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Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals1 Mark
Question
Integrate the function $x \cos^{-1} x$
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Answer

Let I = $\int x .cos^{-1} x$
Now, integrating by parts, we get,
$I=\cos ^{-1} x \int x d x-\int\left\{\left(\frac{d}{d x} \cos ^{-1} x\right) \int x d x\right\} d x$ 
$= \cos ^{-1} x \cdot \frac{x^{2}}{2}-\int \frac{-1}{\sqrt{1-x^{2}}} \cdot \frac{x^{2}}{2} d x$ 
$= \frac{x^{2} \cos ^{-1} x}{2}-\frac{1}{2} \int \frac{1-x^{2}-1}{\sqrt{1-x^{2}}} d x$ 
$= \frac{x^{2} \cos ^{-1} x}{2}-\frac{1}{2} \int\left\{\sqrt{1-x^{2}}+\left(\frac{-1}{\sqrt{1-x^{2}}}\right)\right\} d x$ 
 $=\frac{x^{2} \cos ^{-1} x}{2}-\frac{1}{2} \int \sqrt{1-x^{2}} d x-\frac{1}{2} \int\left(\frac{-1}{\sqrt{1-x^{2}}}\right) d x$ 
$= \frac{x^{2} \cos ^{-1} x}{2}-\frac{1}{2} I_{1}-\frac{1}{2} \cos ^{-1} x ...(i)$
Now, $I_{1}=\int \sqrt{1-x^{2}} d x$ 
$I_{1}=x \sqrt{1-x^{2}}-\int \frac{d}{d x} \sqrt{1-x^{2}} \int d x$ 
$I_{1}=x \sqrt{1-x^{2}}-\int \frac{-2 x}{2 \sqrt{1-x^{2}}} x \cdot d x$ 
$= x \sqrt{1-x^{2}}-\int \frac{-x^{2}}{\sqrt{1-x^{2}}} d x$ 
$= x \sqrt{1-x^{2}}-\int \frac{1-x^{2}-1}{\sqrt{1-x^{2}}} d x$ 
$= x \sqrt{1-x^{2}}-\left\{\int \sqrt{1-x^{2}} d x+\int \frac{-d x}{\sqrt{1-x^{2}}}\right\}$ 
$\therefore I_{1}=x \sqrt{1-x^{2}}-\left\{I_{1}+\cos ^{-1} x\right\}$ 
$\Rightarrow 2I_{1}=x \sqrt{1-x^{2}}-\cos ^{-1} x$
$\Rightarrow I_{1}=\frac{x}{2} \sqrt{1-x^{2}}-\frac{1}{2} \cos ^{-1} x$ 
Now, substituting in $(i),$ we get,
$I=\frac{x^{2} \cos ^{-1} x}{2}-\frac{1}{2}\left(\frac{x}{2} \sqrt{1-x^{2}}-\frac{1}{2} \cos ^{-1} x\right)-\frac{1}{2} \cos ^{-1} x$ 
$= \frac{\left(2 x^{2}-1\right)}{4} \cos ^{-1} x-\frac{x}{4} \sqrt{1-x^{2}}+C$ 

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