Bihar BoardHindi Mediumकक्षा 12 साइन्सगणितसमाकलन5 Marks
Question
सिद्ध कीजिए $\int_{0}^{1}$ sin-1 x dx = $\frac{\pi}{2}$ - 1
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Answer
माना I = $\int_{0}^{1}$ sin-1 x dx = $\int_{0}^{1}$ sin-1 x $\cdot$ 1 dx sin-1 x को पहला तथा 1 को दूसरा फलन लेकर खण्डशः समाकलन का नियम लगाने पर, $I=\left[\left(\sin ^{-1} x\right) x\right]_{0}^{1}$ $-\int_{0}^{1} \frac{x}{\sqrt{1-x^{2}}} d x $ 1 - x2 = t रखने पर, $\Rightarrow$ -2x dx = dt जब x = 0 $\Rightarrow$ t = 1 और जब x = 1 $\Rightarrow$ 0 $\therefore$ $I=\left[x \sin ^{-1} x\right]_{0}^{1}+\frac{1}{2} \int_{1}^{0} \frac{d t}{\sqrt{t}}$ $=\left[x \sin ^{-1} x\right]_{0}^{1}+\frac{1}{2}\left[\frac{t^{\frac1 2}}{\frac1 2}\right]_{1}^{0}$ $=1 \sin ^{-1}(1)+[-\sqrt{1}]$ $=\frac{\pi}{2}$ - 1
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