Question

If ∫ tan (x - α) tan (x + α) · tan 2x dx
= p log |sec 2x| + q log |log (x - α)| + r log |sec (x - α)| + c then p + q + r = ______.

विकल्प

• `(-3)/2`

• `(- 5)/2`

• `5/2`

• `3/2`

Answer 1

If ∫ tan (x - α) tan (x + α) · tan 2x dx
= p log |sec 2x| + q log |log (x - α)| + r log |sec (x - α)| + c then p + q + r = `underline((-3)/2)`.

Explanation:

We have,

∫ tan (x - α) tan (x + α) · tan 2x dx 

= p log |sec 2x| + q log |log (x - α)| + r log |sec (x - α)| + c

HEre, tan 2x = tan ((x - α) + (x + α))

= tan (x - α) + tan (x + α)

1 - tan (x - α) tan (x + α)

⇒ tan 2x - tan 2x tan (x - α) tan (x + α) = tan (x - α) + tan (x + α)

⇒ tan (x - α) tan (x + α) tan 2x

⇒ tan 2x - tan (x - α) - tan (x + α)

∴ ∫ tan (x - α) tan (x + α) tan (2x) dx

= ∫ (tan 2x - tan (x - α)) - tan (x + α)) dx

= `(log |sec 2x|)/2` - log |sec (x - α)| - log |sec (x + α)| + C

`= 1/2` log |sec 2x| + (- 1) log |sec (x - α)| + (- 1) log |sec (x - α)| + C

∴ p = `1/2`, g = - 1 and r = - 1

∴ p + q + r = `1/2 + (- 1) + (- 1)`

`= 1/2 - 2 = (- 3)/2`