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Question: \[\int_{}^{}{\sqrt{\mathbf{1 +}\mathbf{\sin}\left( \frac{\mathbf{x}}{\mathbf{4}} \right)}\mathbf{dx ...

∫1+sin⁡(x4)dx=\int_{}^{}{\sqrt{\mathbf{1 +}\mathbf{\sin}\left( \frac{\mathbf{x}}{\mathbf{4}} \right)}\mathbf{dx =}}

A

8(sin⁡x8−cos⁡x8)+c8\left( \sin\frac{x}{8} - \cos\frac{x}{8} \right) + c

B

(sin⁡x8+cos⁡x8)+C\left( \sin\frac{x}{8} + \cos\frac{x}{8} \right) + C

C

18(sin⁡x8−cos⁡x8)+c\frac{1}{8}\left( \sin\frac{x}{8} - \cos\frac{x}{8} \right) + c

D

8(cos⁡x8−sin⁡x8)+c8\left( \cos\frac{x}{8} - \sin\frac{x}{8} \right) + c

Answer

8(sin⁡x8−cos⁡x8)+c8\left( \sin\frac{x}{8} - \cos\frac{x}{8} \right) + c

Explanation

Solution

∫1+sin⁡(x4)dx=∫[(sin⁡2x8+cos⁡2x8)+(2sin⁡x8.cos⁡x8)]dx\int_{}^{}\sqrt{1 + \sin\left( \frac{x}{4} \right)dx} = \int_{}^{}\sqrt{\left\lbrack \left( \sin^{2}\frac{x}{8} + \cos^{2}\frac{x}{8} \right) + \left( 2\sin\frac{x}{8}.\cos\frac{x}{8} \right) \right\rbrack}dx =∫(sin⁡x8+cos⁡x8)2\mathbf{=}\int_{}^{}\sqrt{\left( \mathbf{\sin}\frac{\mathbf{x}}{\mathbf{8}}\mathbf{+}\mathbf{\cos}\frac{\mathbf{x}}{\mathbf{8}} \right)^{\mathbf{2}}}

=∫(sin⁡x8+cos⁡x8)dx=−cos⁡x/81/8+sin⁡x/81/8\mathbf{=}\int_{}^{}\left( \mathbf{\sin}\frac{\mathbf{x}}{\mathbf{8}}\mathbf{+}\mathbf{\cos}\frac{\mathbf{x}}{\mathbf{8}} \right)\mathbf{dx}\mathbf{=}\frac{\mathbf{-}\mathbf{\cos}\mathbf{x}\mathbf{/8}}{\mathbf{1/8}}\mathbf{+}\frac{\mathbf{\sin}\mathbf{x}\mathbf{/8}}{\mathbf{1/8}}

=8(sin⁡x8−cos⁡x8)+c\mathbf{= 8}\left( \mathbf{\sin}\frac{\mathbf{x}}{\mathbf{8}}\mathbf{-}\mathbf{\cos}\frac{\mathbf{x}}{\mathbf{8}} \right)\mathbf{+ c}