Question
यदि $f(x)=\int \limits_{0}^{x} t(\sin x-\sin t) d t$ है, तो
$ = \sin x\int_0^x {t.dt} - \int_0^x {t\sin t.dt} $
$ = \frac{{{x^2}}}{2}\sin x + \left[ {t\cos t_0^x} \right] + \sin x$
$ \Rightarrow f\left( x \right) = \frac{{{x^2}}}{2}\sin x + x\cos x + \sin x$
$f'\left( x \right) = \frac{{{x^2}}}{2}\cos x + 2\,\cos x$
$f''\left( x \right) = x\cos x - \frac{{{x^2}}}{2}\sin x - 2\sin x$
$f'''\left( x \right) = \cos x - 2x\sin x - \frac{{{x^2}}}{2}\cos x - 2\cos x$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.