39the first one can be done by integration by parts.
10 Answers
39sinx/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)
1for second put x=t2+2 think this should be enough for this problem then it reduces to a std integral I think.
11isnt the second one discontinuous
can its indefinite integral still be found
111 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????
62subash you are right....
But I think indefinite integral can still be found.
When we put the limtis, we have to be careful.
take for example y=√x(1-x)
these can be integratd but are not defined for all R.
39ya bhaiyan sorry. i took the soln in wrong way. we cant solve it by that.
