant to the solution

= \frac{4}{3} ln tan (\frac{1}{2} (x^{3} - 5)) + C

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What is the integral of (4x^2)/(sin(x^3-5)) ?
The integral of (4x^2)/(sin(x^3-5)) is 4/3 ln|tan(1/2 (x^3-5))|+C