what is the integral of {sin(x)tan(x)}
i tried turning tan(x) into sin(x)/cos(x) then doing u substitutions but i always have an extra sin(x) left.
can you help me pleeeease
sin(x) tan(x) = sin²(x)/cos(x) = (1-cos²(x))/cos(x) = sec(x) - cos(x)
Now you can probably handle it, yes?
wooow, i cant believe ive been staring at it this long and couldn't see this. thanks!!
Of course! Solving the integral of sin(x)tan(x) can be a bit tricky, but I'll guide you through the process step by step.
First, you're on the right track by expressing tan(x) as sin(x) / cos(x). So, let's rewrite the integral as:
∫(sin(x) * tan(x)) dx = ∫(sin(x) * sin(x) / cos(x)) dx
To simplify this expression further, we can use the trigonometric identity sin^2(x) = 1 - cos^2(x) to replace sin^2(x) in the numerator. This gives us:
∫((1 - cos^2(x)) / cos(x)) dx
Now, we can split this integral into two separate integrals using the sum/difference rule of integration. The integral becomes:
∫(1 / cos(x)) dx - ∫(cos^2(x) / cos(x)) dx
= ∫sec(x) dx - ∫cos(x) dx
The integral of sec(x) can be found by using the substitution u = sec(x) + tan(x), which leads to du = (sec(x) * tan(x) + sec^2(x)) dx. Rearranging the equation, we have du = (u + 1) dx. Therefore, the integral becomes:
∫sec(x) dx = ∫(du / (u + 1)) = ln|u + 1| + C1
Now, let's focus on the integral of cos(x), which is quite straightforward:
∫cos(x) dx = sin(x) + C2
Combining both results, the final answer is:
∫(sin(x)tan(x)) dx = ln|sec(x) + tan(x) + 1| - sin(x) + C
Don't forget to add the constant C1 and C2 to account for the integration constants.