Integrate by the function 2cosx/3sin²x
To integrate the function 2cosx/3sin²x, you can use a method called substitution. Here's how to do it:
Step 1: Identify a suitable substitution. In this case, we can let u = sinx. This substitution will help simplify the integral.
Step 2: Find the derivative of u with respect to x. Taking the derivative of u = sinx gives du = cosxdx.
Step 3: Rewrite the integral in terms of the new variable u. Using the substitution, we have:
2cosx/3sin²x = 2(du)/(3u²)
Step 4: Simplify the integral using the substitution. Now we can rewrite the integral in terms of the new variable:
∫2(du)/(3u²)
Step 5: Integrate the simplified function. The integral becomes:
(2/3) ∫du/u² = (2/3) * (-1/u) + C = -2/(3u) + C
Step 6: Convert back to the original variable. Remember that we substituted u = sinx. So, replacing u in terms of x:
-2/(3u) + C = -2/(3sinx) + C
That's the final answer. The integral of 2cosx/3sin²x is -2/(3sinx) + C.
∫ 2cosx/3sin²x dx
= ∫ (2/3) cosx (sinx)^-2 dx
= (-2/3) (sinx)^-1 + c
or -2/(3sinx) + c
or (-2/3)csc x + c
I did it by inspection, noting the the derivative of the power sin^2 x showed up as the correct cosx in the product