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प्रश्न
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उत्तर
\[\text{ Let I } = \int\frac{\cos x - \sin x}{\sqrt{8 - \sin2x}}dx\]
\[ = \int\frac{\cos x - \sin x}{\sqrt{9 - 1 - \sin2x}}dx\]
\[ = \int\frac{\cos x - \sin x}{\sqrt{9 - \sin^2 x - \cos^2 x - 2\sin x\cos x}}dx\]
\[ = \int\frac{\cos x - \sin x}{\sqrt{9 - \left( \sin x + \cos x \right)^2}}dx\]
\[ Let \left( \sin x + \cos x \right) = t\]
\[ \text{On differentiating both sides, we get}\]
\[ \left( \cos x - \sin x \right)dx = dt\]
\[ \therefore I = \int\frac{1}{\sqrt{\left( 3 \right)^2 - \left( t \right)^2}}dt\]
\[ = \sin^{- 1} \left( \frac{t}{3} \right) + c\]
\[ = \sin^{- 1} \left( \frac{\sin x + \cos x}{3} \right) + c\]
\[Hence, \int\frac{\cos x - \sin x}{\sqrt{8 - \ sin2x}}dx = \sin^{- 1} \left( \frac{\sin x + \cos x}{3} \right) + c\]
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संबंधित प्रश्न
The value of \[\int\frac{\sin x + \cos x}{\sqrt{1 - \sin 2x}} dx\] is equal to
If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then
\[\int\frac{x + 3}{\left( x + 4 \right)^2} e^x dx =\]
