Question
Evaluate the following integrals:
$\int\frac{1+\cos\text{x}}{(\text{x}+\sin\text{x})^3}\text{dx}$
$\int\frac{1+\cos\text{x}}{(\text{x}+\sin\text{x})^3}\text{dx}$
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Answer
$\int\frac{(1+\cos\text{x})}{(\text{x}+\sin\text{x})^3}\text{dx}$
$\text{Let x}+\sin\text{x}=\text{t}$
$\Rightarrow(1+\cos\text{x})=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(1+\cos\text{x})\text{dx}={\text{dt}}$
$\text{Now,}\int\frac{(1+\cos\text{x})}{(\text{x}+\sin\text{x})^3}\text{dx}$
$=\int\frac{\text{dt}}{\text{t}^3}$
$=\int\text{t}^{-3}\text{dt}$
$=\frac{\text{t}^{-3+1}}{-3+1}+\text{C}$
$=\frac{-1}{2\text{t}^2}+\text{C}$
$=\frac{-1}{2(\text{x}+\sin\text{x})^2}+\text{C}$
$\text{Let x}+\sin\text{x}=\text{t}$
$\Rightarrow(1+\cos\text{x})=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(1+\cos\text{x})\text{dx}={\text{dt}}$
$\text{Now,}\int\frac{(1+\cos\text{x})}{(\text{x}+\sin\text{x})^3}\text{dx}$
$=\int\frac{\text{dt}}{\text{t}^3}$
$=\int\text{t}^{-3}\text{dt}$
$=\frac{\text{t}^{-3+1}}{-3+1}+\text{C}$
$=\frac{-1}{2\text{t}^2}+\text{C}$
$=\frac{-1}{2(\text{x}+\sin\text{x})^2}+\text{C}$
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