What is the integral of sin(x)*tan^2(x)? thanks
One approach is to try to convert trig functions into sin and cos, and then see if the function is odd or even in sin and cos. This approach pays off here.
I=∫sin(x)tan²(x)dx
=∫(sec²(x)-1) sin(x)dx
=∫(1/cos²(x) - 1) sin(x)dx
Now apply the substitution
u=cos(x)
du=-sin(x)dx
I=∫(1/u²-1) -du
=1/u + u + C
=1/cos(x) + cos(x) + C
great, thanks!
You are welcome!
To find the integral of sin(x) * tan^2(x), we can use a technique called substitution.
Let's start by letting u = tan(x).
Next, we need to find the derivative of u with respect to x.
The derivative of u = tan(x) is du = sec^2(x) dx.
Now we can rewrite the original integral in terms of u:
∫sin(x) * tan^2(x) dx = ∫sin(x) * (tan^2(x))(sec^2(x)) dx
= ∫u^2 * (1/ sec^2(x)) du
= ∫u^2 du
= (1/3)u^3 + C
Finally, substitute back u = tan(x) to find the answer:
= (1/3)tan^3(x) + C.
Therefore, the integral of sin(x) * tan^2(x) is (1/3)tan^3(x) + C, where C represents the constant of integration.