Chapter 3 Indefinite Integration Ex 3.3 Solutions

Let I = ∫x3 tan-1 x dx = ∫(tan-1 x). x3 dx

Let I = ∫(log x)2 dx Put log x = t

Let I = ∫sec3 x dx = ∫sec x sec2 x dx = sec x ∫sec2 x dx – ∫[(sec x) ∫sec2 x dx] dx = sec x tan x – ∫(sec x tan x)(tan x) dx = sec x tan x – ∫sec x tan2 x dx = sec x tan x – ∫sec x (sec2 x – 1) dx = sec x tan x – ∫sec3 x dx + ∫sec x dx ∴ I = sec x tan x – I + log|sec x + tan x| ∴ 2I = sec x tan x + log|sec x + tan x| ∴ I = [sec x tan x + log|sec x + tan x|] + c.

= t(log t – 1) + c = (log x). [log(log x) – 1] + c.

Let I = ∫cos√x dx Put √x = t ∴ x = t2 ∴ dx = 2t dt ∴ I = ∫(cos t) 2t dt = ∫2t cos t dt = 2t ∫cos t dt – ∫[(2t) ∫cos t dt]dt = 2t sin t – ∫2 sin t dt = 2t sin t + 2 cos t + c = 2[√x sin√x + cos√x] + c.

Let I = ∫sin θ. log (cos θ) dθ = ∫log(cos θ). sin θ dθ Put cos θ = t ∴ -sin θ dθ = dt ∴ sin θ dθ = -dt = -t log t + t + c = -cos θ. log(cos θ) + cos θ + c = -cos θ [log(cos θ) – 1] + c.

cos 3x = 4 cos3 x – 3 cos x ∴ cos3 x + 3 cos x = 4cos3x ∴ cos3 x = cos 3x + cos x

Let I = Put log x = t dx = dt ∴ I = ∫t dt = t2 + c = (log x)2 + c

Let I = ∫x sin 2x cos 5x dx sin 2x cos 5x = [2 sin 2x cos 5x] = [sin(2x + 5x) + sin(2x – 5x)] = [sin 7x – sin 3x] ∴ ∫sin 2x cos 5x dx = [∫sin 7x dx – ∫sin 3x dx]

Let I =

Let I = dx

Let I = ∫(x + 1) dx Let x + 1 = A[(2x2 + 3)] + B = A(4x) + B = 4Ax + B Comparing the coefficients of x and constant term on both the sides, we get 4A = 1, B = 1 ∴ A = , B = 1

Let I = ∫ dx Let x = A[(5 – 4x – x2)] + B = A[-4 – 2x] + B = -2Ax + (B – 4A) Comparing the coefficients of x and the constant term on both sides, we get -2A = 1, B – 4A = 0

Let I = ∫ex [2 + cot x – cosec2 x] dx Put f(x) = 2 + cot x ∴ f'(x) = (2 + cot x) = (2) + (cot x) = 0 – cosec2 x = -cosec2 x ∴ I = ∫ex [f(x) + f'(x)] dx = ex f(x) + c = ex (2 + cot x) + c.

Let I = ∫ Let f(x) = , f'(x) = ∴ I = ∫ex [f(x) + f'(x)] dx = ex f(x) + c = ex. + c

Let I = ∫