Question

23. integrate
dx by
1-sin x​

Answer 1

EXPLANATION.

⇒ ∫dx/1 - sin(x).

As we know that,

Rationalize the equation, we get.

⇒ ∫dx/1 - sin(x) x (1 + sin(x))/(1 + sin(x)).

⇒ ∫(1 + sin x)dx/(1 - sin²x).

⇒ ∫(1 + sin x)dx/cos²x.

⇒ ∫dx/cos²x + ∫sin x/cos²x dx.

⇒ ∫sec²xdx + ∫tan x . sec x dx.

⇒ tan(x) + sec(x) + c.

                                                                                                                             

MORE INFORMATION.

Standard integrals.

(1) = ∫sin x dx = - cos x + c.

(2) = ∫cos x dx = sin x + c.

(3) = ∫tan x dx = ㏒(sec x) + c = -㏒(cos x) + c.

(4) = ∫cot x dx = ㏒(sin x) + c.

(5) = ∫sec x dx = ㏒(sec x + tan x) + c = ㏒ tan(π/4 + x/2) + c.

(6) = ∫cosec x dx = -㏒(cosec x + cot x) + c = ㏒(cosec x - cot x) + c = ㏒ tan(x/2) + c.

(7) = ∫sec x tan x dx = sec x + c.

(8) = ∫cosec x cot x dx = - cosec x + c.

(9) = ∫sec²xdx = tan x + c.

(10) = ∫cosec²dx = - cot x + c.