MHT CET · Maths · Indefinite Integration
\(\int \cos ^{-1} x d x=\)
- A \(x \cos ^{-1} x+\sqrt{1-x^2}+c\)
- B \(-x \cos ^{-1} x+\sqrt{1+x^2}+c\)
- C \(x \cos ^{-1} x-\sqrt{1+x^2}+c\)
- D \(x \cos ^{-1} x-\sqrt{1-x^2}+c\)
Answer & Solution
Correct Answer
(D) \(x \cos ^{-1} x-\sqrt{1-x^2}+c\)
Step-by-step Solution
Detailed explanation
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 \(\sqrt{1-\mathrm{x}^2}=\mathrm{t} \Rightarrow \frac{(1)(-2 \mathrm{x})}{2 \sqrt{1-\mathrm{x}^2}} \mathrm{dx}=\mathrm{dt}\)
\(
=x \cos ^{-1} x-\int d t=x \cos ^{-1} x-t=\) \(x \cos ^{-1} x-\sqrt{1-x^2}+c
\)
\(
=x \cos ^{-1} x-\int \frac{-x}{\sqrt{1-x^2}} d x=x \cos ^{-1} x-\) \(\frac{1}{2} \int \frac{-2 x}{\sqrt{1-x^2}} d x
\)
Put \(\sqrt{1-\mathrm{x}^2}=\mathrm{t} \Rightarrow \frac{(1)(-2 \mathrm{x})}{2 \sqrt{1-\mathrm{x}^2}} \mathrm{dx}=\mathrm{dt}\)
\(
=x \cos ^{-1} x-\int d t=x \cos ^{-1} x-t=\) \(x \cos ^{-1} x-\sqrt{1-x^2}+c
\)
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