To integrate the given function, we can use a u-substitution. Let's let u = sin(x), then du = cos(x)dx.
The integral becomes:
∫cos(x)dx / (4 + 3u)= ∫du / (4 + 3u)= (1/3)∫du / (u + 4/3)
Now we can use partial fractions to further simplify the integral:
(1/3)∫du / (u + 4/3)= (1/3)∫ [A / (u + 4/3)] du
Solving for A, we get:
A = 1/3
Now we can rewrite the integral as:
(1/3)∫ [1 / 3(u + 4/3)] du= (1/3)∫ [1 / 3u + 4] du= (1/3) (1/3) ln|3u + 4| + C= ln|3sin(x) + 4| / 9 + C
Therefore, ∫cos(x)dx / (4 + 3sin(x)) = ln|3sin(x) + 4| / 9 + C
To integrate the given function, we can use a u-substitution. Let's let u = sin(x), then du = cos(x)dx.
The integral becomes:
∫cos(x)dx / (4 + 3u)
= ∫du / (4 + 3u)
= (1/3)∫du / (u + 4/3)
Now we can use partial fractions to further simplify the integral:
(1/3)∫du / (u + 4/3)
= (1/3)∫ [A / (u + 4/3)] du
Solving for A, we get:
A = 1/3
Now we can rewrite the integral as:
(1/3)∫ [1 / 3(u + 4/3)] du
= (1/3)∫ [1 / 3u + 4] du
= (1/3) (1/3) ln|3u + 4| + C
= ln|3sin(x) + 4| / 9 + C
Therefore, ∫cos(x)dx / (4 + 3sin(x)) = ln|3sin(x) + 4| / 9 + C