Question

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Answer 1

The definite integral $\int\limits^1_0 {(\cos ^{-1} x)^2} \, dx$ can be evaluated by using integration by parts.

$\int\limits^1_0 {(\cos ^{-1} x)^2} \, dx=\int\limits^1_0 {1.(\cos ^{-1} x)^2} \, dx\\\int\limits^1_0 {(\cos ^{-1} x)^2} \, dx=[x.(\cos ^{-1} x)^2} ]_0^1+\int\limits^1_0 {\frac{x}{\sqrt{1-x^2}}.(\cos ^{-1} x)} \, dx \\\int\limits^1_0 {(\cos ^{-1} x)^2} \, dx=0-\int\limits^1_0 {\frac{-2x}{\sqrt{1-x^2}}.(\cos ^{-1} x)} \, dx \\\int\limits^1_0 {(\cos ^{-1} x)^2} \, dx=-\int\limits^1_0 {\frac{-2x}{\sqrt{1-x^2}}.(\cos ^{-1} x)} \, dx$

Again integrating by parts,

$\int\limits^1_0 {(\cos ^{-1} x)^2} \, dx=-2[\sqrt{1-x^2}(\cos ^{-1} x)]_0^1-2\int\limits^1_0 {\sqrt{1-x^2}.\frac{1}{\sqrt{1-x^2}}} \, dx\\ \int\limits^1_0 {(\cos ^{-1} x)^2} \, dx=\pi-2\int\limits^1_0 {1} \, dx\\ \int\limits^1_0 {(\cos ^{-1} x)^2} \, dx=\pi-2$

Answer 2

Answer:-

Use the differentiation method for this purpose

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