Trig Integrals with Linear Inner Functions & arctan Forms (Intro Applications)

When integrating -cos x, we need to find a function whose derivative is -cos x. We know that the derivative of sin x is cos x. Using the constant multiple rule for differentiation, the derivative of -sin x is -cos x. Therefore, -sin x is an antiderivative of -cos x. Adding the constant of integration C gives us -sin x + C. To verify, we differentiate -sin x + C. The derivative of -sin x is -cos x, and the derivative of C is 0. This gives us -cos x + 0 = -cos x, which matches our original integrand. The other options are incorrect because differentiating them does not yield -cos x. The derivative of sin x + C is cos x, not -cos x. The derivative of -cos x + C is sin x, not -cos x. The derivative of cos x + C is -sin x, not -cos x.

This integral is of the form ∫ sin(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 ∫ sin(u) * (du/5) = (1/5) ∫ sin(u) du. The integral of sin(u) is -cos(u) + C. Therefore, we get (1/5) * (-cos(u)) + C = -1/5 cos(u) + C. Substituting back u = 5x - 3 gives -1/5 cos(5x - 3) + C. To verify, we differentiate -1/5 cos(5x - 3) + C. Using the chain rule, the derivative is -1/5 * (-sin(5x - 3)) * 5 = sin(5x - 3). The factor of 5 from the chain rule cancels with the -1/5 coefficient, confirming our answer is correct.

For integrals of sec²(ax + b), we use the substitution u = 2x + 7. The derivative du/dx = 2, so du = 2 dx, which means dx = du/2. Substituting into the integral gives ∫ sec²(u) * (du/2) = (½) ∫ sec²(u) du. We know that ∫ sec²(u) du = tan(u) + C. Therefore, we get (½) * tan(u) + C = (½) tan(u) + C. Substituting back u = 2x + 7 gives (½) tan(2x + 7) + C. To verify, we differentiate (½) tan(2x + 7) + C. Using the chain rule, the derivative is (½) * sec²(2x + 7) * 2 = sec²(2x + 7), which exactly matches our integrand. The coefficient of 1/2 is crucial to cancel the factor of 2 from the chain rule.

We use substitution for this composite function. Let u = 1/2 x + 4. Then du/dx = 1/2, so du = (½) dx, which means dx = 2 du. Substituting into the integral gives ∫ cos(u) * (2 du) = 2 ∫ cos(u) du. The integral of cos(u) is sin(u) + C. Therefore, we get 2 * sin(u) + C = 2 sin(u) + C. Substituting back u = 1/2 x + 4 gives 2 sin(1/2 x + 4) + C. To verify, we differentiate 2 sin(1/2 x + 4) + C. Using the chain rule, the derivative is 2 * cos(1/2 x + 4) * (½) = cos(1/2 x + 4), which matches our integrand. The coefficient 2 compensates for the 1/2 factor in the chain rule.

First, we complete the square for the denominator: x² + 4x + 13 = (x² + 4x + 4) + 9 = (x + 2)² + 9 = (x + 2)² + 3². Now the integral is ∫ dx/[(x + 2)² + 3²]. Let u = x + 2, then du = dx. The integral becomes ∫ du/(u² + 3²). This matches the standard form ∫ du/(a² + u²) = (1/a) arctan(u/a) + C, where a = 3. Applying this formula gives (⅓) arctan(u/3) + C. Substituting back u = x + 2 gives (⅓) arctan((x + 2)/3) + C. To verify, we differentiate this result using the chain rule: d/dx[(⅓) arctan((x + 2)/3)] = (⅓) * 1/[1 + ((x + 2)/3)²] * (⅓) = 1/[(x + 2)² + 9] = 1/(x² + 4x + 13), confirming our answer.

This integral is of the form ∫ f'(x) sec² f(x) dx. Notice that the derivative of (5x - 1) is 5, which appears in the integrand. Let u = 5x - 1. Then du/dx = 5, so du = 5 dx. Our integral has exactly 5 dx, so we can substitute: ∫ sec²(u) du. The integral of sec²(u) is tan(u) + C. Substituting back u = 5x - 1 gives tan(5x - 1) + C. To verify, we differentiate tan(5x - 1) + C. Using the chain rule, the derivative is sec²(5x - 1) * 5 = 5 sec²(5x - 1), which matches our integrand. The coefficient 5 is absorbed into the derivative of the inner function, making the antiderivative simpler than one might expect.

This integral demonstrates the pattern ∫ f'(x) cos f(x) dx. Let u = x² + 1. Then du/dx = 2x, so du = 2x dx. Our integral has x dx, not 2x dx. We can rewrite x dx as (½) * 2x dx = (½) du. Substituting gives ∫ cos(u) * (½) du = (½) ∫ cos(u) du. The integral of cos(u) is sin(u) + C. Therefore, we get (½) * sin(u) + C = (½) sin(u) + C. Substituting back u = x² + 1 gives (½) sin(x² + 1) + C. To verify, we differentiate (½) sin(x² + 1) + C. Using the chain rule, the derivative is (½) * cos(x² + 1) * 2x = x cos(x² + 1), confirming our answer.

