Integrate $\sin(101x)\cdot \sin^{99}x$

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

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.