Mathematics

 $$\displaystyle \int e^{2x-3}+7^{4-3(x/2)}+\sin \left ( 3x-\frac{1}{2} \right )+\cos \left ( \frac{2}{5}x-2 \right )+a^{3x+2}dx$$  is 


SOLUTION
Let $$\displaystyle I=\int  e^{ 2x-3 }+7^{ 4-3(x/2) }+\sin  \left( 3x-\frac { 1 }{ 2 }  \right) +\cos  \left( \frac { 2 }{ 5 } x-2 \right) +\alpha ^{ 3x+2 }dx\\ ={ I }_{ 1 }+{ I }_{ 2 }+{ I }_{ 3 }+{ I }_{ 4 }+{ I }_{ 5 }$$
Where
$$\displaystyle { I }_{ 1 }=\int { e^{ 2x-3 }dx } =\frac { 1 }{ 2 } e^{ 2x-3 }$$

$$\displaystyle { I }_{ 2 }=\int { 7^{ 4-3(x/2) }dx } $$
Put $$\displaystyle t=4-\frac { 3x }{ 2 } \Rightarrow dt=-\frac { 3 }{ 2 }dx $$

$$\displaystyle { I }_{ 2 }=-\frac { 2 }{ 3 } \int { 7^{ t }dt }=-\frac{2}{3}\frac{7^t}{\log 7} =-\frac { 2 }{ 3 } \cdot 7^{ 4-3(x/2) }\cfrac{1}{\log 7}$$

$$\displaystyle { I }_{ 3 }=\int { \sin  \left( 3x-\frac { 1 }{ 2 }  \right) dx } =-\frac { 1 }{ 3 } \cos  \left( 3x-\frac { 1 }{ 2 }  \right) $$

$$\displaystyle { I }_{ 4 }=\int { \cos  \left( \frac { 2 }{ 5 } x-2 \right) dx } =\frac { 5 }{ 2 } \sin  \left( \frac { 2 }{ 5 } x-2 \right) $$

$$\displaystyle { I }_{ 5 }=\int { a ^{ 3x+2 }dx } =\frac { 1 }{ 3 } \frac { a^{ 3x+2 } }{ \log\: a } $$

Therefore
$$\displaystyle I=\frac { 1 }{ 2 } e^{ 2x-3 }-\frac { 2 }{ 3 } \cdot 7^{ 4-3(x/2) }\frac { 1 }{ \log 7 } -\frac { 1 }{ 3 } \cos  \left( 3x-\frac { 1 }{ 2 }  \right) +\frac { 5 }{ 2 } \sin  \left( \frac { 2 }{ 5 } x-2 \right) +\frac { 1 }{ 3 } \frac { a^{ 3x+2 } }{ \log\: a } $$
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Subjective Medium Published on 17th 09, 2020
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