Question

step by step explaination​

Answer 1

integration of

1/√(1 - sin x)

= 1/√(cos^2(x/2) + sin^2(x/2) - 2 sin(x/2) cos(x/2)

= 1/√(cos(x/2) - sin(x/2))^2

= 1/(cos (x/2) - sin (x/2))

= (1/√2)/((1/√2) cos(x/2) - (1/√2) sin(x/2))

= (1/√2) / (cos(π/4) cos(x/2) - sin(π/4) sin(x/2))

= (1/√2) / cos (x/2 + π/4)

= (1/√2) sec (x/2 + π/4)

integration will be

(1/√2) *2 * log ( sec(x/2 + π/4) + tan (x/2 + π/4)) + c

- √2 log(sec(x/2 + π/4) - tan (x/2 - π/4)

- √2 log (1/cos(x/2 + π/4) - sin(x/2+π/4)/cos(x/2+π/4))+c

-√2 log ((1 - sin(x/2+π/4)) / cos(x/2 + π/4))+c

since argument is long, I replace it with A so as to avoid repetitive typing

A = x/2 + π/4

A/2 = x/4 + π/8

our integral becomes

- √2 log (1 - sinA)/cosA) + c

- √2 log (sin^2(A/2) + cos^2(A/2) - 2 sin(A/2)cos(A/2)) / (cos^2(A/2) - sin^2(A/2) + c

- √2 log (( cos(A/2) - sin(A/2))^2 / (cos(A/2)+sin(A/2)) (cos(A/2) - sin(A/2)) + c

- √2 log ((cos(A/2) - sin(A/2)) / (cos(A/2) + sin(A/2)) + c

- √2 log ( (1 - tan(A/2)) / ( 1 + tan (A/2)) + c

- √2 log ((tan(π/4) - tan(A/2)) / (1+ tan(π/4)(tan(A/2)) + c

- √2 log tan (π/4 - A/2) + c

√2 log tan (-(π/4 -A/2)) + c

√2 log tan (A/2 - π/4) + c

putting value of A/2 back

√2 log tan (x/4 + π/8 - π/4) + c

√ log tan (x/4 - π/8) + c

proved