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Introduction to integration

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Question

$\int \sin 101 x \sin^{99} x d x = \frac{\sin (100 x) (\sin x)^{2}}{μ} then \frac{λ + μ}{100} equal to$

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Solution

$\begin{array}{l} \int \sin (100 + 1) x - \sin^{99} x = \int (\sin 100 x \cos x + \cos 100 x \sin x) \sin^{99} x \\ = \sin 100 x \frac{\sin^{100} x}{100} - \int \frac{1}{100} \sin^{100} x \cos 100 x (100) + \int \cos 100 x \sin^{100} x d x \\ = \frac{\sin 100 x \cdot \sin^{100} x}{100} λ = 100, μ = 100 \end{array}$

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