Question

`int(sin2x)/(5sin^2x+3cos^2x)  dx=` ______.

विकल्प

• `1/2log abs(5sin^2x + 3cos^2x) +c`

• `1/4log abs(3sin^2x + 5cos^2x) +c`

• `1/4log abs(5sin^2x + 3cos^2x) +c`

• `1/2log abs(3sin^2x + 5cos^2x) +c`

Answer 1

`int(sin2x)/(5sin^2x+3cos^2x)  dx= underline(1/2log abs(5sin^2x + 3cos^2x) +c)`.

Explanation:

Put 5sin²x + 3cos²x = t

⇒ (5 × 2 sinx cosx - 3 × 2 sinx cosx) dx = dt

⇒ 4 sinx cosx dx = dt

⇒ 2sin 2x dx = dt

`therefore int(sin2x)/(5sin^2x+3cos^2x)  dx=1/2intdt/t`

`=1/2log abs(t)+c`

`=1/2log abs(5sin^2x + 3cos^2x) +c`