KCET · Maths · Indefinite Integration
When \(x>0\), then \(\int \cos ^{1}\left(\frac{1-x^{2}}{1+x^{2}}\right) d x\) is
- A \(2\left[x \tan { }^{1} x-\log \left(1+x^{2}\right)\right]+C\)
- B \(2\left[x \tan ^{1} x+\log \left(1+x^{2}\right)\right]+C\)
- C \(2 x \tan ^{-1} x+\log \left(1+x^{2}\right)+C\)
- D \(2 x \tan ^{-1} x-\log \left(1+x^{2}\right)+C\)
Answer & Solution
Correct Answer
(D) \(2 x \tan ^{-1} x-\log \left(1+x^{2}\right)+C\)
Step-by-step Solution
Detailed explanation
Given, when \(x>0\)
\[
=\int \cos ^{1}\left(\frac{1-x^{2}}{1+x^{2}}\right) d x
\]
\(\left[\begin{array}{l}\because x>0 \\ \therefore 2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\end{array}\right]\)
\(=\int 2 \tan ^{-1} x d x\)
\(=2 \int \frac{1}{\text { II }} \tan ^{-1} x d x\)
\(=2\left\{x \cdot \tan ^{-1} x-\int \frac{1}{1+x^{2}} \cdot x d x\right\}\)
\(=2 x \tan ^{-1} x-\int \frac{2 x}{1+x^{2}} d x\)
\(=2 x \tan ^{-1} x-\log \left(1+x^{2}\right)+C\)
\[
=\int \cos ^{1}\left(\frac{1-x^{2}}{1+x^{2}}\right) d x
\]
\(\left[\begin{array}{l}\because x>0 \\ \therefore 2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\end{array}\right]\)
\(=\int 2 \tan ^{-1} x d x\)
\(=2 \int \frac{1}{\text { II }} \tan ^{-1} x d x\)
\(=2\left\{x \cdot \tan ^{-1} x-\int \frac{1}{1+x^{2}} \cdot x d x\right\}\)
\(=2 x \tan ^{-1} x-\int \frac{2 x}{1+x^{2}} d x\)
\(=2 x \tan ^{-1} x-\log \left(1+x^{2}\right)+C\)
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