Bihar BoardHindi Mediumकक्षा 12 साइन्सगणितसमाकलन2 Marks
Question
सिद्ध कीजिए $\int_{0}^{\frac\pi 2} $sin3 x dx = $\frac{2}{3}$
✓
Answer
माना I = $\int_{0}^{\frac\pi 2} $ sin3 x dx = $\int_{0}^{\frac\pi 2} $ sin2 x sin x dx = $\int_{0}^{\frac\pi 2} $(1 - cos2 x) sin x dx ($\because$ sin2 x = 1 - cos2x) cos x = t रखने पर, $\Rightarrow$ -sin x dx = dt जब x = 0 $\Rightarrow$ t = cos 0 = 1, जब x = $\frac{\pi}{2}$ $\Rightarrow$ t = cos $\frac{\pi}{2}$ = 0 $\therefore$ I $=\int_{0}^{\frac\pi 2}$ (1 - cos2 x) sin x dx = $\int_{1}^{0}$(1 - t2) (-dt) $=-\left[t-\frac{t^{3}}{3}\right]_{1}^{0}$ $=-\left\{(0-0)-\left(1-\frac{1}{3}\right)\right\}=\frac{2}{3}$
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