Question

If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of $$f^{\prime }\left(x\right)$$

Answer 1

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

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

Differentiating both sides with respect to x, we get

$$f^{\prime }\left(x\right)=-[x\times (-\sin x)+\cos x\times 1]+\cos x$$
$$\Rightarrow f^{\prime }\left(x\right)=-(-x\sin x)-\cos x+\cos x$$
$$\Rightarrow f^{\prime }\left(x\right)=x\sin x$$

Thus, the value of $$f^{\prime }\left(x\right)$$ is `x sinx`.