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प्रश्न
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उत्तर
\[\int \text{cosec}^2 x .\text{ cos}^2 \text{ 2x dx} \]
\[ \Rightarrow \int\text{ cosec}^2 x \left( 1 - 2 \sin^2 x \right)^2 dx\]
\[ \Rightarrow \int \text{ cosec}^2 x \left( 1 + 4 \sin^4 x - 4 \sin^2 x \right)dx\]
\[ \Rightarrow \int\left( {cosec}^2 x + 4 \sin^2 x - 4 \right)dx\]
\[ \Rightarrow \int {cosec}^2 x \text{ dx} + 4\int\left( \frac{1 - \cos 2x}{2} \right)dx - 4\int dx\]
\[ \Rightarrow - \cot x + 2 \left[ x - \frac{\sin 2x}{2} \right] - 4x + C\]
\[ \Rightarrow - \cot x + 2x - \sin 2x - 4x + C\]
\[ \Rightarrow - \cot x - \sin 2x - 2x + C\]
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संबंधित प्रश्न
` ∫ sin x \sqrt (1-cos 2x) dx `
` ∫ {sec x "cosec " x}/{log ( tan x) }` dx
` = ∫1/{sin^3 x cos^ 2x} dx`
\[\int\frac{x}{\sqrt{8 + x - x^2}} dx\]
If \[\int\frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx\]
\[\int\frac{1}{2 + \cos x} \text{ dx }\]
