Given f(x)=∫0xt(sinx−sint)dt
⇒f(x)=sinx∫0xtdt−∫0xtsintdt
On differentiating, we get
⇒f′(x)=(sinx)x+cosx∫0xtdt−xsinx
⇒f′(x)=cosx∫0xtdt
Differentiating the above equation again, we get
⇒f′′(x)=(cosx)x−(sinx)∫0xtdt
⇒f′′′(x)=x(−sinx)+cosx−(sinx)x−(cosx)∫0xtdt
⇒f′′′(x)+f′(x)=cosx−2xsinx.