Q60 / 150 · MHT CET 2021 (22 Sep Shift 2) · Mathematics

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

  • A
    $x \cos ^{-1} x+\sqrt{1-x^2}+c$
  • B
    $-x \cos ^{-1} x+\sqrt{1+x^2}+c$
  • C
    $x \cos ^{-1} x-\sqrt{1+x^2}+c$
  • Correct answer
    D
    $x \cos ^{-1} x-\sqrt{1-x^2}+c$

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

Hence, the correct answer is option: 'D'.

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