- ✓$\sin x - \cos x$
- 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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If the mean and variance of a binomial distribution are 4 and 3, respectively, the probability of getting exactly six successes in this distribution is:
$\text{ }^{16}\text{C}_6\big(\frac{1}{4}\big)^{10}\big(\frac{3}{4}\big)^6$
$\text{ }^{16}\text{C}_6\big(\frac{1}{4}\big)^{6}\big(\frac{3}{4}\big)^{10}$
$\text{ }^{12}\text{C}_6\big(\frac{1}{20}\big)\big(\frac{3}{4}\big)^6$
$\text{ }^{12}\text{C}_6\big(\frac{1}{20}\big)^6\big(\frac{3}{4}\big)^6$
$\pi$
$\frac{\pi}{2}$
$0$
$2\pi$