Question

Integration of sqrt(cosx)/sinx???

Answer 1

Hey

Hope this helps ^_^

Wait... we're substituting cosx = u

Answer 2

we have to find integration of √cosx/sinx

i.e., I = ∫ √(cosx)/sinx dx

let cosx = P

differentiating both sides we get,

-sinx dx = dP

so, dx = -dP/sinx

now putting value of dx and cosx = P in I.

I = ∫ √P/sinx × -dP/sinx

= ∫-√P/sin²x dP

= -∫ √P/(1 - cos²x) dP [ we get sin²x =1 - cos²x ]

= -∫ √P/(1 - P²) dP

now let P = t²

after differentiating we get,

dP = 2t dt

so, I = -∫ t (2t dt)/(1 - t⁴)

= -2∫t² dt/(1 - t²)(1 + t²)

= -∫{(1 + t²) - (1 - t²) }dt/(1 - t²)(1 + t²)

= -∫dt/(1 - t²) + ∫dt/(1 + t²)

= -1/2 ln|1 - t|/|1 + t| + tan-¹t + C

now putting, t = √P = √cosx

= -1/2 ln|1 - √cosx|/|1 + √cosx| + tan-¹√(cosx) + C

hence I = ∫ √(cosx)/sinx dx = -1/2 ln|1 - √cosx|/|1 + √cosx| + tan-¹√(cosx) + C