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Question: \[\int_{}^{}\left( \mathbf{\sin}^{\mathbf{4}}\mathbf{x}\mathbf{-}\mathbf{\cos}^{\mathbf{4}}\mathbf{x...

∫(sin⁡4x−cos⁡4x)dx=\int_{}^{}\left( \mathbf{\sin}^{\mathbf{4}}\mathbf{x}\mathbf{-}\mathbf{\cos}^{\mathbf{4}}\mathbf{x} \right)\mathbf{dx =}

A

−cos⁡2x2+c- \frac{\cos 2x}{2} + c

B

−sin⁡2x2+c- \frac{\sin 2x}{2} + c

C

sin⁡2x2+c\frac{\sin 2x}{2} + c

D

cos⁡2x2+c\frac{\cos 2x}{2} + c

Answer

−sin⁡2x2+c- \frac{\sin 2x}{2} + c

Explanation

Solution

∫(sin⁡4x−cos⁡4x)dx=∫(sin⁡2x−cos⁡2x)(sin⁡2x+cos⁡2x)dx=∫(sin⁡2x−cos⁡2x)dx=−∫(cos⁡2x−sin⁡2x)dx=−∫cos⁡2xdx\int_{}^{}{(\sin^{4}x - \cos^{4}x)dx} = \int_{}^{}{(\sin^{2}x - \cos^{2}x)(\sin^{2}x + \cos^{2}x)dx} = \int_{}^{}{(\sin^{2}x - \cos^{2}x)dx}\mathbf{= -}\int_{}^{}{\mathbf{(}\mathbf{\cos}^{\mathbf{2}}\mathbf{x}}\mathbf{-}\mathbf{\sin}^{\mathbf{2}}\mathbf{x}\mathbf{)dx = -}\int_{}^{}{\mathbf{\cos}\mathbf{2}\mathbf{xdx}} =−sin⁡2x2+C\mathbf{=}\frac{\mathbf{-}\mathbf{\sin}\mathbf{2}\mathbf{x}}{\mathbf{2}}\mathbf{+ C}