CBSE Boardहिन्दी माध्यमकक्षा 12 साइन्सगणितसमाकलन2 Marks
Question
सिद्ध कीजिए $\int_{0}^{\frac\pi 2} \sin^3 x dx = \frac{2}{3}$
✓
Answer
माना I = $\int_{0}^{\frac\pi 2} \sin^3 x dx = \int_{0}^{\frac\pi 2} \sin^2x \sin x dx$
= $\int_{0}^{\frac\pi 2} (1 - \cos^2 x) sin\ x\ dx (\because \sin^2 x = 1 - \cos^2x$)
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 $=\int_{0}^{\frac\pi 2} (1 - cos^2 x)$ sin x dx = $\int_{1}^{0}(1 - t^2) (-dt)$
$=-\left[t-\frac{t^{3}}{3}\right]_{1}^{0}$ $=-\left\{(0-0)-\left(1-\frac{1}{3}\right)\right\}=\frac{2}{3}$
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