Basic Trig Antiderivatives & u-Substitution (sin, cos, sec², ax+b)

To find the integral of sin x, we need to determine which function has sin x as its derivative. Let's test each candidate by differentiating it. The derivative of cos x is -sin x. The derivative of -cos x is -(-sin x) which equals sin x. The derivative of sin x is cos x, not sin x. The derivative of -sin x is -cos x, not sin x. Therefore, -cos x is the antiderivative of sin x. The constant C is added because the derivative of any constant is zero, so any constant could be present in the original function. To verify, we differentiate -cos x + C, which gives us sin x + 0 = sin x, confirming our answer is correct.

Integration is the reverse process of differentiation. We need to find a function whose derivative is cos x. When we differentiate sin x with respect to x, we get cos x. This tells us that sin x is an antiderivative of cos x. The constant C is included in indefinite integrals because constants disappear during differentiation. To check our answer, we differentiate sin x + C, which gives cos x + 0 = cos x. This matches the integrand, confirming that the integral of cos x dx is sin x + C. The other options are incorrect because differentiating them does not yield cos x.

We recall from differentiation rules that the derivative of tan x is sec² x. Since integration reverses differentiation, the antiderivative of sec² x must be tan x. The constant C is added because any constant would be lost during differentiation. To verify this result, we differentiate tan x + C. The derivative of tan x is sec² x, and the derivative of C is 0. This gives us sec² x + 0 = sec² x, which matches our original integrand. The other options are derivatives of different functions: sec x tan x is the derivative of sec x, sec x is the derivative of ln|sec x + tan x|, and cot x has a derivative of -csc² x.

For integrals of the form sin(ax + b), we use substitution. Let u = 3x + 2, then du/dx = 3, which means du = 3 dx. Our original integral has dx, not 3 dx, so we solve for dx = du/3. Substituting gives us ∫ sin(u) * (du/3) = (⅓) ∫ sin(u) du. The integral of sin(u) is -cos(u) + C. So we get (⅓) * (-cos(u)) + C = -1/3 cos(u) + C. Substituting back u = 3x + 2 gives -1/3 cos(3x + 2) + C. We can verify this by differentiating: the derivative of -1/3 cos(3x + 2) is -1/3 * (-sin(3x + 2)) * 3 = sin(3x + 2). The factor of 3 comes from the chain rule, which cancels with the -1/3 factor.

We use the substitution method for this integral. Let u = 4x - 1, then the derivative du/dx = 4, which means du = 4 dx. Solving for dx gives dx = du/4. Substituting into the integral, we get ∫ cos(u) * (du/4) = (1/4) ∫ cos(u) du. The integral of cos(u) is sin(u) + C. Therefore, we have (1/4) * sin(u) + C = (1/4) sin(u) + C. Substituting back u = 4x - 1 gives (1/4) sin(4x - 1) + C. To verify, we differentiate (1/4) sin(4x - 1) + C. Using the chain rule, the derivative is (1/4) * cos(4x - 1) * 4 = cos(4x - 1), which matches our integrand.

This integral follows the pattern ∫ sec²(ax + b) dx. We use substitution with u = 5x + 3. Then du/dx = 5, so du = 5 dx, which means dx = du/5. Substituting gives ∫ sec²(u) * (du/5) = (1/5) ∫ sec²(u) du. We know the integral of sec²(u) is tan(u) + C. Therefore, we get (1/5) * tan(u) + C = (1/5) tan(u) + C. Substituting back u = 5x + 3 gives (1/5) tan(5x + 3) + C. Verification: differentiate (1/5) tan(5x + 3) + C. The derivative is (1/5) * sec²(5x + 3) * 5 = sec²(5x + 3), which matches the original integrand.

The integral of dx/(a² + x²) follows the standard formula (1/a) arctan(x/a) + C. Here, x² + 9 matches the form x² + 3², where a = 3. Applying the formula directly gives (⅓) arctan(x/3) + C. To derive this, we can factor out 9 from the denominator: ∫ dx/(9 * (1 + x²/9)) = (1/9) ∫ dx/(1 + (x/3)²). Now substitute u = x/3, so du = dx/3, which means dx = 3 du. The integral becomes (1/9) * ∫ 3 du/(1 + u²) = (⅓) ∫ du/(1 + u²) = (⅓) arctan(u) + C = (⅓) arctan(x/3) + C. The other options have incorrect coefficients or arguments.

