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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals3 Marks
Question
Evaluate the following integrals:

$\int\text{x}^2\sin^2\text{x dx}$

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Answer

Let $\text{I}=\int\text{x}^2\sin^2\text{x dx}$
$=\int\text{x}^2\Big(\frac{1-\cos2\text{x}}{2}\Big)\text{dx}$
$=\int\frac{\text{x}^2}{2}\text{dx}-\int\Big(\frac{\text{x}^2\cos2\text{x}}{2}\Big)\text{dx}$
$=\frac{\text{x}^3}{6}-\frac{1}{2}[\int\text{x}^2\cos2\text{x dx}]$
$=\frac{\text{x}^3}{6}-\frac{1}{2}[\text{x}^2\int\cos2\text{x dx}-\int(2\text{x}\int\cos2\text{x dx})\text{dx]}$
$=\frac{\text{x}^3}{6}-\frac{1}{2}\Big(\text{x}^2\frac{\sin2\text{x}}{2}\Big)+\frac{1}{2}\times2\int\Big(\text{x}\frac{\sin2\text{x}}{2}\Big)\text{dx}$
$=\frac{\text{x}^3}{6}-\frac{1}{4}\text{x}^2\sin2\text{x}+\frac{1}{2}[\text{x}\int\sin2\text{x dx}-\int(1\int\sin2\text{x dx})\text{dx}]$
$=\frac{\text{x}^3}{6}-\frac{1}{4}\text{x}^2\sin2\text{x}+\frac{1}{2}\Big[\text{x}\Big(-\frac{\cos2\text{x}}{2}\Big)-\int\Big(-\frac{\cos2\text{x}}{2}\Big)\text{dx}\Big]$
$=\frac{\text{x}^3}{6}-\frac{1}{4}\text{x}^2\sin2\text{x}-\frac{1}{4}\text{x}\cos2\text{x}+\frac{1}{4}\frac{\sin2\text{x}}{2}+\text{C}$
$\text{I}=\frac{\text{x}^3}{6}-\frac{1}{4}\text{x}^2\sin2\text{x}-\frac{1}{4}\text{x}\cos2\text{x}+\frac{1}{8}\sin2\text{x}+\text{C}$

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