write in parametric form! denominator becomes sin(x+aplha)
take x+alpha=t and write x=t-alpha.......expand numerator n integrate!
write in parametric form! denominator becomes sin(x+aplha)
take x+alpha=t and write x=t-alpha.......expand numerator n integrate!
1. integrate
2.
FOR PRCTC
as alwayz pritish babu has done a nice work !!!!!!!!!!!!!!!!!
2) ∫(sinxlogx−xcosx)dx=∫sinxlogxdx−∫xcosxdx .......(1)
let I1 = ∫sinxlogxdx
considering sin x as 2nd and log x as first function we have BY USING BY PARTS,
I1 = logx∫sinx+∫xcosx
now substituting the value of I1 in (1) we get,
∫(sinxlogx−xcosx)dx=logx∫sinx+∫xcosx - ∫cos xx = logx∫sinx
= - logx cosx
LONG ONE THOUGH.
∫(√tanx+√cotx)dx=∫√sinxcosxsinx+cosxdx now put sinx - cosx = t → (cosx + sinx)dx = dt
integral reduces to √2∫√1−t2dt=√2sin−1t+c=√2sin−1(sinx−cosx)+c
similarly we find ∫(√tanx−√cotx)dx=−√2ln∣∣sinx+cosx+√sin2x∣∣+c
now ∫√cotx=21∫2√cotxdx=21∫(√tanx+√cotx)−(√tanx−√cotx)dx
subtituting the values of ∫(√tanx+√cotx)dx and ∫(√tanx−√cotx)dx we get ∫√cotx = 21[√2sin−1(sinx−cosx)+√2ln∣∣(sinx+cosx)+√sin2x∣∣]+k,k=integ.constant
whats the solution for 1st question ,
iam getting in the for of
K ∫√1- tan α. cos t dt
where t is x - α