This integral shows the pattern ∫ f'(x) sin f(x) dx with a negative sign. Let u = x³. Then du/dx = 3x², so du = 3x² dx. Our integral has -3x² dx, which is -du. Substituting gives ∫ sin(u) * (-du) = -∫ sin(u) du. The integral of sin(u) is -cos(u) + C. Therefore, -∫ sin(u) du = -(-cos(u)) + C = cos(u) + C. Substituting back u = x³ gives cos(x³) + C. To verify, we differentiate cos(x³) + C. Using the chain rule, the derivative is -sin(x³) * 3x² = -3x² sin(x³), which matches our integrand. The negative signs cancel appropriately, leaving a simple result.

This integral demonstrates ∫ f'(x) [f(x)]ⁿ dx. Notice that sec² x is the derivative of tan x. Let u = tan x. Then du/dx = sec² x, so du = sec² x dx. Our integral becomes ∫ u³ du. The integral of u³ is (1/4) u⁴ + C. Substituting back u = tan x gives (1/4) tan⁴ x + C. To verify, we differentiate (1/4) tan⁴ x + C. Using the chain rule, the derivative is (1/4) * 4 tan³ x * sec² x = tan³ x sec² x, which matches our integrand. This shows how recognizing that sec² x is the derivative of tan x allows us to treat this as a simple power rule integration.

This is a conceptual question about equivalent forms of antiderivatives. We can solve ∫ sin x cos x dx in two ways. First, let u = sin x, then du = cos x dx. The integral becomes ∫ u du = (½) u² + C = (½) sin² x + C. Second, let u = cos x, then du = -sin x dx, so -du = sin x dx. The integral becomes ∫ (-du) * u = -∫ u du = -(½) u² + C = -(½) cos² x + C. These appear different, but they differ only by a constant: (½) sin² x = (½)(1 - cos² x) = (½) - (½) cos² x. The constant (½) can be absorbed into C. Therefore, both forms represent the same family of antiderivatives. This illustrates that antiderivatives can look different but be equivalent up to an additive constant.

The substitution u = ax + b is most directly applicable when the integrand is a composite function whose inner function is linear (ax + b) and the outer function is a basic trigonometric function. In ∫ sin(4x + 1) dx, the inner function is 4x + 1 and the outer function is sin. Letting u = 4x + 1 transforms the integral to ∫ sin(u) du (up to a constant factor), which is a basic integral. Option A requires integration by parts because we have x multiplied by a composite trig function. Option C has two different trig functions multiplied together, requiring a different approach. Option D has the linear factor outside the trig function. The key is recognizing when the inner function is linear and the outer function is basic, making u = ax + b the appropriate substitution.

The integral ∫ dx/(x² + 9) is already in the standard arctan form ∫ dx/(a² + x²) with a = 3, giving (⅓) arctan(x/3) + C. However, ∫ dx/(x² + 6x + 10) has a quadratic denominator that isn't a perfect sum of squares. We must complete the square: x² + 6x + 10 = (x² + 6x + 9) + 1 = (x + 3)² + 1². This transforms it to ∫ dx/[(x + 3)² + 1²], which matches the arctan form after substituting u = x + 3. Option B is incorrect because both are arctan forms after transformation. Option C is incorrect because one requires completing the square while the other doesn't. Option D is false as both are integrable. Recognizing when algebraic manipulation is needed is a key skill in integration.

This integral is of the form ∫ f'(x) cos f(x) dx. Notice that eˣ is the derivative of eˣ. Let u = eˣ. Then du/dx = eˣ, which means du = eˣ dx. Our integral contains eˣ dx, so we can substitute directly: ∫ cos(u) du. The integral of cos(u) is sin(u) + C. Substituting back u = eˣ gives sin(eˣ) + C. To verify, we differentiate sin(eˣ) + C. Using the chain rule, the derivative is cos(eˣ) * eˣ = eˣ cos(eˣ), which matches our integrand. This is a classic example where the derivative of the inner function (eˣ) is present, enabling a straightforward substitution. Options B, C, and D are less efficient or incorrect approaches.

Displacement is the definite integral of velocity: ∫ from 0 to π of 3 sin(2t) dt. First, find the antiderivative: ∫ 3 sin(2t) dt = 3 (-½) cos(2t) + C = -3/2 cos(2t) + C. Now evaluate from 0 to π: [-3/2 cos(2t)] from 0 to π = -3/2 cos(2π) - (-3/2 cos(0)) = -3/2 * 1 + 3/2 * 1 = -3/2 + 3/2 = 0. The displacement is zero because the particle moves forward and backward symmetrically over this interval. The positive and negative areas under the velocity curve cancel out. This demonstrates that definite integrals represent net change, not total distance traveled. The total distance would be the integral of the absolute value of velocity, which would be positive.

We can solve this using the constant multiple rule and substitution. First, factor out the constant 4: 4 ∫ cos(3x) dx. For ∫ cos(3x) dx, let u = 3x. Then du = 3 dx, so dx = du/3. Substituting gives ∫ cos(u) (du/3) = (⅓) ∫ cos(u) du = (⅓) sin(u) + C. Substituting back u = 3x gives (⅓) sin(3x) + C. Now multiply by the constant 4 we factored out earlier: 4 (⅓) sin(3x) + C = (4/3) sin(3x) + C. To verify, we differentiate (4/3) sin(3x) + C. The derivative is (4/3) cos(3x) * 3 = 4 cos(3x), which matches our integrand. The chain rule multiplies by 3, which cancels with the 1/3 factor, leaving the original coefficient 4.