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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS4 Marks
Question
Integrate the function in Exercise:
$\text{x}\ \cos^{-1}\text{x}$
✓

Answer

Let $\text{I}=\int\text{x}\cos^{-1}\text{x dx}$
Taking $\cos^{-1}\text{x}$ as first function and x as second function and integrating by parts, we obtain.
$=\cos^{-1}\text{x}\int\text{x} \ \text{dx}-\int\Bigg\{\Big(\frac{\text{d}}{\text{dx}}\cos^{-1}\text{x}\Big)\int\text{x} \ \text{dx}\Bigg\}\text{dx}$
$=\cos^{-1}\text{x}.\frac{\text{x}^2}{2}-\int\frac{-1}{\sqrt{1-\text{x}^2}}.\frac{\text{x}^2}{2}\text{dx}$
$=\frac{\text{x}^2\cos^{-1}\text{x}}{2}-\frac{1}{2}\int\frac{1-\text{x}^2-1}{\sqrt{1-\text{x}^2}}\text{dx}$
$=\frac{\text{x}^2\cos^{-1}\text{x}}{2}-\frac{1}{2}\int\Bigg\{\sqrt{1-\text{x}^2}+\Bigg(\frac{-1}{\sqrt{1-\text{x}^2}}\Bigg)\Bigg\}\text{dx}$
$=\frac{\text{x}^2\cos^{-1}\text{x}}{2}-\frac{1}{2}\int\sqrt{1-\text{x}^2}\text{dx}-\frac{1}{2}\int\Bigg(\frac{-1}{\sqrt{1-\text{x}^2}}\Bigg)\text{dx}$
$=\frac{\text{x}^2\cos^{-1}\text{x}}{2}-\frac{1}{2}\text{I}_1-\frac{1}{2}\cos^{-1}\text{x}\dots(\text{i})$
Where, $\text{I}_1=\int\sqrt{1-\text{x}^2}\text{dx}$
$\Rightarrow\ \text{I}_1=\text{x}\sqrt{1-\text{x}^2}-\int\frac{\text{d}}{\text{dx}}\sqrt{1-\text{x}^2}\int\text{x dx}$
$\Rightarrow\ \text{I}_1=\text{x}\sqrt{1-\text{x}^2}-\int\frac{-2\text{x}}{2\sqrt{1-\text{x}^2}}. \ \text{x dx}$
$\Rightarrow\ \text{I}_1=\text{x}\sqrt{1-\text{x}^2}-\int\frac{-\text{x}^2}{\sqrt{1-\text{x}^2}}.\text{dx}$
$\Rightarrow\ \text{I}_1=\text{x}\sqrt{1-\text{x}^2}-\int\frac{1-\text{x}^2-1}{\sqrt{1-\text{x}^2}}.\text{dx}$
$\Rightarrow\ \text{I}_1=\text{x}\sqrt{1-\text{x}^2}-\Bigg\{\int\sqrt{1-\text{x}^2}\text{dx}+\int\frac{-\text{dx}}{\sqrt{1-\text{x}^2}}\Bigg\}$
$\Rightarrow\ \text{I}_1=\text{x}\sqrt{1-\text{x}^2}-\{\text{I}_1+\cos^{-1}\text{x}\}$
$\Rightarrow\ 2\text{I}_1=\text{x}\sqrt{1-\text{x}^2}-\cos^{-1}\text{x}$
$\therefore \ \text{I}_1=\frac{\text{x}}{2}\sqrt{1-\text{x}^2}-\frac{1}{2}\cos^{-1}\text{x}$
Substituting in (i), we obtain
$\text{I}_1=\frac{\text{x}\cos^{-1}\text{x}}{2}-\frac{1}{2}\Bigg(\frac{\text{x}}{2}\sqrt{1-\text{x}^2}-\frac{1}{2}\cos^{-1}\text{x}\Bigg)-\frac{1}{2}\cos^{-1}\text{x}$
$=\frac{(2\text{x}^2-1)}{4}\cos^{-1}\text{x}-\frac{\text{x}}{4}\sqrt{1-\text{x}^2}+\text{C}$

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