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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals3 Marks
Question
Find the integral of the function sin4 x
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Answer

We can write, sin4x = sin2xsin2x
$= \left(\frac{1-\cos 2 x}{2}\right)\left(\frac{1-\cos 2 x}{2}\right)$ 
$=\frac{1}{4}(1-\cos 2 x)^{2}$ 
$= \frac{1}{4}\left[1+\cos ^{2} 2 x-2 \cos 2 x\right]$ 
$= \frac{1}{4}\left[1+\left(\frac{1+\cos 4 x}{2}\right)-2 \cos 2 x\right]$ 
$= \frac{1}{4}\left[1+\frac{1}{2}+\frac{1}{2} \cos 4 x-2 \cos 2 x\right]$ 
$= \frac{1}{4}\left[\frac{3}{2}+\frac{1}{2} \cos 4 x-2 \cos 2 x\right]$ 
Now, $\int \sin ^{4} x d x=\frac{1}{4} \int\left[\frac{3}{2}+\frac{1}{2} \cos 4 x-2 \cos 2 x\right] d x$ 
$= \frac{1}{4}\left[\frac{3}{2} x+\frac{1}{2}\left(\frac{\sin 4 x}{4}\right)-\frac{2 \sin 2 x}{2}\right]+C$ 
$= \frac{1}{8}\left[3 \mathrm{x}+\left(\frac{\sin 4 \mathrm{x}}{4}\right)-2 \sin 2 \mathrm{x}\right]+\mathrm{C}$ 
$= \frac{3 \mathrm{x}}{8}-\frac{1}{4} \sin 2 \mathrm{x}+\frac{1}{32} \sin 4 \mathrm{x}+\mathrm{C}$

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