Chapter 3 Indefinite Integration Ex 3.2(A) Solutions

Let I =

Let I = Dividing numerator and denominator by cos2x, we get

Let I = Dividing numerator and denominator of cos2x, we get

Let I = Put, Numerator = A (Denominator) + B [ (Denominator)] ∴ sin x+ 2 cos x = A(3 sin x + 4 cos x) + B [ (3 sin x + 4 cos x)] = A(3 sin x + 4 cos x) + B (3 cos x – 4 sin x) ∴ sin x + 2 cos x = (3A – 4B) sin x + (4A + 3B) cos x Equating the coefficients of sin x and cos x on both the sides, we get 3A – 4B = 1 …… (1) and 4A + 3B = 2 …… (2) Multiplying equation (1) by 3 and equation (2) by 4, we get 9A – 12B = 3 16A + 12B = 8 On adding, we get

Let I = Numerator = A (Denominator) + B [ (Denominator)] ∴ cos x = A(2 cos x + 3 sin x) + B [ (2 cos x + 3 sin x)] = A (2 cos x + 3 sin x) + B (-2 sin x + 3 cos x) ∴ cos x = (2A + 3B) cos x + (3A – 2B) sin x Equating the coefficients of cosx and sinx on both the sides, we get 2A + 3B = 1 …… (1) and 3A – 2B = 0 ……. (2) Multiplying equation (1) by 2 and equation (2) by 3, we get 4A + 6B = 2 9A – 6B = 0 On adding, we get

Let I = Put, Numerator = A (Denominator) + B [ (Denominator)] ∴ 4ex – 25 = A(2ex – 5) + B[ (2ex – 5)] = A(2ex – 5) + B(2ex – 0) ∴ 4ex – 25 = (2A + 2B) ex – 5A Equating the coefficient of ex and constant on both sides, we get 2A + 2B = 4 …….(1) and 5A = 25 ∴ A = 5 from (1), 2(5) + 2B = 4 ∴ 2B = -6 ∴ B = -3

Let I = Put, Numerator = A (Denominator) + B [ (Denominator)] ∴ 3e2x + 5 = A(4e2x – 5) + B [ (4e2x – 5)] = A(4e2x – 5) + B(4. e2x × 2 – 0) ∴ 3e2x + 5 = (4A + 8B) e2x – 5A Equating the coefficient of e2x and constant on both sides, we get 4A + 8B = 3 …….. (1) and -5A = 5 ∴ A = -1 ∴ from (1), 4(-1) + 8B = 3 ∴ 8B = 7 ∴ B =

Let I = ∫ tan5x dx

Let I = ∫ tan 3x tan 2x tan x dx Consider tan 3x = tan (2x + x) = tan 3x (1 – tan 2x tan x) = tan 2x + tan x tan 3x – tan 3x tan 2x tan x = tan 2x + tan x tan 3x – tan 2x – tan x = tan 3x tan 2x tan x I = ∫(tan 3x – tan 2x – tan x) dx = ∫tan3x dx – ∫tan 2x dx – ∫tan x dx = log | sec 3x| – log |sec 2x| – log |sec x| + c.