int sin ^4 x cos ^4 x , dx = | vedclass

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(B) $I = \int \sin ^4 x \cos ^4 x \, dx = \int \left(\frac{\sin 2x}{2}\right)^4 \, dx = \frac{1}{16} \int \sin ^4 2x \, dx$ $I = \frac{1}{16} \int \le

Vedclass · Accepted Answer

$\frac{1}{256}\left(-2 \sin ^3 2 x \cos 2 x-3 \sin 2 x \cos 2 x+6 x\right)+c$

Vedclass · Answer

(B) $I = \int \sin ^4 x \cos ^4 x \, dx = \int \left(\frac{\sin 2x}{2}\right)^4 \, dx = \frac{1}{16} \int \sin ^4 2x \, dx$ $I = \frac{1}{16} \int \left(\frac{1 - \cos 4x}{2}\right)^2 \, dx = \frac{1}{64} \int (1 - 2\cos 4x + \cos^2 4x) \, dx$ $I = \frac{1}{64} \int \left(1 - 2\cos 4x + \frac{1 + \cos 8x}{2}\right) \, dx = \frac{1}{128} \int (2 - 4\cos 4x + 1 + \cos 8x) \, dx$ $I = \frac{1}{128} \int (3 - 4\cos 4x + \cos 8x) \, dx = \frac{1}{128} \left(3x - \sin 4x + \frac{\sin 8x}{8}\right) + c$ Using $\sin 8x = 2 \sin 4x \cos 4x$ and $\sin 4x = 2 \sin 2x \cos 2x$,$\cos 4x = 1 - 2\sin^2 2x$: $I = \frac{1}{128} \left(3x - 2\sin 2x \cos 2x + \frac{2 \sin 4x \cos 4x}{8}\right) + c$ Simplifying leads to: $I = \frac{1}{256} (-2 \sin^3 2x \cos 2x - 3 \sin 2x \cos 2x + 6x) + c$

$\int \sin ^4 x \cos ^4 x \, dx =$

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