मराठी

Π / 2 ∫ 0 √ Sin ϕ Cos 5 ϕ D ϕ

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प्रश्न

\[\int\limits_0^{\pi/2} \sqrt{\sin \phi} \cos^5 \phi\ d\phi\]

 

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उत्तर

\[\int_0^\frac{\pi}{2} \sqrt{\sin \phi} \cos^5 \phi\ d \phi\]

\[Let\ \sin \phi = t . Then, \cos \phi\ d\phi = dt\]

\[When\ \phi = 0, t = 0\ and\ \phi = \frac{\pi}{2}, t = 1\]

\[Also, \cos^5 \phi = \cos^4 \phi \cos \phi = \left( 1 - \sin^2 \phi \right)^2 \cos \phi\]

\[ \therefore I = \int_0^\frac{\pi}{2} \sqrt{\sin \phi} \cos^5 \phi d \phi\]

\[ \Rightarrow I = \int_0^1 \sqrt{t} \left( 1 - t^2 \right)^2 dt\]

\[ \Rightarrow I = \int_0^1 \sqrt{t}\left( 1 + t^4 - 2 t^2 \right) dt\]

\[ \Rightarrow I = \int_0^1 \left( \sqrt{t} + t^\frac{9}{2} - 2 t^\frac{5}{2} \right) dt\]

\[ \Rightarrow I = \left[ \frac{2 t^\frac{3}{2}}{3} + \frac{2 t^\frac{11}{2}}{11} - \frac{4 t^\frac{7}{2}}{7} \right]_0^1 \]

\[ \Rightarrow I = \frac{2}{3} + \frac{2}{11} - \frac{4}{7}\]

\[ \Rightarrow I = \frac{64}{231}\]

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पाठ 19: Definite Integrals - Exercise 20.2 [पृष्ठ ३९]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Exercise 20.2 | Q 11 | पृष्ठ ३९

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