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