Question

integration of sin(log x)dx

Answer 1

Answer :

Rule -

∫ u v dx

= u ∫ v dx - ∫ { d/dx (u) ∫ v dx } dx

Now,

∫ sin (logx) dx

= sin (logx) × ∫ dx - ∫ { d/dx sin (logx) ∫dx } dx

= sin (logx) × x - ∫ { cos (logx) × 1/x × x } dx + c,
c = integral constant

= x sin (logx) - ∫ cos (logx) dx + c

= x sin (logx) - [ cos (logx) ∫ dx - ∫ { d/dx cos (logx) ∫ dx } ] + c

= x sin (logx) - x cos (logx)
+ ∫ { - sin (logx) × 1/x × x } dx + c

= x {sin (logx) - cos (logx)}
- ∫ sin (logx) dx + c

⇒ 2 ∫ sin (logx) dx
= x {sin (logx) - cos (logx)} + c

⇒ ∫ sin (logx) dx
= x/2 {sin (logx) - cos (logx) } + c/2

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Answer 2

Answer:

Explanation:I=∫sin(logx)×1dx

=sin(logx)×x−∫cos(logx)×(

x

1

​

)×xdx

=xsin(logx)−∫cos(logx)×1dx

=xsin(logx)−[cos(logx)×x−∫sin(logx)×(

x

1

​

)×xdx]

∴I=xsin(logx)−cos(logx)×x−∫sin(logx)dx

or2I=x[sin(logx)−cos(logx)]

∴I=(

2

x

​

)[sin(logx)−cos(logx)]