$\int(\sin 100 x \cos x+\cos 100 x \sin x) \cdot \sin ^{99} x d x$
$ = \int {\underbrace {\sin 100x}_I} \underbrace {\cos x{{\sin }^{99}}x}_{II}dx + \int {\cos } (100x){(\sin x)^{100}}dx$
$\Rightarrow I=\frac{\sin (100 x)(\sin x)^{100}}{100}-\frac{100}{100}$
$\int \cos (100 x)(\sin x)^{100} d x+\int \cos (100 x)(\sin x)^{100} d x$
$\Rightarrow \mathrm{I}=\frac{\sin (100 \mathrm{x})(\sin \mathrm{x})^{100}}{100}+\mathrm{c}$
$\lambda=100, \quad \mu=100 \Rightarrow \frac{\lambda}{\mu}=1$
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$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad k \quad, \quad x=0$
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\frac{\cos ^{2} x-\sin ^{2} x-1}{\sqrt{x^{2}+1}-1} ,\,\,\, x>0$
is continuous at $x=0$, then $\frac{1}{a}+\frac{1}{b}+\frac{4}{k}$ is equal to :