$\int \cos ^{-1} x d x=$

$\int \cos ^{-1} x d x=$
  1. $x \cos ^{-1} x+\sqrt{1-x^2}+c$
  2. $-x \cos ^{-1} x+\sqrt{1+x^2}+c$
  3. $x \cos ^{-1} x-\sqrt{1+x^2}+c$
  4. $x \cos ^{-1} x-\sqrt{1-x^2}+c$

Solution

Let $I=\int \cos ^{-1} x d x=\int\left(\cos ^{-1} x\right) \cdot(1) d x$ $=x \cos ^{-1} x-\int \frac{-x}{\sqrt{1-x^2}} d x=x \cos ^{-1} x-\frac{1}{2} \int \frac{-2 x}{\sqrt{1-x^2}} d x$ Put $\sqrt{1-\mathrm{x}^2}=\mathrm{t} \Rightarrow \frac{(1)(-2 \mathrm{x})}{2 \sqrt{1-\mathrm{x}^2}} \mathrm{dx}=\mathrm{dt}$ $=x \cos ^{-1} x-\int d t=x \cos ^{-1} x-t=x \cos ^{-1} x-\sqrt{1-x^2}+c$

Asked in: MHT CET 2021 (22 Sep Shift 2)

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