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प्रश्न
\[\int\limits_0^{\pi/2} \cos^5 x\ dx\]
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उत्तर
\[Let\ I = \int_0^\frac{\pi}{2} \cos^5 x\ d\ x . Then, \]
\[I = \int_0^\frac{\pi}{2} \cos^4 x \cos\ x\ d x\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \left( 1 - \sin^2 x \right)^2 \cos x\ dx\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \left( 1 - 2 \sin^2 x + \sin^4 x \right) \cos x dx \]
\[Let\ \sin x = t . Then, \cos dx = du\]
\[When\ x = 0, t = 0\ and\ x = \frac{\pi}{2}, t = 1\]
\[ \therefore I = \int_0^1 \left( 1 - 2 t^2 + t^4 \right) dt\]
\[ \Rightarrow I = \left[ t - \frac{2 t^3}{3} + \frac{t^5}{5} \right]_0^1 \]
\[ \Rightarrow I = 1 - \frac{2}{3} + \frac{1}{5}\]
\[ \Rightarrow I = \frac{8}{15}\]
\[I = \int_0^\frac{\pi}{2} \cos^4 x \cos\ x\ d x\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \left( 1 - \sin^2 x \right)^2 \cos x\ dx\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \left( 1 - 2 \sin^2 x + \sin^4 x \right) \cos x dx \]
\[Let\ \sin x = t . Then, \cos dx = du\]
\[When\ x = 0, t = 0\ and\ x = \frac{\pi}{2}, t = 1\]
\[ \therefore I = \int_0^1 \left( 1 - 2 t^2 + t^4 \right) dt\]
\[ \Rightarrow I = \left[ t - \frac{2 t^3}{3} + \frac{t^5}{5} \right]_0^1 \]
\[ \Rightarrow I = 1 - \frac{2}{3} + \frac{1}{5}\]
\[ \Rightarrow I = \frac{8}{15}\]
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