To find the integral of sin(x)tan(x) dx, we first need to express tan(x) in terms of sine and cosine:
tan(x) = rac{sin(x)}{cos(x)}
This allows us to rewrite the integral as:
∫ sin(x)tan(x) dx = ∫ sin(x) * rac{sin(x)}{cos(x)} dx = ∫ rac{sin^2(x)}{cos(x)} dx
Now, let’s denote this integral as:
I = ∫ rac{sin^2(x)}{cos(x)} dx
To solve this integral, we can use a substitution. Let:
u = cos(x)
Then the derivative of this substitution is:
du = -sin(x) dx
From this, we can express sin(x) dx as:
-du = sin(x) dx
Now, we notice that:
sin^2(x) = 1 - cos^2(x) = 1 - u^2
Thus, we can rewrite the integral:
I = ∫ rac{1 - u^2}{u} (-du)
This simplifies to:
I = - ∫ rac{1 - u^2}{u} du = - ∫ rac{1}{u} du + ∫ u du
Now we can compute the two separate integrals:
I = -ln|u| + rac{u^2}{2} + C
Substituting back for u = cos(x), we get:
I = -ln|cos(x)| + rac{cos^2(x)}{2} + C
Putting the result in concise integral form:
∫ sin(x)tan(x) dx = -ln|cos(x)| + rac{cos^2(x)}{2} + C
In conclusion, we’ve solved the integral by using substitution and simplification techniques to arrive at our final answer.