Question

`int1/(4 + 3cos^2x)dx` = ______ 

Options

• `1/sqrt7 tan^-1((2tanx)/sqrt7) + c`

• `1/(2sqrt7) tan^-1((2tanx)/sqrt7) + c`

• `1/(5sqrt7) tan^-1((2tanx)/sqrt7) + c`

• `1/(2sqrt7) tan^-1((sqrt7tanx)/2) + c`

Answer 1

`int1/(4 + 3cos^2x)dx` = `underline(1/(2sqrt7) tan^-1((2tanx)/sqrt7) + c)`

Explanation:

Let I = `int1/(4 + 3cos^2x)dx`

= `intsec^2x/(4sec^2x + 3)dx`

= `intsec^2x/(4(1 + tan^2x) + 3)dx`

= `intsec^2x/(4tan^2x + 7)dx`

Put tan x = t ⇒ sec²x dx = dt

∴ I = `intdt/(4t^2 + 7)`

= `1/4int dt/(t^2 + (sqrt7/2)^2)`

= `1/4 . 1/(sqrt7/2) tan^-1(1/(sqrt7/2)) + c`

= `1/(2sqrt7) tan^-1((2t)/sqrt7) + c`

∴ I = `1/(2sqrt7) tan^-1((2tanx)/sqrt7) + c`