Question

Integrate: sin inverse root x - cos inverse root x divided by sin inverse root x + cos inverse root x dx

Answer 1

Let I=∫(sin-¹√x-cos-¹√x/sin-¹√x+cos-¹√x)dx

=Sin-¹√x+cos-¹√x=π/2

=sin-¹√x=x/2-cos-¹√x

i= [∫Sin-¹√x=-x/2-cos-¹√x)/π/2]dx...1

=2π∫(2sin-¹√x-π/2)dx

=4/π∫sin-¹√xdx-∫dx

i=4/π∫sin-¹√xdx-x+c₁

i₁=∫sin-¹√xdx

=x=sin²Q

=dx=2sinQcosdQ

=i₁=∫sin-¹(√sinQ).2sinQcosQdQ

=∫sin-¹(sinQ).2sinQcos dQ

=i₁∫Q.sin2QdQ

taking Qas the first and sin2Q in second

i₁=Q∫sin2QdQ=∫d/dQ(Q).∫sin2QdQ)dQ+c₂

Q(-cos2Q/2)+∫1.(-cos2Q)/2dQ+c

=-Q(2sinQ-1)/2sinQcosQ/2+C₂

from 2 we get

I=4/π(2x-1)/2sin-¹√x+√x-x²/2+c₂)-x+4

i=2/π(2x-1)sin-¹√x+2/π√x-x²+4/πC₂-x+C₁

i=2/π(2x-1)sin-¹√x+2/π√x-x²-x+C

where,

= c=c₁+4/πc₂