1. [integration] (3x+7sin(x))^2 dx
i tried 3 different methods of integration but all of my answers are wrong.
2. [Integration]from 0 to 5 (3w-3)/(w+6)dw
i divided the numerator by the denominator and then tried to solve the question but im getting the wrong answer.
1. Ah, integrals can be quite tricky sometimes. It seems like you've been on a wild goose chase with this one. Maybe it's time to call in the integration circus! *cue circus music*
Let's take a different approach and break it down step by step. We have (3x + 7sin(x))^2 dx. First, let's expand the square inside: (9x^2 + 42xsin(x) + 49sin^2(x)) dx.
Now, the integral of 9x^2 dx is pretty straightforward, right? You got this! It's (3/1)x^3 + C. Next, we have 42xsin(x) dx. This one requires integration by parts. Remember the "uv" formula? Give it a go!
Lastly, we have 49sin^2(x) dx. Hmm, this looks familiar. Do you recall the half-angle identity for sin^2(x)? It might just save the day!
Keep on clowning around with those integrals, and you'll surely find the right answer. Good luck!
2. Integrating can be a real balancing act, can't it? For this one, let's hop on the integration roller coaster and see where it takes us! Wheeeeeeee!
You mentioned dividing the numerator by the denominator, which is a good start. That gives you (3w - 3) / (w + 6) dw. Now, we need to simplify the expression further.
Dividing (3w - 3) by (w + 6) gives you 3 - (27 / (w + 6)). But hold on, we're not done yet! Now, we can split the integral into two parts: the integral of 3 dw and the integral of (27 / (w + 6)) dw.
The integral of 3 dw is pretty simple. It just gives you 3w. As for (27 / (w + 6)) dw, it requires a little dance with the natural logarithm function. If you remember the integral of 1/x, you're on the right track! Apply that knowledge and you'll get it.
Finally, evaluate both integrals from 0 to 5 and subtract the values. Voila! You've completed the integration circus. Keep going, my comedic friend!
1. To integrate (3x+7sin(x))^2 dx, we can expand the expression using the binomial theorem:
(3x+7sin(x))^2 = (3x)^2 + 2(3x)(7sin(x)) + (7sin(x))^2
= 9x^2 + 42x sin(x) + 49 sin^2(x)
Now, we can integrate term by term:
∫ (9x^2 + 42x sin(x) + 49 sin^2(x)) dx
= 3∫ x^2 dx + 42∫ x sin(x) dx + 49∫ sin^2(x) dx
Let's integrate each term separately:
∫ x^2 dx: This is a standard power rule integration. By applying the power rule, we get:
= (1/3) x^3 + C
∫ x sin(x) dx: We can use integration by parts. Let u = x (choose this as the first function) and dv = sin(x) dx (choose this as the second function). Then, du = dx and v = -cos(x). Applying the formula, we have:
= -x cos(x) - ∫ -cos(x) dx
= -x cos(x) + sin(x) + C
∫ sin^2(x) dx: We can use the double-angle identity for sine. sin^2(x) = (1/2) - (1/2) cos(2x). Thus:
= ∫ (1/2 - (1/2) cos(2x)) dx
= (1/2) x - (1/4) sin(2x) + C
Now, we can combine the results for each term:
∫ (9x^2 + 42x sin(x) + 49 sin^2(x)) dx
= (1/3) x^3 + C - x cos(x) + sin(x) + C + (1/2) x - (1/4) sin(2x) + C
= (1/3) x^3 + (3/2) x - x cos(x) + sin(x) - (1/4) sin(2x) + 3C
So, the integral of (3x+7sin(x))^2 dx is (1/3) x^3 + (3/2) x - x cos(x) + sin(x) - (1/4) sin(2x) + C, where C represents the constant of integration.
2. To integrate (3w-3)/(w+6)dw from 0 to 5, we can use the method of partial fractions:
First, divide the numerator by the denominator and express it as a sum of fractions:
(3w-3)/(w+6) = (3w)/(w+6) - (3)/(w+6)
Now, we can integrate each fraction separately:
∫ (3w)/(w+6) dw: We can use u-substitution. Let u = w+6, then du = dw. We have:
= ∫ (3(u-6))/u du
= 3∫ (u-6)/u du
= 3(∫ (u/u) du - ∫ (6/u) du)
= 3(∫ 1 du - 6∫ (1/u) du)
= 3(u - 6 ln|u|) + C
= 3(w+6 - 6 ln|(w+6)|) + C
∫ (-3)/(w+6) dw: This is a straightforward logarithmic integration. By applying the natural logarithm rule for integration, we get:
= -3 ln|w+6| + C
Now, we can integrate the original equation from 0 to 5:
∫ (3w-3)/(w+6) dw from 0 to 5
= [3(w+6 - 6 ln|(w+6)|)] - 3 ln|w+6| | from 0 to 5
= [3(5+6 - 6 ln|(5+6)|)] - 3 ln|5+6| - [3(0+6 - 6 ln|(0+6)|)] + 3 ln|0+6|
= [3(11 - 6 ln|11|)] - 3 ln|11| - [3(6 - 6 ln|6|)] + 3 ln|6|
= 33 - 18 ln|11| - 18 + 18 ln|6|
= -18 + 18 ln|6| - 18 ln|11|
Hence, the solution for the integral from 0 to 5 of (3w-3)/(w+6)dw is -18 + 18 ln|6| - 18 ln|11|.
1. Integrating the expression (3x+7sin(x))^2 dx requires the use of multiple integration techniques. In this case, it is best to use the method of substitution.
To start, let's make a substitution: let u = 3x + 7sin(x). This means that du/dx = 3 + 7cos(x), and solving for dx, we get dx = du/(3 + 7cos(x)).
Substituting u and dx into the original expression, we now have the integral of u^2 * (du/(3 + 7cos(x))).
Next, we can simplify the expression by substituting cos(x) with (u - 3x)/7. This gives us the integral of u^2 * (du/(3 + 7(u - 3x)/7)).
Simplifying further, we have the integral of u^2/(3 + u - 3x) du, which can be separated into two separate integrals: the integral of u^2/(u + 3) du and the integral of -3u^2/(u + 3) du.
These individual integrals can then be solved by using techniques like polynomial long division or partial fractions to break them down into more manageable forms.
Finally, evaluate the limits of integration to obtain the final answer.
2. Integrating the expression (3w-3)/(w+6)dw requires the method of partial fractions.
First, divide the numerator by the denominator to get the quotient and remainder. In this case, we have (3w-3)/(w+6) = 3 - 21/(w+6).
Next, rewrite the expression using partial fractions: 3 - 21/(w+6) = A + B/(w+6), where A and B are constants to be determined.
To find the values of A and B, we can equate the numerators on both sides of the equation.
From A + B(w+6) = 3w - 3, we can solve for A and B by comparing the coefficients of like terms.
Once we find the values of A and B, we can integrate each term separately. The integral of A dw is simply A*w, and the integral of B/(w+6) dw can be solved using the substitution method.
Finally, evaluate the limits of integration to get the final answer.