MCQ
$\int_{}^{} {{{\sin }^3}x\;.\;\cos x\;dx = } $
- A$\frac{{{{\sin }^4}x{{\cos }^2}x}}{8} + c$
- ✓$\frac{{{{\sin }^4}x}}{4} + c$
- C$\frac{{{{\sin }^2}x}}{2} + c$
- D$4{\sin ^4}x + c$
✓
Answer
Correct option: B.
$\frac{{{{\sin }^4}x}}{4} + c$
b
(b)$\int_{}^{} {{{\sin }^3}x\,.\,\cos x\,dx} $. Put $\sin x = t,$ then
$\cos x\,dx = dt$; $\int_{}^{} {{t^3}dt} = \frac{{{t^4}}}{4} = \frac{{{{\sin }^4}x}}{4} + c$.
(b)$\int_{}^{} {{{\sin }^3}x\,.\,\cos x\,dx} $. Put $\sin x = t,$ then
$\cos x\,dx = dt$; $\int_{}^{} {{t^3}dt} = \frac{{{t^4}}}{4} = \frac{{{{\sin }^4}x}}{4} + c$.
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