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प्रश्न
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उत्तर
` ∫ sin x \sqrt{ 1 + \text{cos 2 x dx} `
` = ∫ sin x . \sqrt{ 2cos ^2 x} dx ` ` [∴ 1 + cos 2x = 2 cos^2 X]`
\[ = \sqrt{2}\int\text{sin x }\text{cos x dx}\]
\[ = \frac{\sqrt{2}}{2}\int\text{2 }\text{sin x } \text{cos x dx}\]
\[ = \frac{1}{\sqrt{2}}\int\text{sin 2x dx}\]
\[ = \frac{1}{\sqrt{2}}\left[ \frac{- \cos 2x}{2} \right] + C\]
\[ = \frac{- 1}{2\sqrt{2}}\cos 2x + C\]
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संबंधित प्रश्न
Evaluate the following integrals:
The value of \[\int\frac{\sin x + \cos x}{\sqrt{1 - \sin 2x}} dx\] is equal to
\[\int\frac{x^3}{\sqrt{1 + x^2}}dx = a \left( 1 + x^2 \right)^\frac{3}{2} + b\sqrt{1 + x^2} + C\], then
\[\int\frac{1 - x^4}{1 - x} \text{ dx }\]
\[\int\frac{1 + \sin x}{\sin x \left( 1 + \cos x \right)} \text{ dx }\]
Find : \[\int\frac{e^x}{\left( 2 + e^x \right)\left( 4 + e^{2x} \right)}dx.\]
