Test: Indefinite Integration: Integration by Parts (24 Sep) - JEE MCQ

 ∫ex[tan x + log sec x] dx
=∫ex . tan x dx + ∫ex . log sec x dx
= ex .∫ tan x dx − ∫[ddx(ex)×∫ tan x dx] dx + ∫ex . log sec x dx
= ex . log sec x − ∫ex . log sec x dx + ∫ex . log sec x dx + C
= ex . log sec x + C

 q = √x, dq = dx/2√x
⇒ dx = 2q dq
so the integral is 2∫qcosqdq
integration by parts using form 
∫uv' = uv − ∫u'v
here u = q, u'= 1 and v'= cosq, v=sinq
so we have 2(qsinq −∫sinqdq)
= 2(qsinq + cosq + C)
= 2(√xsin√x + cos√x + C)

Let  x=tanθ , so that  θ=tan⁻¹x ,  dx=sec2θdθ 
Then the given integral is equivalent to
∫tan⁻¹(2x/(1−x²)dx
=∫tan⁻¹ (2tanθ/(1−tan²θ))⋅sec²θdθ 
=∫tan⁻¹tan2θ⋅sec²θdθ 
=2×∫θsec²θdθ 
(integrate by parts)
=2θ⋅∫sec²θdθ −2⋅∫1⋅(∫sec²θdθ)dθ 
=2θtanθ−2⋅∫tanθdθ 
=2θtanθ−2loge secθ+c 
=2xtan⁻¹x−2loge√1+x²+c 
=2xtan⁻¹x−loge(1+x²)+c 

Let I = ∫e^x . cos 3x dx
⇒ I = cos 3x × ∫e^x dx − ∫[d/dx(cos 3x) × ∫e^x dx]dx
⇒ I = e^x cos 3x − ∫(− 3 sin 3x . e^x)dx
⇒ I = e^x cos 3x + 3∫sin 3x . e^x dx
⇒ I = e^x cos 3x + 3[sin 3x × ∫e^x dx − ∫{ddx(sin 3x) × ∫e^x dx}dx]
⇒ I = e^x cos 3x + 3[sin 3x . e^x − ∫3 cos 3x . ex dx]
⇒ I = e^x cos 3x + 3 ex . sin 3x − 9∫e^x . cos 3x dx
⇒ I =  e^x cos 3x + 3 e^x . sin 3x − 9I
⇒ 10I =  e^x cos 3x + 3 e^x . sin 3x
⇒ I = 1/10[e^x cos 3x + 3 e^x . sin 3x] + C

I=∫sin(logx)×1dx
= sin(logx) × x−∫cos(logx) × (1/x)×xdx
= xsin(logx)−∫cos(logx) × 1dx
= xsin(logx)−[cos(logx) × x−∫sin(logx) × (1/x) × xdx]
∴ I=xsin(logx)−cos(logx) × x−∫sin(logx)dx
2I=x[sin(logx)−cos(logx)]
∴ I=(x/2)[sin(logx)−cos(logx)]

(tan^-1 x)x dx
= tan^-1 ∫x dx - ∫d(tan^-1 x)/dx ∫x.dx)dx
= tan^-1 x . (x²)/2 - ∫1/(1+x²). (x²)/2 dx
=(x²)/2 tan^-1 x - 1/2∫(x²)/(x² + 1) dx
= (x²)/2 tan^-1 x - 1/2∫(x² + 1 - 1)/(x² + 1) dx
= (x²)/2 tan^-1 x - ½[∫(x² + 1)/(x² + 1)dx - dx/(x² + 1)]
= (x²)/2 tan^-1 x - ½[dx - ∫dx/(x² + 1)]
By applying formula, ∫dx/(a² + x²) = 1/a tan^-1x/a + c, we get
= (x²)/2 tan^-1 x - x/2 + ½ * 1/1 tan^-1 (x) + c
= (x²)/2 tan^-1 x - x/2 + ½ tan^-1 (x) + c