Integration of Trig Composites: Identities, Chain Rule & Integration by Parts (ax+b)

The integral of sin(x) is -cos(x) + C. For sin(ax+b), we divide by a. Here a = 2, so we divide by 2. With the constant factor 7, we have 7 × (-½)cos(2x+π/3) + C, which simplifies to (-7/2)cos(2x+π/3) + C. The phase shift by π/3 doesn't affect the integration constant, but it does appear in the trigonometric function.

The integral of cos(x) is sin(x) + C. For cos(ax+b), we divide by a. Here a = 5, and we have a constant factor of 3. Therefore, the integral is 3 × (1/5)sin(5x-2) + C = (3/5)sin(5x-2) + C. This follows the standard formula for integrating cos(ax+b).

The integral of sec²(x) is tan(x) + C. For sec²(ax+b), we divide by a. Here a = 4, and we have a constant factor of 1/3. So the integral is (⅓) × (1/4)tan(4x+1) + C = (1/12)tan(4x+1) + C. This follows the pattern where the constant factor and the division by a are applied together.

This requires substitution. Let u = x²-x+3, then du = (2x-1) dx. The integral becomes ∫sin(u) du = -cos(u) + C. Substituting back u = x²-x+3 gives us -cos(x²-x+3) + C. This follows the pattern ∫f'(x)sin(f(x)) dx = -cos(f(x)) + C.

First, we find f'(x) by differentiating f(x) = sin(3x), which gives f'(x) = 3cos(3x). Substituting into the integral, we get ∫3cos(3x) × cos(3x) dx = ∫3cos²(3x) dx. Using the identity cos²(θ) = (1+cos(2θ))/2, we have ∫3(1+cos(6x))/2 dx = (3/2)∫(1+cos(6x)) dx = (3/2)(x + (1/6)sin(6x)) + C = (3/2)x + (1/4)sin(6x) + C. Alternatively, using the identity sin(2θ) = 2sin(θ)cos(θ), we can write this as (1/6)sin²(3x) + C.

This requires substitution. Let u = x³+3x, then du = (3x²+3) dx = 3(x²+1) dx. Therefore, (x²+1) dx = (⅓)du. The integral becomes ∫cos(u) × (⅓) du = (⅓)sin(u) + C. Substituting back u = x³+3x gives us (⅓)sin(x³+3x) + C. This follows the pattern ∫f'(x)cos(f(x)) dx = sin(f(x)) + C.

Using the double-angle identity sin(2θ) = 2sin(θ)cos(θ), we can write sin(4x)cos(4x) = (½)sin(8x). So the integral becomes ∫(½)sin(8x) dx = (½) × (-1/8)cos(8x) + C = -(1/8)cos(8x) + C. This approach uses the product-to-sum identity to simplify the integral before applying the standard integration formula.

The integral of sec²(x) is tan(x) + C. For sec²(kx) where k is a constant, we divide by k. Here k = 3/2, so we divide by 3/2, which means multiplying by 2/3. With the constant factor 5, we have 5 × (⅔)tan(3x/2) + C = (10/3)tan(3x/2) + C. This follows the formula for integrating sec²(kx) with a constant multiplier.

This requires integration by parts twice. First, let u = (2x+3)² and dv = sin(2x+3) dx. Then du = 2(2x+3) × 2 dx = 4(2x+3) dx and v = -cos(2x+3)/2. Using integration by parts: ∫u dv = uv - ∫v du = (2x+3)²(-cos(2x+3)/2) - ∫(-cos(2x+3)/2) × 4(2x+3) dx = -(2x+3)²cos(2x+3)/2 + 2∫(2x+3)cos(2x+3) dx. Now apply integration by parts again to the remaining integral with u = 2x+3 and dv = cos(2x+3) dx, giving us du = 2 dx and v = sin(2x+3)/2. This results in -(2x+3)²cos(2x+3)/2 + (2x+3)sin(2x+3)/2 + C.

First, we find f'(x) by differentiating f(x) = x², which gives f'(x) = 2x. Substituting into the integral, we get ∫2x sin(2x²) dx. This requires substitution. Let u = 2x², then du = 4x dx, so (½)du = 2x dx. The integral becomes ∫sin(u) × (½) du = (½)(-cos(u)) + C = (-½)cos(u) + C = (-½)cos(2x²) + C. This type of integral follows the pattern ∫f'(x)g(f(x)) dx, where we can use substitution with g being any function.

This requires substitution. Let u = x²-x, then du = (2x-1) dx = -(1-2x) dx. Therefore, (1-2x) dx = -du. The integral becomes ∫sec²(u) × (-du) = -∫sec²(u) du = -tan(u) + C. Substituting back u = x²-x gives us -tan(x²-x) + C. This follows the pattern ∫f'(x)sec²(f(x)) dx = tan(f(x)) + C, with an additional negative sign from the substitution.

This requires substitution. Let u = x²+1, then du = 2x dx, so 2x dx = du. The integral becomes ∫2 × sec²(u) du = 2∫sec²(u) du = 2tan(u) + C. Substituting back u = x²+1 gives us 2tan(x²+1) + C. This follows the pattern ∫f'(x)sec²(f(x)) dx = tan(f(x)) + C, with the constant factor 2 being incorporated appropriately.

Using the identity cos²(θ) = (1+cos(2θ))/2, we have cos²(3x) = (1+cos(6x))/2. Therefore, the integral becomes ∫(1+cos(6x))/2 dx = (½)∫dx + (½)∫cos(6x) dx = (½)x + (½)(1/6)sin(6x) + C = x/2 + sin(6x)/12 + C. This uses the power-reduction identity to convert the square of cosine into a form that's easier to integrate.

This requires substitution. Let u = x³+3x²+3x+1, then du = (3x²+6x+3) dx. The integral becomes ∫cos(u) du = sin(u) + C. Substituting back u = x³+3x²+3x+1 gives us sin(x³+3x²+3x+1) + C. Notice that the polynomial in the differential exactly matches the derivative of the polynomial inside the cosine function, making this a perfect case for substitution following the pattern ∫f'(x)cos(f(x)) dx = sin(f(x)) + C.

Using the identity sin(x)cos(x) = (½)sin(2x), we can rewrite the integral as ∫(x+2) × (½)sin(2x) dx = (½)∫(x+2)sin(2x) dx. Now we use integration by parts. Let u = x+2 and dv = sin(2x) dx. Then du = dx and v = -cos(2x)/2. Using integration by parts: (½)[(x+2)(-cos(2x)/2) - ∫(-cos(2x)/2) dx] = (½)[-(x+2)cos(2x)/2 + (½)∫cos(2x) dx] = (½)[-(x+2)cos(2x)/2 + (½)(sin(2x)/2)] + C = -(x+2)cos(2x)/4 + sin(2x)/8 + C. Using the identity cos(2x) = 2cos²(x)-1 and sin(2x) = 2sin(x)cos(x), we can rewrite this as (-x-2)cos²(x)/2 + sin(x)/2 + (x+2)sin(x)cos(x)/2 + C, which shows the same result in terms of simpler trigonometric functions.