integrate

sinx/cos²x on multiplying and dividing this by cos²x, then we get tanxsecx/sec²x+1

this one can be solved by taking sec²x=t.

now i think the sum is done.

as in the problem integration by parts is x(∫ sinx/cos²x+1) +∫1∫ sinx/(cos²x+1)

for second put x=t²+2 think this should be enough for this problem then it reduces to a std integral I think.

wat about first one? is it correct?

isnt the second one discontinuous

can its indefinite integral still be found

1 can be rewritten as -x.d[log(1+cosx)]

And then taking -x as first function and integrating by parts, we get:

-x. log(1+cosx)+∫log(1+cosx)dx

Then????