I=∫(cos⁡x)2n+1dx=∫(cos⁡x)2n⋅cos⁡xdx Put sin⁡x=tcos⁡xdx=dt=∫1−sin2⁡xn⋅cos⁡xdx=∫1−t2ndt=∫1−nc1t2+nc2t4+−−−−(−1)nt2ndt=t−nc1t33+nc255+−−−−+(−1)nt2n+12n+1+C=sin⁡x−nqsin3⁡x3+nc2sin5⁡x5+−−−+(−1)nsin2n+1⁡x(2n+1)+C

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$\int {(\cos x)}^{2 n + 1} d x$

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$\frac{{(\sin x)}^{2 n + 2}}{2 n + 2} + C$

$(\cos x) - n_{c_{1}} \frac{\cos^{3} x}{3} + n_{c_{2}} \frac{\cos^{5} x}{5} + - - - + \frac{{(- 1)}^{n} \cos^{2 n + 1} (x)}{(2 n + 1)} + C$

$\sin x - n_{c_{1}} \frac{s i n^{2} x}{3} + n_{c_{2}} \frac{\sin^{5} x}{5} + - - - + \frac{{(- 1)}^{n} \sin^{2 n + 1} (x)}{(2 n + 1)} + C$

$(\sin x) - n_{c_{1}} \frac{s i n^{3} x}{3} + n_{c_{2}} \frac{\sin^{5} x}{5} + - - - + {(- 1)}^{n} \sin (n + 1) x + C$

answer is C.

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Detailed Solution

$\begin{array}{l} I = \int (\cos x)^{2 n + 1} d x \\ = \int (\cos x)^{2 n} \cdot \cos x d x \\ Put \sin x = t \\ \cos x d x = d t \\ = \int {(1 - \sin^{2} x)}^{n} \cdot \cos x d x \\ = \int {(1 - t^{2})}^{n} d t \end{array}$

$\begin{array}{l} = \int \{1 - n_{c_{1}} t^{2} + n_{c_{2}} t^{4} + - - - - (- 1)^{n} t^{2 n}\} d t \\ = t - n_{c_{1}} \frac{t^{3}}{3} + n_{c_{2}} \frac{5}{5} + - - - - + (- 1)^{n} \frac{t^{2 n + 1}}{2 n + 1} + C \\ = \sin x - n_{q} \frac{\sin^{3} x}{3} + n_{c_{2}} \frac{\sin^{5} x}{5} + - - - + \frac{(- 1)^{n} \sin^{2 n + 1} x}{(2 n + 1)} + C \end{array}$