Question

integral of (sec x-1)^1/2

Answer 1

Let I=∫sec(x)−1−−−−−−−−√dxI=∫sec⁡(x)−1dx

=∫sec2(x)−1√sec(x)+1√dx=∫sec2⁡(x)−1sec⁡(x)+1dx(Multiplied numerator and denominator by sec(x)+1−−−−−−−−√sec⁡(x)+1)

=∫tan(x)sec(x)+1√dx=∫tan⁡(x)sec⁡(x)+1dx

Let y=sec(x)+1−−−−−−−−√y=sec⁡(x)+1

⟹y2=sec(x)+1⟹y2=sec⁡(x)+1

⟹2ydy=sec(x)tan(x)dx⟹2ydy=sec⁡(x)tan⁡(x)dx

⟹2yy2−1dy=tan(x)dx⟹2yy2−1dy=tan⁡(x)dx

Substituting the assumed values into I, we get,

I=∫2yy(y2−1)dyI=∫2yy(y2−1)dy

=∫2y2−1dy=∫2y2−1dy

Applying the partial decomposition technique, we have,

I=∫1y−1dy−∫1y+1dyI=∫1y−1dy−∫1y+1dy

=ln(y−1)−ln(y+1)=ln⁡(y−1)−ln⁡(y+1)

=ln(y−1y+1)=ln⁡(y−1y+1)

Substituting yy in II, we get,

I=ln(sec(x)+1√−1sec(x)+1√+1)+C