Evaluate the following integrals: - 8- 2x dx - Vidyadip

Question

Evaluate the following integrals:

$\int\frac{\cos\text{x}-\sin\text{x}}{\sqrt{8-\sin2\text{x}}}\text{ dx}$

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Answer

Let $\text{I}=\int\frac{\cos\text{x}-\sin\text{x}}{\sqrt{8-\sin2\text{x}}}\text{ dx}$
$=\int\frac{\cos\text{x}-\sin\text{x}}{\sqrt{9-1-\sin2\text{x}}}\text{ dx}$
$=\int\frac{\cos\text{x}-\sin\text{x}}{\sqrt{9-\sin^2\text{x}-\cos^2\text{x}-2\sin\text{x}\cos\text{x}}}\text{ dx}$
$=\int\frac{\cos\text{x}-\sin\text{x}}{\sqrt{9-(\sin\text{x}+\cos\text{x})^2}}\text{ dx}$
Let $(\sin\text{x}+\cos\text{x})=\text{t}$
On differentiating both sides, we get
$(\cos\text{x}-\sin\text{x})\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{1}{(3)^2-(\text{t})^2}\text{ dt}$
$=\sin^{-1}\Big(\frac{\text{t}}{3}\Big)+\text{C}$
$=\sin^{-1}\Big(\frac{\sin\text{x}-\cos\text{x}}{3}\Big)+\text{C}$
Hence, $\int\frac{\cos\text{x}-\sin\text{x}}{\sqrt{8-\sin2\text{x}}}\text{ dx}=\sin^{-1}\Big(\frac{\cos\text{x}-\sin\text{x}}{3}\Big)+\text{C}$

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