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∫ E 2 X Cos ( 3 X + 4 ) D X

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Question

\[\int e^{2x} \cos \left( 3x + 4 \right) \text{ dx }\]
Sum
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Solution

\[\text{ Let I } = \int e^{2x} \cos \left( 3x + 4 \right)dx\]
`\text{Considering cos  (3x + 4  ) as first function and` `\text{ e}^{2x}`   ` \text{ as second function} `
\[I = \text{ cos }\left( 3x + 4 \right)\frac{e^{2x}}{2} - \int - \text{ sin }\left( 3x + 4 \right) \times 3\frac{e^{2x}}{2}dx\]
\[ \Rightarrow I = \frac{e^{2x} \text{ cos }\left( 3x + 4 \right)}{2} + \frac{3}{2}\int e^{2x} \text{ sin }\left( 3x + 4 \right)dx\]
\[ \Rightarrow I = \frac{e^{2x} \text{ cos } \left( 3x + 4 \right)}{2} + \frac{3}{2} I_1 . . . . . \left( 1 \right)\]
\[\text{ where I}_1 = \int e^{2x} \text{ sin } \left( 3x + 4 \right)dx\]
`\text{Considering cos  (3x + 4  ) as first function and` `\text{ e}^{2x}`   ` \text{ as second function} `
\[ I_1 = \text{ sin } \left( 3x + 4 \right)\frac{e^{2x}}{2} - \int 3 \text{ cos }\left( 3x + 4 \right)\frac{e^{2x}}{2}dx\]
\[ \Rightarrow I_1 = \frac{e^{2x} \text{ sin } \left( 3x + 4 \right)}{2} - \frac{3}{2}\int e^{2x} \text{ cos} \left( 3x + 4 \right)dx\]
\[ \Rightarrow I_1 = \frac{e^{2x} \text{ sin }\left( 3x + 4 \right)}{2} - \frac{3}{2} I . . . . . \left( 2 \right)\]
` \text{ From ( 1 ) and ( 2 ) } `
\[I = \frac{e^{2x} \text{ cos }\left( 3x + 4 \right)}{2} + \frac{3}{4} \text{ e}^{2x} \text{ sin }\left( 3x + 4 \right) - \frac{9}{4}I\]
\[ \Rightarrow I + \frac{9}{4}I = \frac{2 e^{2x} \cos\left( 3x + 4 \right) + 3 e^{2x} \text{ sin }\left( 3x + 4 \right)}{4}\]
\[ \Rightarrow I = \frac{e^{2x}}{13}\left[ 2 \text{ cos } \left( 3x + 4 \right) + 3 \text{ sin }\left( 3x + 4 \right) \right] + C\]

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Chapter 18: Indefinite Integrals - Exercise 19.27 [Page 149]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 18 Indefinite Integrals
Exercise 19.27 | Q 4 | Page 149

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