In=∫sinn⁡xdx, then 5I5−4I3 isequal to

I5=∫sin5xdxI5=∫sin4⁡xsin⁡xdx=sin4⁡x∫sin⁡xdx−∫ddxsin4⁡x∫sin⁡xdxdx Using ∫fgdx=f∫gdx−∫f1∫gdxdx=−sin4⁡x(cos⁡x)−∫4sin3⁡x⋅cos⁡x(−cos⁡x)dx=−sin4⁡xcos⁡x+4∫sin3⁡x1−sin2⁡xdxI5=−sin4⁡xcos⁡x+4∫sin3⁡xdx−4∫sin5⁡xdxI5=−sin4⁡xcos⁡x+4I3−4I55I5−4I3=−sin4⁡xcos⁡x+C

In=∫sinn⁡xdx, then 5I5−4I3 isequal to

I5=∫sin5xdxI5=∫sin4⁡xsin⁡xdx=sin4⁡x∫sin⁡xdx−∫ddxsin4⁡x∫sin⁡xdxdx Using ∫fgdx=f∫gdx−∫f1∫gdxdx=−sin4⁡x(cos⁡x)−∫4sin3⁡x⋅cos⁡x(−cos⁡x)dx=−sin4⁡xcos⁡x+4∫sin3⁡x1−sin2⁡xdxI5=−sin4⁡xcos⁡x+4∫sin3⁡xdx−4∫sin5⁡xdxI5=−sin4⁡xcos⁡x+4I3−4I55I5−4I3=−sin4⁡xcos⁡x+C

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$I_{n} = \int \sin^{n} x d x, then 5 I_{5} - 4 I_{3} isequal to$

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$\sin^{4} x \cos x + C$

$\cos^{4} x \sin x + C$

$- \sin^{4} x \cos x + C$

$- \cos^{4} x \sin x + C$

answer is C.

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Detailed Solution

$I_{5} = \int \sin^{5} x d x$

$\begin{array}{l} I_{5} = \int \sin^{4} x \sin x d x \\ = \sin^{4} x \int \sin x d x - \int [\frac{d}{d x} (\sin^{4} x) \int \sin x d x] d x \\ Using \int f g d x = f \int g d x - \int (f^{1} \int g d x) d x) \\ = - \sin^{4} x (\cos x) - \int 4 \sin^{3} x \cdot \cos x (- \cos x) d x \end{array}$

$\begin{array}{l} = - \sin^{4} x \cos x + 4 \int \sin^{3} x (1 - \sin^{2} x) d x \\ I_{5} = - \sin^{4} x \cos x + 4 \int \sin^{3} x d x - 4 \int \sin^{5} x d x \\ I_{5} = - \sin^{4} x \cos x + 4 I_{3} - 4 I_{5} \\ 5 I_{5} - 4 I_{3} = - \sin^{4} x \cos x + C \end{array}$