Question

If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]

Answer 1

\[f\left( x \right) = \int_0^x t\sin\ tdt\]
\[ \Rightarrow f\left( x \right) = \left.{t\left( - \cos t \right)}\right|_0^x - \int_0^x \frac{d}{dt}\left( t \right) \times \left( - \cos t \right)dt\]
\[ \Rightarrow f\left( x \right) = - \left( x\cos x - 0 \right) + \int_0^x \cos t dt\]
\[ \Rightarrow f\left( x \right) = - x\cos x + \left.\sin t\right|_0^x\]

\[\Rightarrow f\left( x \right) = - x\cos x + \left( \sin x - 0 \right)\]
\[ \Rightarrow f\left( x \right) = - x\cos x + \sin x\]

Differentiating both sides with respect to x, we get

\[f'\left( x \right) = - \left[ x \times \left( - \sin x \right) + \cos x \times 1 \right] + \cos x\]
\[ \Rightarrow f'\left( x \right) = - \left( - x\sin x \right) - \cos x + \cos x\]
\[ \Rightarrow f'\left( x \right) = x\sin x\]

Thus, the value of \[f'\left( x \right)\] is `x sinx`.