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In=sinnxdx, then 5I54I3 isequal to 

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a
sin4xcosx+C
b
cos4xsinx+C
c
−sin4xcosx+C
d
−cos4xsinx+C

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detailed solution

Correct option is C

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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