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प्रश्न
उत्तर
\[Let\ I = \int_0^1 \left( \cos^{- 1} x \right)^2 d x . Then, \]
\[I = \int_0^1 1 \left( \cos^{- 1} x \right)^2 d x\]
\[\text{Integrating by parts}\]
\[ \Rightarrow I = \left[ x \left( \cos^{- 1} x \right)^2 \right]_0^1 - \int_0^1 2x \cos^{- 1} x \frac{- 1}{\sqrt{1 - x^2}} dx\]
\[\text{Again, integrating second term by parts}\]
\[ \Rightarrow I = \left[ x \left( \cos^{- 1} x \right)^2 \right]_0^1 + \left\{ 2 \left[ \sqrt{1 - x^2} \cos^{- 1} x \right]_0^1 - 2 \int_0^1 \frac{1}{\sqrt{1 - x^2}}\sqrt{1 - x^2} dx \right\}\]
\[ \Rightarrow I = \left[ x \left( \cos^{- 1} x \right)^2 \right]_0^1 + 2 \left[ \sqrt{1 - x^2} \cos^{- 1} x \right]_0^1 - 2 \left[ x \right]_0^1 \]
\[ \Rightarrow I = 0 + \frac{2\pi}{2} - 2\]
\[ \Rightarrow I = \pi - 2\]
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