To find the integral of ∫ sin(x) tan(x) dx, we can start by rewriting tan(x) in terms of sine and cosine. Recall that:
tan(x) = sin(x) / cos(x)
Substituting this into the integral gives us:
∫ sin(x) tan(x) dx = ∫ sin(x) (sin(x) / cos(x)) dx = ∫ sin²(x) / cos(x) dx
Now, to evaluate this integral, we can use a substitution. Let:
u = cos(x), then du = -sin(x) dx
This implies that sin(x) dx = -du. Moreover, when x is 0, u = cos(0) = 1, and as x approaches π/2, u approaches 0. The integral now becomes:
∫ (1 – u²) / u (-du) = -∫ (1 – u²)/u du = -∫ (1/u – u) du
This can be split into two separate integrals:
-∫ (1/u) du + ∫ u du
The first integral, -∫ (1/u) du, gives us -ln|u|, and the second integral, ∫ u du, gives us (1/2)u². Putting it all back together, we get:
-ln|cos(x)| + (1/2)(cos²(x)) + C
Therefore, the integral of ∫ sin(x) tan(x) dx is:
∫ sin(x) tan(x) dx = -ln|cos(x)| + (1/2)cos²(x) + C
where C is the constant of integration.