Advanced Trig Integrals: Chain Rule Patterns, Composites & Mixed Techniques

The integral of sin(x) is -cos(x) + C. When we have a constant multiple like 3sin(x), we multiply the antiderivative by that constant. So the integral becomes 3 × (-cos(x)) + C, which simplifies to -3cos(x) + C. The constant of integration C must be included since this is an indefinite integral.

The integral of cos(x) is sin(x) + C. However, when we have cos(ax) where a is a constant other than 1, we need to divide by that constant. This is because when we differentiate (1/a)sin(ax), we get cos(ax) by the chain rule. Here a = 4, so the integral is (1/4)sin(4x) + C.

The integral of sec²(x) is tan(x) + C. When we have sec²(ax+b), we divide by the coefficient a. This accounts for the chain rule in differentiation. The derivative of (½)tan(2x+3) is (½) × 2sec²(2x+3) = sec²(2x+3), which confirms our result. Therefore, the correct antiderivative is (½)tan(2x+3) + C.

The integral of cos(x) is sin(x) + C. For cos(ax+b), we divide by a. For cos(3x-π), a = 3, so we divide by 3. We also have a constant factor of 5. Therefore, the integral is 5 × (⅓)sin(3x-π) + C, which equals (5/3)sin(3x-π) + C. Note that shifting by π doesn't affect the integration process.

The integral of sin(x) is -cos(x) + C. When we have sin(ax+b), we divide by a. For sin(5x+1), a = 5, so we need (1/5) multiplied by our result. The constant factor 2 gives us 2 × (-1/5)cos(5x+1) + C, which simplifies to (-2/5)cos(5x+1) + C.

First, we find f'(x) by differentiating f(x) = cos(2x). Using the chain rule, f'(x) = -2sin(2x). Substituting, we get ∫(-2sin(2x))sin(2x) dx = ∫(-2sin²(2x)) dx. Using the identity sin²(θ) = (1-cos(2θ))/2, we have ∫(-2 × (1-cos(4x))/2) dx = ∫(-1 + cos(4x)) dx = -x + (1/4)sin(4x) + C. However, since f(x) = cos(2x), we can write this as (-½)cos²(2x) + C by recognizing that cos(2x) = 2cos²(x) - 1.

This requires substitution. Let u = x², then du = 2x dx, so (½)du = x dx. The integral becomes ∫4 × (½)cos(u) du = ∫2cos(u) du = 2sin(u) + C. Substituting back u = x² gives us 2sin(x²) + C. Notice that 4x dx becomes 2 × 2x dx, and the 2x dx matches perfectly with our substitution.

We can use substitution here. Let u = tan(5x), then du = 5sec²(5x) dx, so (1/5)du = sec²(5x) dx. The integral becomes ∫u × (1/5)du = (1/5)∫u du = (1/5) × (u²/2) + C = (1/10)u² + C. Substituting back u = tan(5x) gives us (1/10)tan²(5x) + C. We can verify this by differentiating: d/dx[(1/10)tan²(5x)] = (1/10) × 2tan(5x) × 5sec²(5x) = tan(5x)sec²(5x).

We integrate term by term. The integral of 3sin(x) is 3 × (-cos(x)) = -3cos(x). The integral of 2cos(x) is 2 × sin(x) = 2sin(x). Adding these together with the constant of integration gives us -3cos(x) + 2sin(x) + C.

The integral of sec²(x) is tan(x) + C. When we have sec²(kx) where k is a constant, we divide by k. Here k = 1/4, so we divide by 1/4, which means multiplying by 4. The (1/4) factor and the division by (1/4) cancel each other. Therefore, the integral is simply tan(x/4) + C.

This requires integration by parts. Let u = 2x+3 and dv = sin(2x) dx. Then du = 2 dx and v = -(½)cos(2x). Using the integration by parts formula ∫u dv = uv - ∫v du, we get: (2x+3)(-½)cos(2x) - ∫(-½)cos(2x) × 2 dx = -(½)(2x+3)cos(2x) + ∫cos(2x) dx = -(½)(2x+3)cos(2x) + (½)sin(2x) + C, which simplifies to -(x+3/2)cos(2x) + (½)sin(2x) + C.

The integral of sec²(x) is tan(x) + C. For sec²(ax+b), we divide by a. Here a = 3, and we have a constant factor of 6. So the integral is 6 × (⅓)tan(3x-π/2) + C = 2tan(3x-π/2) + C. The phase shift by π/2 doesn't affect the integration process, and the constant factor and the division work together to give us the final result.

We can use the double-angle identity sin(2θ) = 2sin(θ)cos(θ), but here we need cos(2x)sin(2x). Using the identity sin(2θ) = 2sin(θ)cos(θ), we can write cos(2x)sin(2x) = (½)sin(4x). So the integral becomes ∫(½)sin(4x) dx = (½) × (-1/4)cos(4x) + C = -(1/8)cos(4x) + C. However, using the substitution u = sin(2x), du = 2cos(2x) dx, we get (½)∫u du = (½) × (u²/2) + C = (1/4)sin²(2x) + C. Both methods give equivalent results.

First, we find f'(x) by differentiating f(x) = 2x, which gives f'(x) = 2. Substituting into the integral, we get ∫2sec²(2x) dx. The integral of sec²(2x) is (½)tan(2x), so we have 2 × (½)tan(2x) + C = tan(2x) + C. This type of integral, ∫f'(x)sec²(f(x)) dx, follows the pattern where the antiderivative is tan(f(x)) + C.

This requires substitution. Let u = x²+2x+5, then du = (2x+2) dx = 2(x+1) dx. Therefore, (x+1) dx = (½)du. The integral becomes ∫sec²(u) × (½) du = (½)∫sec²(u) du = (½)tan(u) + C. Substituting back u = x²+2x+5 gives us (½)tan(x²+2x+5) + C. The pattern ∫f'(x)sec²(f(x)) dx = tan(f(x)) + C is at work here.