MCQ
જો $\int_{}^{} {{e^x}\sin x\;dx = \frac{1}{2}{e^x}\;.\;a + c} $, તો $a = $
- ✓$\sin x - \cos x$
- B$\cos x - \sin x$
- C$ - \cos x - \sin x$
- D$\cos x + \sin x$
Let $I = \int_{}^{} {{e^x}\sin x\,dx} = - {e^x}\cos x + \int_{}^{} {{e^x}\cos x\,dx + c} $
$ = - {e^x}\cos x + {e^x}\sin x - \int_{}^{} {{e^x}\sin x\,dx + c} $
$ \Rightarrow 2I = {e^x}( - \cos x + \sin x) + c$. Now from $(i),$
we get $\frac{1}{2}{e^x}a = \frac{1}{2}{e^x}(\sin x - \cos x) \Rightarrow a = \sin x - \cos x.$
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$\begin{array}{|l|l|l|l|l|l|} \hline X=x & 0 & 1 & 2 & 3 & 4 \\ \hline P(X=x) & \frac{1}{3} & \frac{1}{2} & 0 & \frac{1}{6} & 0 \\ \hline \end{array}$
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