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
What is the integral of sin(x)*tan^2(x)? thanks
3 answers
great, thanks!
You are welcome!