MCQ
$\int_{}^{} {\sin (\log x)dx = } $
- A$\frac{1}{2}x[\cos (\log x) - \sin (\log x)]$
- B$\cos (\log x) - x$
- ✓$\frac{1}{2}x[\sin (\log x) - \cos (\log x)]$
- D$ - \cos \log x$
✓
Answer
Correct option: C.
$\frac{1}{2}x[\sin (\log x) - \cos (\log x)]$
c
(c) Let $I = \int_{}^{} {\sin (\log x)\,dx} $
Put $\log x = t \Rightarrow x = {e^t} \Rightarrow dx = {e^t}dt,$ then
$I = \int_{}^{} {\sin t\,.\,{e^t}\,dt} = \sin t\,.\,{e^t} - \int_{}^{} {{e^t}.\,\cos t\,dt} $
$ = \sin t\,.\,{e^t} - [\cos t\,.\,{e^t} + \int_{}^{} {{e^t}.\sin t\,dt} ]$
$ \Rightarrow 2I = \sin t\,.\,{e^t} - \cos t\,.\,{e^t}$
$ \Rightarrow I = \int_{}^{} {\sin \,(\log x)\,dx} $$ = \frac{1}{2}x\,[\sin \,(\log x) - \cos \,(\log x)]$.
(c) Let $I = \int_{}^{} {\sin (\log x)\,dx} $
Put $\log x = t \Rightarrow x = {e^t} \Rightarrow dx = {e^t}dt,$ then
$I = \int_{}^{} {\sin t\,.\,{e^t}\,dt} = \sin t\,.\,{e^t} - \int_{}^{} {{e^t}.\,\cos t\,dt} $
$ = \sin t\,.\,{e^t} - [\cos t\,.\,{e^t} + \int_{}^{} {{e^t}.\sin t\,dt} ]$
$ \Rightarrow 2I = \sin t\,.\,{e^t} - \cos t\,.\,{e^t}$
$ \Rightarrow I = \int_{}^{} {\sin \,(\log x)\,dx} $$ = \frac{1}{2}x\,[\sin \,(\log x) - \cos \,(\log x)]$.
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