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STD 12 - 7.1 inderfinite integral — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 7.1 inderfinite integral4 Marks
MCQ
If $\int_{}^{} {\sin 5x\cos 3x\;dx = - \frac{{\cos 8x}}{{16}}} + A$, then $A = $
  • A
    $\frac{{\sin 2x}}{{16}} + $constant
  • ✓
    $ - \frac{{\cos 2x}}{4} + $constant
  • C
    Constant
  • D
    None of these
✓

Answer

Correct option: B.
$ - \frac{{\cos 2x}}{4} + $constant
b
(b)$\int_{}^{} {\sin 5x\cos 3x\;dx} = \frac{1}{2}\int_{}^{} {(\sin 8x + \sin 2x)} dx$
$ = \frac{{ - \cos 8x}}{{16}} - \frac{{\cos 2x}}{4} + c$
Equating to the given value, we get $A = \frac{{ - \cos 2x}}{4} + c$.

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