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Mathematics Question on Integrals of Some Particular Functions

If f(x) = ∫x0 t(sin x-sin t)dt then

A

f′′′(x)+f′′(x)=sin⁡xf '''(x)+f ''(x)=\sin x

B

f′′′(x)+f′′(x)−f(x)=cos⁡xf '''(x)+f''(x)-f (x)=\cos x

C

f′′′(x)+f′(x)=cos⁡x−2xsin⁡xf'''(x)+f '(x)=\cos x-2x \sin x

D

f′′′(x)−f′′(x)=cos⁡x−2xsin⁡xf '''(x)-f ''(x)=\cos x-2x \sin x

Answer

f′′′(x)+f′(x)=cos⁡x−2xsin⁡xf'''(x)+f '(x)=\cos x-2x \sin x

Explanation

Solution

f(x)=∫0xt(sin x−sin t)dtf\left(x\right) = \int^{^x}_{_0}t\left(sin\,x - sin\,t\right)dt
f(x)=sinx∫0xt dt−∫0xt sin tdtf\left(x\right) = sinx \int^{^x}_{_0}t\,dt - \int^{^x}_{_0} t \,sin\,tdt
f′(x)=(sinx)x+cosx∫0xt dt−x sinxf'\left(x\right) = \left(sinx\right) x +cosx \int^{^x}_{_0} t \,dt-x\,sinx
f′(x)=cosx∫0xtdtf'\left(x\right) = cosx \int^{^x}_{_0} tdt = xcos⁡xx\cos x
f′′(x)=(cosx)x−(sinx)∫0xtdtf''\left(x\right) = \left(cosx\right) x - \left(sinx\right) \int^{^x}_{_0} tdt
f′′′(x)=x(−sinx)+cosx−(sinx)x−(cosx)∫0xtdtf'''\left(x\right) = x\left(-sinx\right) + cosx-\left(sinx\right)x-\left(cosx\right) \int^{^x}_{_0} tdt
f′′′(x)+f′(x)=cosx−2x sinxf'''\left(x\right)+f'\left(x\right) = cosx-2x\,sinx