मराठी

∫ 4 X 3 √ 5 − X 2 D X

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प्रश्न

\[\int4 x^3 \sqrt{5 - x^2} dx\]
बेरीज
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उत्तर

\[\int 4 x^3 \sqrt{5 - x^2} dx\]
\[ = 4\int x^2 \times x \sqrt{5 - x^2} \text{ dx }\]
\[\text{Let 5} - x^2 = t \]
\[ \Rightarrow x^2 = 5 - t\]
\[ \Rightarrow 2x = - \frac{dt}{dx}\]
\[ \Rightarrow \text{x dx} = - \frac{dt}{2}\]
\[Now, 4\int x^2 \times x \sqrt{5 - x^2} \text{ dx }\]
\[ = \frac{4}{- 2} \int\left( 5 - t \right) . \sqrt{t}  \text{ dt } \]
\[ = - 2\int5 t^\frac{1}{2} + 2 \int t^\frac{3}{2}  \text{ dt }\]
\[ = - 10 \left[ \frac{t^\frac{1}{2} + 1}{\frac{1}{2} + 1} \right] + 2 \left[ \frac{t^\frac{3}{2} + 1}{\frac{3}{2} + 1} \right] + C\]
\[ = - \frac{20}{3} t^\frac{3}{2} + \frac{4}{5} t^\frac{5}{2} + C\]
\[ = - \frac{20}{3} \left( 5 - x^2 \right)^\frac{3}{2} + \frac{4}{5} \left( 5 - x^2 \right)^\frac{5}{2} + C\]

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पाठ 18: Indefinite Integrals - Exercise 19.09 [पृष्ठ ५९]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 18 Indefinite Integrals
Exercise 19.09 | Q 69 | पृष्ठ ५९
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