Question

Plz solve it. . . . . . .I need proper ans.. . . ​

Answer 1

EXPLANATION.

∫sin2x dx/a²sin²x + b²cos²x.

As we know that,

Let we assume that,

⇒ a²sin²x + b²cos²x = t.

Differentiate w.r.t x, we get.

⇒ (a²2sinxcosx + b²2cosx(-sinx))dx = dt.

⇒ (a²2sinxcosx - b²2cosxsinx)dx = dt.

⇒ (a² - b²)2sinxcosx dx = dt.

⇒ (a² - b²)sin2x dx = dt.

⇒ sin2x dx = dt/(a² - b²).

Put the value in equation, we get.

⇒ ∫dt/t(a² - b²).

⇒ 1/(a² - b²) ∫dt/t.

⇒ 1/(a² - b²) ㏑|t| + c.

Put the value of t in equation, we get.

⇒ 1/(a² - b²)㏑|a²sin²x + b²cos²x| + c.

                                                                                                                         

MORE INFORMATION.

Some standard integrals.

(1) = ∫0.dx = c.

(2) = ∫1.dx = x + c.

(3) = ∫k dx = kx + c, (k ∈ R).

(4) = ∫xⁿdx = xⁿ⁺¹/n + 1 + c, (n ≠ -1).

(5) = ∫dx/x = ㏒(x) + c.

(6) = ∫eˣdx = eˣ + c.

(7) = ∫aˣdx = (aˣ)/㏒(a) + c = aˣ㏒(e) + c.