Question

Find the value of the integral \(\displaystyle\int x^2\tan(x^3+2)\,dx\).

Hint:

Put u = x^3 + 2 so du = 3x^2 dx, then integrate tan u.

Answer 1

Step 1: Choose a substitution:
The argument of tan is \(x^3+2\), and its derivative \(3x^2\) is present (up to a constant) in the integrand as \(x^2\).
So let \(u = x^3 + 2\), which gives \(du = 3x^2\,dx\), or \(x^2\,dx = \dfrac{du}{3}\).

Step 2: Rewrite the integral in terms of u:
Substituting into the original integral:
\[ \int x^2\tan(x^3+2)\,dx = \int \tan u \cdot \frac{du}{3} = \frac{1}{3}\int \tan u \, du \]

Step 3: Integrate tan u:
Use the standard result \(\int \tan u\,du = \ln|\sec u| + C\).
\[ \frac{1}{3}\int \tan u\, du = \frac{1}{3}\ln|\sec u| + C \]

Step 4: Substitute back u = x^3+2:
Replace \(u\) by its original expression to get the final antiderivative.

Final Answer:
The integral evaluates to one third times log of sec of x cubed plus 2. \[ \boxed{\int x^2\tan(x^3+2)\,dx = \frac{1}{3}\ln\left|\sec(x^3+2)\right| + C} \]