Question

What is the of integration of sinx.

Answer 1

dy = cos x dx

By trigonometric identities, cos x = √1 - sin²x. Then the above equation becomes,

dy = √1 - sin²x dx

dy = √1 - y² dx

dy / √1 - y² = dx

Multiplying both sides by sin x,

(sin x dy) / √1 - y² = sin x dx

Again substitute sin x = y on the left side.

(y dy) / √1 - y² = sin x dx

Integrating on both sides,

∫ (y dy) / √1 - y² = ∫ sin x dx

Let 1 - y² = u. Then -2y dy = du (or) y dy = -1/2 du.

Then the above left-hand side integral becomes,

(-1/2) ∫ 1/√u du = ∫ sin x dx

(-1/2) ∫ u^-1/2 du = ∫ sin x dx

Using one of the integration formulas, ∫ xⁿ dx = (xⁿ⁺¹)/(n+1) + C. So we get

(-1/2) (u^1/2 / (1/2)) + C = ∫ sin x dx

-u^1/2 + C = ∫ sin x dx

Substituting u = 1 - y² back here,

-(1 - y²)^1/2 + C = ∫ sin x dx

Substitute y = sin x back here,

-(1 - sin²x)^1/2 + C = ∫ sin x dx

-(cos²x)^1/2 + C = ∫ sin x dx

-cos x + C = ∫ sin x dx