Question
Evaluate the following integrals:
$\int(\text{e}^{\log\text{x}}+\sin\text{x})\cos\text{x dx}$
$\int(\text{e}^{\log\text{x}}+\sin\text{x})\cos\text{x dx}$
✓
Answer
Let $\text{I}=\int(\text{e}^{\log\text{x}}+\sin\text{x})\cos\text{x dx}$
$=\int(\text{x}+\sin\text{x})\cos\text{x dx}$
$=\int\text{x}\cos\text{x dx}+\int\sin\text{x}\cos\text{x dx}$
$=\big[\text{x}\int\cos\text{x dx}-\int(1\int\cos\text{x dx})\text{dx}\big]+\frac{1}{2}\int\sin2\text{x dx}$
$=\big[\text{x}\sin\text{x}-\int\sin\text{x dx}\big]+\frac{1}{2}\Big(-\frac{\cos2\text{x}}{2}\Big)+\text{C}$
$\text{I}=\text{x}\sin\text{x}+\cos\text{x}-\frac{1}{4}\cos2\text{x}+\text{C}$
$\text{I}=\text{x}\sin\text{x}+\cos\text{x}-\frac{1}{4}\big[1-2\sin^2\text{x}\big]+\text{C}$
$\text{I}=\text{x}\sin\text{x}+\cos\text{x}-\frac{1}{4}+\frac{1}{2}\sin^2\text{x}+\text{C}$
$\text{I}=\text{x}\sin\text{x}+\cos\text{x}+\frac{1}{2}\sin^2\text{x}+\text{C}-\frac{1}{4}$
$\text{I}=\text{x}\sin\text{x}+\cos\text{x}+\frac{1}{2}\sin^2\text{x}+\text{k},$ where $\text{k}=\text{C}-\frac{1}{4}$
$=\int(\text{x}+\sin\text{x})\cos\text{x dx}$
$=\int\text{x}\cos\text{x dx}+\int\sin\text{x}\cos\text{x dx}$
$=\big[\text{x}\int\cos\text{x dx}-\int(1\int\cos\text{x dx})\text{dx}\big]+\frac{1}{2}\int\sin2\text{x dx}$
$=\big[\text{x}\sin\text{x}-\int\sin\text{x dx}\big]+\frac{1}{2}\Big(-\frac{\cos2\text{x}}{2}\Big)+\text{C}$
$\text{I}=\text{x}\sin\text{x}+\cos\text{x}-\frac{1}{4}\cos2\text{x}+\text{C}$
$\text{I}=\text{x}\sin\text{x}+\cos\text{x}-\frac{1}{4}\big[1-2\sin^2\text{x}\big]+\text{C}$
$\text{I}=\text{x}\sin\text{x}+\cos\text{x}-\frac{1}{4}+\frac{1}{2}\sin^2\text{x}+\text{C}$
$\text{I}=\text{x}\sin\text{x}+\cos\text{x}+\frac{1}{2}\sin^2\text{x}+\text{C}-\frac{1}{4}$
$\text{I}=\text{x}\sin\text{x}+\cos\text{x}+\frac{1}{2}\sin^2\text{x}+\text{k},$ where $\text{k}=\text{C}-\frac{1}{4}$
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