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1--50
2-- 30

t=1/z;
dt=-1/z^2
putting in previous eqn.
dt/(1-t^2)1/2 (t^-1)

-z^3/{z^2}{(z^-1)^1/2}{ (1-z^2)} dz
put z^2-1=k; 2zdz=dk
-1/2(dk/(k^1/2)*(-k)

1/2 i
i=dk/k^3/2 (integrate it and put the values)

hope now you can understand it
please rate me if i am correct

obviously cbse

integral of sin^1/2 is undefined

read in tmh for iit

cos^1/2 can be solved using by parts

will show the answer later

remind you,i had asked for commplete soln

post your qustns yaar

goiitians answer soon plzzzzzzzz

konic be fast in repyling,

konichiva, i had left that question to stndrd integral type yaar,

dx/(a){(b)^.5}

a is quadratic,

b is quadratic,

you can solve further by

putting t=1/z and simplifying

then put

1+z^2=m^2 and got the answer.

qustn which i had given is my query but you can accept it as challenge too,it a tricky one ,cant solve easily,i had already posted it in coomunity q&a

jaggy i think you had misread the question
its like this one

I = dx/(a+btanx)^2

its square of (a+btanx)

raul rag i had tried this way and ihad myself solved this one upto this,but i am unable yo solve it further using partial fraction or other ways too.

plzz solve it further

y(1+xy)dx+x(1--xy)dy==0

got this one in pre boards

plz ans. with full solution to be rated