e^{4 x}(\;3 x - 3 x) dx =

Correct Answer is : e4x25(sin⁡ 3x−7cos⁡3x)+C\;{e^{4 x}}{25}(\;3 x - 7 3 x) + C25e4x(sin3x−7cos3x)+C

\;{e^{4 x}}{25}(7 \;3 x - 3 x) + C

\;{e^{4 x}}{25}(\;3 x - 7 3 x) + C

\;{e^{4 x}}{5}(7 \;3 x + 3 x) + C

\;{e^{4 x}}{5}(\;3 x + 7 3 x) + C

Correct Answer is : e4x25(sin⁡ 3x−7cos⁡3x)+C\;{e^{4 x}}{25}(\;3 x - 7 3 x) + C25e4x(sin3x−7cos3x)+C

Test Index

TG EAMCET 2 May 2025 Shift 1 Paper

Question : 73 of 160

Marks: +1, -0

\int e^{4 x}(\sin\;3 x - \cos 3 x) dx =

Solution:

\int e^{4 x}(\sin\;3 x - \cos 3 x) dx

= \int e^{4 x} \sin\;3 x dx - \int e^{4 x} \cos 3 x dx

Using the standard integrals

\;\int e^{a x} \sin\;b x dx = \;\frac{e^{a x}}{a^2+b^2}(a \sin\;b x - b \cos b x)

\;\int e^{a x} \cos b x dx = \;\frac{e^{a x}}{a^2+b^2}(a \cos b x + b \sin\;b x)

Here,

a=4,\, b=3

So,

\int e^{4 x}(\sin\;3 x - \cos 3 x) dx

\;=\;\frac{e^{4 x}}{4^2+3^2}(4 \sin\;3 x - 3 \cos 3 x) - \;\frac{e^{4 x}}{4^2+3^2}(4 \cos 3 x + 3 \sin\;3 x) + C

\Rightarrow \;\frac{e^{4 x}}{25}[4 \sin\;3 x - 3 \cos 3 x - 4 \cos 3 x - 3 \sin\;3 x] + C

\Rightarrow \;\frac{e^{4 x}}{25}[\sin\;3 x - 7 \cos 3 x] + C

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