int dx/3sinx+4cosx dx =
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How do you get two dx's ?
|(4-cosx)dx
To integrate the given expression ∫(dx/3sinx + 4cosx), we can use a technique called trigonometric substitution.
Let's start by manipulating the integral to simplify it. We can rewrite the expression as ∫(1/3sinx + 4cosx)dx, which is the same as ∫((1/3)sin^(-1)x + 4cosx)dx.
Now, let's use the substitution u = sin(x). Taking the derivative of both sides, we have du = cos(x)dx. Rearranging, we can express dx in terms of du: dx = du/cos(x).
Substituting this into our integral, we have ∫((1/3)(1/u) + 4cosx)(du/cos(x)).
Simplifying further, we get ∫(1/3u + 4)du.
Now, we can integrate each term separately:
∫(1/3u)du = (1/3)ln|u| + C1,
∫4du = 4u + C2.
Plugging these results back into our original integral, we have:
∫((1/3)(1/u) + 4)du = (1/3)ln|u| + C1 + 4u + C2.
Now, we need to substitute back u for sin(x):
(1/3)ln|sin(x)| + C1 + 4sin(x) + C2.
In conclusion, the integral of dx/3sinx + 4cosx is (1/3)ln|sin(x)| + 4sin(x) + C, where C represents the constant of integration.