Integral of (2x+5sin(x))^2 dx
i didn't know how to start this question so i expand the squre first and then did the rest but its wrong and i don't think my method is right.
thank you.
To solve this integral, let's use a different approach. Instead of expanding the square, we can utilize a trigonometric identity.
First, let's consider the term (2x)^2. When we expand it out, we get 4x^2.
Next, let's work on the term (5sin(x))^2. To simplify this, we can use the trigonometric identity sin^2(x) = (1 - cos(2x))/2. Applying this identity, we can rewrite (5sin(x))^2 as (25/2)(1 - cos(2x)).
Now, we can rewrite the original integral as:
∫(4x^2 + (25/2)(1 - cos(2x))) dx.
Expanding this out, we get:
∫4x^2 dx + ∫(25/2)(1 - cos(2x)) dx.
Now, let's solve each term separately:
1. ∫4x^2 dx:
To find the integral of x^2, we can use the power rule, which states that ∫x^n dx = (x^(n+1))/(n+1). Applying this rule, we have:
∫4x^2 dx = (4/3)x^3 + C1.
2. ∫(25/2)(1 - cos(2x)) dx:
To integrate this term, we need to use the linearity property of integration:
∫(25/2)(1 - cos(2x)) dx = (25/2)∫(1 - cos(2x)) dx.
Now, let's break down this integral further:
∫(1 - cos(2x)) dx = ∫1 dx - ∫cos(2x) dx.
The integral of 1 with respect to x is simply x:
∫1 dx = x.
To integrate cos(2x), we can use the substitution method. Let u = 2x, then du = 2 dx. Rearranging, we find dx = (1/2) du. Substituting this back into the integral, we have:
∫cos(2x) dx = ∫cos(u) (1/2) du = (1/2)∫cos(u) du.
The integral of cos(u) is sin(u):
(1/2)∫cos(u) du = (1/2)sin(u) + C2.
Substituting u back in terms of x, we have:
∫cos(2x) dx = (1/2)sin(2x) + C2.
Now we can go back to ∫(25/2)(1 - cos(2x)) dx:
∫(25/2)(1 - cos(2x)) dx = (25/2)∫(1 - cos(2x)) dx
= (25/2)(x - (1/2)sin(2x)) + C3.
Putting it all together:
∫(4x^2 + (25/2)(1 - cos(2x))) dx = (4/3)x^3 + (25/2)(x - (1/2)sin(2x)) + C,
where C, C1, C2, and C3 represent constants of integration.
Please note that the constants of integration are added to account for the infinite number of possible solutions when integrating indefinite integrals.