मराठी

∫ E 2 X Sin X D X

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प्रश्न

\[\int e^{2x} \sin x\ dx\]
बेरीज
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उत्तर

\[\text{ Let I }= \int e^{2x} \text{ sin  x  dx }\]
`\text{Considering sin  x  as first function and` `\text{ e}^{2x}`   ` \text{ as second function} `
\[I = \sin x\frac{e^{2x}}{2} - \int \cos x\frac{e^{2x}}{2}dx\]
\[ \Rightarrow I = \text{ sin  x}\frac{e^{2x}}{2} - \frac{1}{2}\int \text{ cos  x  e }^{2x} \text{ dx }\]
\[ \Rightarrow I = \frac{\text{ sin  x  e}^{2x}}{2} - \frac{1}{2}\left[ \cos x\frac{e^{2x}}{2} - \int\left( - \sin x \right)\frac{e^{2x}}{2}dx \right]\]
\[ \Rightarrow I = \frac{\text{ sin  x  e}^{2x}}{2} - \frac{\text{ cos  x  e}^{2x}}{4} - \frac{1}{2}\int\frac{e^{2x} \sin x}{2}dx\]
\[I = \frac{e^{2x} \left( 2 \sin x - \cos x \right)}{4} - \frac{I}{4}\]
\[ \Rightarrow 5I = e^{2x} \left( 2 \sin x - \cos x \right)\]
\[ \Rightarrow I = \frac{e^{2x} \left( 2 \sin x - \cos x \right)}{5} + C\]

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पाठ 18: Indefinite Integrals - Exercise 19.27 [पृष्ठ १४९]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 18 Indefinite Integrals
Exercise 19.27 | Q 6 | पृष्ठ १४९

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