Find the integral of (sinx + cosx) / [3+sin2x] dx.

Integration is exactly the reverse of differentiation. The integral of (sinx + cosx) / [3+sin2x] dx can be found using substitution method.

Answer: The integral of (sinx + cosx) / [3+sin2x] dx is 1/4 log {(2 + sinx - cosx) / (2 - sinx + cosx)} + C.

Let's look into the step by step solution.

Explanation:

⇒ ∫ (sinx+cosx)dx/ 3 + sin2x

⇒ ∫ (sinx+cosx)dx / (4 - 1) + sin2x

⇒ ∫ (sinx+cosx)dx / 4 - (1 - sin2x )

⇒ ∫ (sinx+cosx)dx / 4 - (sin²x + cos²x - sin2x ) [Using the trigonometric identity, sin²x + cos²x = 1 ]

⇒ ∫ (sinx+cosx)dx / 4- (sinx - cosx )² [Using the algebraic identity, (a-b)²] --------------- (1)

Let, sinx - cosx = t ---------- (2)

On differentiating both the sides,

⇒ (cosx + sinx) dx = dt ------------- (3)

Substituting (2) and (3) in (1) we get,

⇒ ∫ dt / (2² - t²)

⇒ 1/2(2) log {(2+t) / (2-t)} + C [ Since, 1 / (a^a- x²) dx = 1/2a log {(a+x) / (a-x)} + C]

⇒ 1/4 log {(2 + sinx - cosx) / (2 - sinx + cosx)} + C

where, C is the constant of integration.