Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIndefinite Integrals3 Marks
Question
Evaluate the following integrals:$\int\frac{\text{x}\cos^{-1}\text{x}}{\sqrt{1-\text{x}^3}}\text{dx}$
✓
Answer
Let $\text{I}=\int\frac{\text{x}\cos^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}$
Let the first function be $\cos^{-1}\text{x}$ and second function be $\frac{\text{x}}{\sqrt{1-\text{x}^2}}\cdot$
First we find the integral of the second function, i.e, $\int\frac{\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}.$
Put $\text{t}=1-\text{x}^2.$ Then $\text{dt}=-2\text{x dx}$
Therefore,
$\int\frac{\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}=-\frac{1}{2}\int\frac{1}{\sqrt{\text{t}}}\text{dt}$
$=-\sqrt{\text{t}}$
$=-\sqrt{1-\text{x}^2}$
Hence, using integration by parts, we get
$\int\frac{\text{x}\cos^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}=(\cos^{-1}\text{x})\int\frac{\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}-\int\bigg[\bigg(\frac{\text{d}(\cos^{-1}\text{x})}{\text{dx}}\bigg)\int\bigg(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}\bigg)\bigg]\text{dx}$
$=(\cos^{-1})\big(-\sqrt{1-\text{x}^2}\big)-\int\Big(\frac{-1}{\sqrt{1-\text{x}^2}}\Big)\big(-\sqrt{1-\text{x}^2}\big)\text{dx}$
$=-\sqrt{1-\text{x}^2}\cos^{-1}\text{x}-\text{x+C}$
Hence, $\int\frac{\text{x}\cos^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}=-\sqrt{1-\text{x}^2}\cos^{-1}\text{x}-\text{x+C}$
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