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D)tauu tant /cot -I

Answer»

Write tanx as sinx/cosx and cotx as cosx/sinx

I= √((sin^2x/sinxcosx) + √(cos^2x/sinxcosx))dx

I= (sinx+cosx dx)/√(sinxcosx)

sinx-cosx=t

(sinx+cosx)dx=dt

(sinx-cosx)^2=t^2

1–2sinxcosx=t^2

sinxcosx=(1-t^2)/2

Therefore,

I=dt/√((1-t^2)/2)

I=(√2)sin^-1(t) + C

I=(√2)sin^-1(sinx-cosx) + C


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