consider the following integral:
⨜cot(7x) dx
Find a substitution to rewrite the integrand as 1/7u du
u=?
du=(?)dx
u = sin(7x)
du = 7cos(7x)
Now you have
∫ 1/7 du/u
Thank you so much!
To rewrite the integrand ⨜cot(7x) dx as 1/7u du, we can use the substitution u = 7x.
Let's find the value of du:
To find du, we need to find the derivative of u with respect to x.
Taking the derivative with respect to x:
du/dx = 7
Now, let's solve for dx:
dx = du/(7)
Therefore, the substitution is:
u = 7x
du = 7 dx
To find a suitable substitution for the integral ⨜cot(7x) dx, we can look for a function that resembles the derivative of cot(7x). In this case, it is helpful to recall the derivative of cot(x), which is -csc^2(x). To get a similar derivative involving 7x, we can divide both the angle and the derivative term by 7.
Let's use the substitution u = 7x. This means that du/dx = 7. Solving for du, we get du = 7 dx. Dividing both sides by 7, we have du/7 = dx.
Therefore, the substitution u = 7x results in:
u = 7x
du = 7 dx
dx = du/7
Now, we can rewrite the original integral using the substitution:
⨜cot(7x) dx = ⨜cot(u) (du/7)
Next, we can simplify the integrand further. The cot(u) term can be expressed as 1/tan(u), which in turn is equal to 1/(sin(u)/cos(u)). Rearranging, we get cos(u)/sin(u).
Therefore, the integral becomes:
⨜cot(7x) dx = ⨜cos(u)/sin(u) (du/7)
Simplifying, we have:
(1/7) ⨜cos(u)/sin(u) du
Finally, we have rewritten the integrand as 1/7u du, where u = 7x and du = 7 dx.