This integral matches the pattern ∫ f'(x) cos f(x) dx. Notice that the derivative of x² is 2x, which appears in the integrand. Let u = x², then du/dx = 2x, so du = 2x dx. Our integral has exactly 2x dx, so we can substitute: ∫ cos(u) du. The integral of cos(u) is sin(u) + C. Substituting back u = x² gives sin(x²) + C. To verify, we differentiate sin(x²) + C. Using the chain rule, the derivative is cos(x²) * 2x = 2x cos(x²), which matches our integrand. This shows the power of recognizing when the derivative of the inner function is present in the integral.

This integral follows the pattern ∫ f'(x) sin f(x) dx (with a negative sign adjustment). Let u = cos x. Then du/dx = -sin x, which means du = -sin x dx. Our integral contains -sin x dx, so we can substitute directly: ∫ cos(u) du. The integral of cos(u) is sin(u) + C. Substituting back u = cos x gives sin(cos x) + C. To verify, we differentiate sin(cos x) + C. Using the chain rule, the derivative is cos(cos x) * (-sin x) = -sin x cos(cos x), which matches our integrand. The key insight is recognizing that -sin x is the derivative of cos x, making this a straightforward substitution problem.

This integral can be solved by recognizing that sec² x is the derivative of tan x. Let u = tan x, then du/dx = sec² x, which means du = sec² x dx. Our integral is ∫ sec² x tan x dx = ∫ u du. The integral of u du is (½) u² + C. Substituting back u = tan x gives (½) tan² x + C. To verify, we differentiate (½) tan² x + C. Using the chain rule, the derivative is (½) * 2 tan x * sec² x = tan x sec² x, which matches our integrand. Alternatively, we could use integration by parts, but substitution is more efficient here.

The fundamental theorem of calculus tells us that integration and differentiation are inverse operations. If we start with a function f(x), differentiate it to get f'(x), and then integrate f'(x), we return to the original function f(x) (plus a constant C). For example, if f(x) = sin x, then f'(x) = cos x. Integrating cos x gives us sin x + C. Option A is incorrect because integration of sin x gives -cos x, not cos x. Option C is incorrect because the derivative of ∫ sin x dx = -cos x + C is sin x, not -cos x. Option D is incorrect because the operations are directly related as inverses. This fundamental relationship is why we can verify integrals by differentiation.

The substitution u = ax + b is used when we have a composite function where the inner function is linear (of the form ax + b). In ∫ sin(2x + 5) dx, the inner function is 2x + 5, which is linear. Setting u = 2x + 5 simplifies the integral to a basic form. Option A requires integration by parts because we have x multiplied by sin x (a product of two different function types). Option B also requires integration by parts for the same reason. Option D can be solved using the double-angle identity or substitution with u = sin x or u = cos x, but the inner function is not linear. Therefore, ∫ sin(2x + 5) dx is the only one that specifically calls for the u = ax + b substitution method.

The standard arctan integration formula is ∫ du/(a² + u²) = (1/a) arctan(u/a) + C. The denominator x² + 2x + 5 doesn't match this form directly. By completing the square, we rewrite it as (x² + 2x + 1) + 4 = (x + 1)² + 2². Now it matches the form a² + u² where a = 2 and u = x + 1. This allows us to apply the arctan formula. Option A is partially true but not the main reason. Option C is incorrect because x² + 2x + 5 doesn't factor over the reals. Option D is incorrect because partial fractions require a factorable denominator. The key insight is recognizing that completing the square transforms the expression into a form that fits a known integration pattern.

The integral ∫ 3x² sin(x³) dx is a perfect candidate for substitution because it contains both a composite function sin(x³) and what appears to be the derivative of its inner function. Let u = x³. Then du/dx = 3x², which means du = 3x² dx. The 3x² factor in our integral is exactly what we need for the substitution du = 3x² dx. This allows us to rewrite the integral as ∫ sin(u) du. Without the 3x² factor (or some multiple of it), we couldn't perform this direct substitution. Option A is incorrect because C represents the constant of integration. Option C is incorrect because we don't remove it; we use it for substitution. Option D is incorrect because integration by parts isn't needed when the derivative of the inner function is present.

Trigonometric functions describe periodic phenomena like waves and oscillations. In physics, simple harmonic motion (like a mass on a spring or a pendulum) has velocity described by v(t) = A sin(ωt + φ) or v(t) = A cos(ωt + φ). The displacement (position) is the integral of velocity. Therefore, ∫ v(t) dt = ∫ A sin(ωt + φ) dt gives us the position function. The area under a velocity-time graph represents displacement, and when velocity is described by a trig function, we must integrate trig functions to find it. Option A involves finding extrema, which uses differentiation, not integration. Option C involves algebra, not calculus. Option D is also algebraic factoring. The connection between trigonometric integration and oscillatory motion in physics is a fundamental application that demonstrates why these techniques are important.