Definite Integration
General
Grade 12

Question:

If $I_{m,n} = \int_{0}^{\pi/2} \sin^m x \cdot \cos^n x dx$, then show that $I_{m,n} = \frac{m-1}{m+n} I_{m-2,n}$

Step-by-Step Solution

Key Concept: General
$$I_{m,n} = \int_{0}^{\pi/2} \sin^{m-1} x (\sin x \cos^n x) dx$$ $$= \left[ -\frac{\sin^{m-1} x \cdot \cos^{n+1} x}{n+1} \right]_0^{\pi/2} + \int_{0}^{\pi/2} \frac{\cos^{n+1} x}{n+1} (m-1) \sin^{m-2} x \cos x dx$$ $$= \left( \frac{m-1}{n+1} \right) \int_{0}^{\pi/2} \sin^{m-2} x \cdot \cos^n x \cdot \cos^2 x dx$$ $$= \left( \frac{m-1}{n+1} \right) \int_{0}^{\pi/2} (\sin^{m-2} x \cdot \cos^n x - \sin^m x \cdot \cos^n x) dx$$ $$= \left( \frac{m-1}{n+1} \right) I_{m-2,n} - \left( \frac{m-1}{n+1} \right) I_{m,n} \Rightarrow \left( 1 + \frac{m-1}{n+1} \right) I_{m,n} = \left( \frac{m-1}{n+1} \right) I_{m-2,n}$$ $$I_{m,n} = \left( \frac{m-1}{m+n} \right) I_{m-2,n}$$
Correct Answer: A

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