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Π / 2 ∫ 0 Cos 4 X D X

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प्रश्न

\[\int\limits_0^{\pi/2} \cos^4\ x\ dx\]

 

योग
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उत्तर

\[Let\ I = \int_0^\frac{\pi}{2} \cos^4 x\ dx\ . Then, \]
\[I = \int_0^\frac{\pi}{2} \left( \cos^2 x \right)^2 dx\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \frac{\left( 1 + \cos 2x \right)^2}{4} dx\]
\[ \Rightarrow I = \frac{1}{4} \int_0^\frac{\pi}{2} \left( 1 + \cos^2 2x + 2 \cos 2x \right) dx\]
\[ \Rightarrow I = \frac{1}{4} \int_0^\frac{\pi}{2} \left( 1 + 2 \cos 2x + \frac{1 + \cos 4x}{2} \right) dx\]
\[ \Rightarrow I = \frac{1}{4} \int_0^\frac{\pi}{2} \left( \frac{3 + 4 \cos 2x + \cos 4x}{2} \right) dx\]
\[ \Rightarrow I = \frac{1}{4} \left[ \frac{3x}{2} + \frac{2 \sin 2x}{2} + \frac{\sin 4x}{8} \right]_0^\frac{\pi}{2} \]
\[ \Rightarrow I = \frac{1}{4}\left[ \frac{3\pi}{4} + 0 \right]\]
\[ \Rightarrow I = \frac{3\pi}{16}\]

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अध्याय 19: Definite Integrals - Exercise 20.1 [पृष्ठ १६]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Exercise 20.1 | Q 22 | पृष्ठ १६

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