Question
Evaluate the following integrals:
$\int\text{e}^{2\text{x}}\cos^2\text{x }\text{dx}$
$\int\text{e}^{2\text{x}}\cos^2\text{x }\text{dx}$
✓
Answer
Let $\text{I}=\int\text{e}^{2\text{x}}\cos^2\text{x }\text{dx}$
$=\frac{1}{2}\int\text{e}^{2\text{x}}2\cos^2\text{x dx}$
$=\frac{1}{2}\int\text{e}^{2\text{x}}(1+\cos2\text{x})\text{dx}$
$=\frac{1}{2}\int\text{e}^{2\text{x}}\text{dx}+\frac{1}{2}\int\text{e}^{2\text{x}}\cos2\text{x }\text{dx}$
$\because\ \int\text{e}^{2\text{x}}\cos\text{bx dx}=\frac{\text{e}^{2\text{x}}}{\text{a}^2+\text{b}^2}\{\text{a}\cos\text{bx}-\text{b}\sin\text{bx}\}+\text{C}$
$\therefore\ \text{I}=\frac{1}{4}\text{e}^{2\text{x}}+\frac{1}{2}\frac{\text{e}^{2\text{x}}}{8}\{2\cos2\text{x}+2\sin2\text{x}\}+\text{C}$
Hence,
$\text{I}=\frac{\text{e}^{2\text{x}}}{4}+\frac{\text{e}^{2\text{x}}}{16}\{2\cos2\text{x}+2\sin2\text{x}\}+\text{C}$
or
$\text{I}=\frac{\text{e}^{2\text{x}}}{4}+\frac{\text{e}^{2\text{x}}}{8}\{\cos2\text{x}+\sin2\text{x}\}+\text{C}$
$=\frac{1}{2}\int\text{e}^{2\text{x}}2\cos^2\text{x dx}$
$=\frac{1}{2}\int\text{e}^{2\text{x}}(1+\cos2\text{x})\text{dx}$
$=\frac{1}{2}\int\text{e}^{2\text{x}}\text{dx}+\frac{1}{2}\int\text{e}^{2\text{x}}\cos2\text{x }\text{dx}$
$\because\ \int\text{e}^{2\text{x}}\cos\text{bx dx}=\frac{\text{e}^{2\text{x}}}{\text{a}^2+\text{b}^2}\{\text{a}\cos\text{bx}-\text{b}\sin\text{bx}\}+\text{C}$
$\therefore\ \text{I}=\frac{1}{4}\text{e}^{2\text{x}}+\frac{1}{2}\frac{\text{e}^{2\text{x}}}{8}\{2\cos2\text{x}+2\sin2\text{x}\}+\text{C}$
Hence,
$\text{I}=\frac{\text{e}^{2\text{x}}}{4}+\frac{\text{e}^{2\text{x}}}{16}\{2\cos2\text{x}+2\sin2\text{x}\}+\text{C}$
or
$\text{I}=\frac{\text{e}^{2\text{x}}}{4}+\frac{\text{e}^{2\text{x}}}{8}\{\cos2\text{x}+\sin2\text{x}\}+\text{C}$
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