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The Value of 2 π ∫ 0 √ 1 + Sin X 2 D X Is(A) 0 (B) 2 (C) 8 (D) 4

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प्रश्न

The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is 

विकल्प

  • 0

  • 2

  • 8

  • 4

MCQ
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उत्तर

8

\[\int_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}} d x\]
\[ = \int_0^{2\pi} \sqrt{\sin^2 \frac{x}{4} + \cos^2 \frac{x}{4} + 2\sin\frac{x}{4}\cos\frac{x}{4}} d x\]
\[ = \int_0^{2\pi} \left( \sin\frac{x}{4} + \cos\frac{x}{4} \right)dx\]
\[ = \left[ \frac{- \cos\frac{x}{4}}{\frac{1}{4}} + \frac{\sin\frac{x}{4}}{\frac{1}{4}} \right]_0^{2\pi} \]
\[ = 4 \left[ \sin\frac{x}{4} - \cos\frac{x}{4} \right]_0^{2\pi} \]
\[ = 4\left[ \sin\frac{2\pi}{4} - \cos\frac{2\pi}{4} - \sin 0 + \cos 0 \right]\]
\[ = 4\left[ \sin\frac{\pi}{2} - \cos\frac{\pi}{2} - 0 + 1 \right]\]
\[ = 4\left[ 1 - 0 - 0 + 1 \right]\]
\[ = 4 \times 2\]
\[ = 8\]

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अध्याय 19: Definite Integrals - MCQ [पृष्ठ ११७]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
MCQ | Q 4 | पृष्ठ ११७

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