Question

\(If ƒ(x)=∫_0^xt\ sin\ t\ dt,\ then  ƒ'(x)is\)

• A. \(cos\ x+x\ sin\ x\)
• B. \(x\ sin\ x\)
• C. \(x\ cos\ x\)
• D. \(sin\ x+x\ cos\ x\)

Answer 1

The Correct Option is B

\( ƒ(x)=∫_0^xt\ sin\ t\ dt\)

\(Integrating\  by\  parts, \ we \ obtain\)

\( ƒ(x)=t∫_0^x sin\ t\ dt-∫_0^x[{(\frac {d}{dt}t)∫sin\ t\ dt}]\ dt\)

= \([t(-cos\ t)]_0^x-∫_0^x(-cos\ t)dt\)

= \([-t\ cos\ t+sin\ t]_0^x\)

= \(-x\ cos\ x+sin\ x\)

\(⇒ ƒ(x)=-[{x(-sin\ x)}+cos\ x]+cos\ x\)

= \(x\ sin\ x-cos\ x+cos\ x\)

= \(x\ sin\ x\)

\(Hence, \ correct\  Answer\  is \ B.\)