If ∫ t a n ⁡ ( x - α ). t a n ⁡ ( x + α ) . t a n 2 x d x = p l o g | s e c 2 x | + q l o g | s e c ⁡ ( x + α ) + r l o g | s e c ⁡ ( x - α ) | + c then p + q + r = ……

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MARKS App · Accepted Answer

We have, t a n ∫ ( x - a ) tan ⁡ x + α . t a n 2 x d x = p l o g s e c 2 x + q l o g sec ⁡ x + a + r l o g sec ⁡ x - a + C t a n 2 x = tan ⁡ x - α + x + α = t a n ⁡ ( x - α ) + t a n ⁡ ( x + α ) ] 1 - tan ⁡ x - α tan ⁡ x + α ⇒ t a n 2 x - t a n 2 x t a n x - α tan ⁡ x + α = tan ⁡ x - α + tan ⁡ x + α ⇒ tan ⁡ x - α tan ⁡ x + α t a n 2 x ⇒ t a n 2 x - tan ⁡ x - α - tan ⁡ x + α ∴ t a n ∫ ( x - a ) tan ⁡ x + α . t a n 2 x d x log ⁡ s e c 2 x 2 - l o g ⁡ | s e c ⁡ ( x - α ) | - l o g ⁡ | s e c ⁡ ( x + α ) + C = 1 2 log ⁡ s e c 2 x + - 1 log ⁡ sec ⁡ x + α + + ( - 1 ) l o g ⁡ | s e c ⁡ ( x - α ) + C ∴ p = 1 2 , g - 1 and r = - 1 ∴ p + q + r = 1 2 + - 1 + - 1 = 1 2 - 2 = - 3 2

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