1/a arctan u/a +c
To find the integral of the expression 1/a arctan(u/a) + c, where u is a variable and a and c are constants, we can use the basic integration rules
To find the integral of the expression 1/a arctan(u/a) + c, where u is a variable and a and c are constants, we can use the basic integration rules.
First, let’s simplify the expression:
∫(1/a arctan(u/a)) du + c
Since arctan(u/a) is a function, we need to use the substitution method. Let’s assume that w = u/a, which means u = aw.
Differentiating both sides with respect to u gives us:
dw/du = 1/a
Now we need to substitute these values into the integral:
∫(1/a arctan(u/a)) du = ∫(1/a arctan(w)) a dw
Simplifying this further:
= ∫arctan(w) dw
The integral of arctan(w) can be found using integration by parts. Recall the formula:
∫u dv = uv – ∫v du
Here, let’s choose:
u = arctan(w) –> du = 1/(1+w^2) dw
dv = dw –> v = w
Applying the formula:
∫arctan(w) dw = w arctan(w) – ∫w/(1+w^2) dw
Now we can proceed with integrating the remaining term:
∫w/(1+w^2) dw
We can use another substitution to simplify this integral. Let’s assume z = 1 + w^2:
dz/dw = 2w –> dw = dz/(2w)
Substituting these values back into the integral:
∫w/(1+w^2) dw = ∫(1/2) dz/z
The integral becomes:
= (1/2) ln|z| + c’
Substituting back z = 1 + w^2:
= (1/2) ln|1+w^2| + c’
Now we can substitute w = u/a:
= (1/2) ln|1+(u/a)^2| + c’
Finally, we substitute back to our original variable u:
∫(1/a arctan(u/a)) du = (1/2) ln(1+(u/a)^2) + c’
Hence, the integral of 1/a arctan(u/a) du is (1/2) ln(1+(u/a)^2) + c’ where c’ represents the constant of integration.
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