Some big numbers

$I=\int { \sin\left( 101x \right) { \sin }^{ 99 }x } dx\\ =\int { \sin\left( 100x+x \right) { \sin }^{ 99 }x } dx\\ =\int { \sin\left( 100x \right) (\cos x{ )(\sin }^{ 99 }x) } dx+\int { \cos(100x{ )\sin }^{ 100 }x } dx$

Now on using Integration By Parts for first integral and leaving the second integral as it is, we get

$I=\frac { \left( \sin100x \right) \left( { \sin }^{ 100 }x \right) }{ 100 } -\int { \cos(100x{ )\sin }^{ 100 }x } dx+\int { \cos(100x{ )\sin }^{ 100 }x } dx+c\\ \therefore \quad I=\frac { \left( \sin 100x \right) \left( { \sin }^{ 100 }x \right) }{ 100 } +\mathcal{C}$

I considered $\sin(100x)$ as frist function and $(\cos x{ )(\sin }^{ 99 }x)$ as second function while using integration by parts.

Another way to solve this question is to directly differentiate the expression $\dfrac{ \sin(ax) (\sin x)^b}{k} + C$ with respect to $x$ , by differentiation - chain rule ,

$\begin{aligned}\dfrac d{dx} \dfrac1k \left [ \sin(ax) (\sin x)^b \right ] &=& \dfrac1k \left [b\sin(ax) (\sin x)^{b-1} \cos x + a \cos (ax) (\sin x)^b \right ] \\ &=& \dfrac1k (\sin x)^{b-1} \left [ b \sin(ax) \cos x + a \cos (ax) \sin x \right ] \\ \end{aligned}$

Since the aforementioned expression is equal to $\sin (101 x) \sin^{99}x$ , this suggest that $(\sin x)^{b-1} = (\sin x)^{99} \Rightarrow b -1 =99 \Leftrightarrow b = 100$ . Continuing with our solution, we have

$\dfrac1k (\sin x)^{b-1} \left [ b \sin(ax) \cos x + a \cos (ax) \sin x \right ] = \dfrac1k (\sin x)^{99} \left [ 100 \sin(ax) \cos x + a \cos (ax) \sin x \right ]$

Recall the compound angle formula, $\sin (A\pm B) = \sin A \cos B \pm \cos A \sin B$ , so to combine $100 \sin(ax) \cos x + a \cos (ax) \sin x$ into a single expression, it is reasonable to assume that $a=b=100$ .

$\begin{aligned}&& \dfrac1k (\sin x)^{99} \left [ 100 \sin(ax) \cos x + a \cos (ax) \sin x \right ] \\ &= &\dfrac1k (\sin x)^{99} \left [ 100 \sin(ax) \cos x + 100 \cos (ax) \sin x \right ] \\ &= & \dfrac{100} k (\sin x)^{99} [ \sin(100x) \cos x + \cos (100 x) \sin x ] \\ &=& \dfrac{100} k (\sin x)^{99} [ \sin(100x + x) ] \\ &=& \dfrac{100} k (\sin x)^{99} [ \sin(1001 x) ] \\ \end{aligned}$

Comparing it with the expression in question suggests that $k = 100$ as well. Double checking our work indeed shows that $a=b=k=100$ works. Thus our answer is $a+b+k=\boxed{300}$ .