Question

∫(1/cos^2⁡x + 6x^2 )dx=...

Answer 1

(tg x)' = 1/cos²x

(xⁿ)' = nx^n-1

∫(1/cos²x + 2•3x²)dx = tg x + 2x³ + C

Answer 2

To solve the integral ∫(1/cos^2(x) + 6x^2) dx, we'll break it down into two separate integrals:

∫(1/cos^2(x)) dx + ∫(6x^2) dx

Let's solve each integral step by step:

1. ∫(1/cos^2(x)) dx: This is the integral of the secant squared function. The integral of sec^2(x) is a common trigonometric integral and can be solved using the tangent half-angle substitution or simply by recognizing its derivative:

∫(1/cos^2(x)) dx = ∫sec^2(x) dx = tan(x) + C₁

2. ∫(6x^2) dx: This is a simple polynomial integral. The integral of x^n with respect to x is (x^(n+1))/(n+1). Applying this rule:

∫(6x^2) dx = 2x^3 + C₂

Now we have both integrals in terms of their antiderivatives. The final result is the sum of these two antiderivatives:

∫(1/cos^2(x) + 6x^2) dx = tan(x) + 2x^3 + C

where C = C₁ + C₂ is the constant of integration.