MCQ
$\int_{}^{} {{{\cos }^5}x\;dx = } $
- ✓$\sin x - \frac{2}{3}{\sin ^3}x + \frac{1}{5}{\sin ^5}x + c$
- B$\sin x + \frac{2}{3}{\sin ^3}x + \frac{1}{5}{\sin ^5}x + c$
- C$\sin x - \frac{2}{3}{\sin ^3}x - \frac{1}{5}{\sin ^5}x + c$
- DNone of these
✓
Answer
Correct option: A.
$\sin x - \frac{2}{3}{\sin ^3}x + \frac{1}{5}{\sin ^5}x + c$
a
(a)$\int_{}^{} {{{\cos }^5}x\,dx} = \int_{}^{} {{{\cos }^4}x\cos x\,dx} = \int_{}^{} {{{(1 - {{\sin }^2}x)}^2}\cos xdx} $
Put $\sin x = t \Rightarrow \cos x\;dx = dt$, then it reduces to
$\int_{}^{} {{{(1 - {t^2})}^2}dt = \int_{}^{} {(1 + {t^4} - 2{t^2})\;dt = \frac{{{t^5}}}{5} - \frac{{2{t^3}}}{3} + t + c} } $
$ = \frac{{{{\sin }^5}x}}{5} - \frac{{2{{\sin }^3}x}}{3} + \sin x + c$.
(a)$\int_{}^{} {{{\cos }^5}x\,dx} = \int_{}^{} {{{\cos }^4}x\cos x\,dx} = \int_{}^{} {{{(1 - {{\sin }^2}x)}^2}\cos xdx} $
Put $\sin x = t \Rightarrow \cos x\;dx = dt$, then it reduces to
$\int_{}^{} {{{(1 - {t^2})}^2}dt = \int_{}^{} {(1 + {t^4} - 2{t^2})\;dt = \frac{{{t^5}}}{5} - \frac{{2{t^3}}}{3} + t + c} } $
$ = \frac{{{{\sin }^5}x}}{5} - \frac{{2{{\sin }^3}x}}{3} + \sin x + c$.
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