MCQ
$\int_{}^{} {\frac{{\sin 3x}}{{\sin x}}\;dx = } $
- ✓$x + \sin 2x + c$
- B$3x + \sin 2x + c$
- C$3x + {\sin ^2}x + c$
- DNone of these
✓
Answer
Correct option: A.
$x + \sin 2x + c$
a
(a)$\int_{}^{} {\frac{{\sin 3x}}{{\sin x}}\,dx} = \int_{}^{} {\frac{{3\sin x - 4{{\sin }^3}x}}{{\sin x}}\,dx} $
$\int_{}^{} {3\,dx} - 4\int_{}^{} {{{\sin }^2}x\,dx} = 3x - 2\int_{}^{} {(1 - \cos 2x)\,dx + c} $
$ = 3x - 2x + \sin 2x + c = x + \sin 2x + c.$
(a)$\int_{}^{} {\frac{{\sin 3x}}{{\sin x}}\,dx} = \int_{}^{} {\frac{{3\sin x - 4{{\sin }^3}x}}{{\sin x}}\,dx} $
$\int_{}^{} {3\,dx} - 4\int_{}^{} {{{\sin }^2}x\,dx} = 3x - 2\int_{}^{} {(1 - \cos 2x)\,dx + c} $
$ = 3x - 2x + \sin 2x + c = x + \sin 2x + c.$
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