વિધેય $\cos^{4} 2x$ નું સંકલન શોધો.

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આપણે જાણીએ છીએ કે $\cos^{2} \theta = \frac{1+\cos 2\theta}{2}$.
$\cos^{4} 2x = (\cos^{2} 2x)^{2} = \left(\frac{1+\cos 4x}{2}\right)^{2}$
$= \frac{1}{4} (1 + 2\cos 4x + \cos^{2} 4x)$
$= \frac{1}{4} \left[1 + 2\cos 4x + \frac{1+\cos 8x}{2}\right]$
$= \frac{1}{4} \left[1 + 2\cos 4x + \frac{1}{2} + \frac{\cos 8x}{2}\right]$
$= \frac{1}{4} \left[\frac{3}{2} + 2\cos 4x + \frac{\cos 8x}{2}\right] = \frac{3}{8} + \frac{1}{2}\cos 4x + \frac{1}{8}\cos 8x$
હવે,$x$ ની સાપેક્ષમાં સંકલન કરતા:
$\int \cos^{4} 2x \, dx = \int \left(\frac{3}{8} + \frac{1}{2}\cos 4x + \frac{1}{8}\cos 8x\right) \, dx$
$= \frac{3}{8}x + \frac{1}{2} \cdot \frac{\sin 4x}{4} + \frac{1}{8} \cdot \frac{\sin 8x}{8} + C$
$= \frac{3}{8}x + \frac{\sin 4x}{8} + \frac{\sin 8x}{64} + C$,જ્યાં $C$ એ સ્વૈર અચળાંક છે.

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નીચે આપેલ વિધાન $(A)$ અને કારણ $(R)$ ધ્યાનમાં લો:
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