So, I am trying to evaluate the following anti-derivative:
$$\int \frac{1}{\sin(x)\cos^3(x)} \,dx$$
I reached a point where I have the following:
$$\int \frac{\sin(x)}{\cos^3(x)} + \frac{1}{\sin(x)\cos(x)} \,dx$$
My idea now is to calculate two separate anti derivatives from here. I am using wolfram alpha to try to help me solve the anti derivative btw. but when I put on Wolfram this anti derivative:
$$\int \frac{1}{\sin(x)\cos(x)} \,dx$$
the result is not the same as:
$$\int \frac{\sin^2(x)+\cos^2(x)}{\sin(x)\cos(x)} \,dx$$
Why is that?? How can I solve this indefinite integral?
Best Answer
Write the integral as $$\int \sec^3(x)\csc(x)\,dx$$ Make the substitution $u=\tan x$, then $du=\sec^2(x) \,dx$, and it is $$\int \sec(x)\csc(x)\,du$$ $$\int u+\frac{1}{u}\,du ~~~~ \text{(why?)}$$ Can you finish?