Advertisements
Advertisements
प्रश्न
उत्तर
\[\int \text{cosec}^2 x .\text{ cos}^2 \text{ 2x dx} \]
\[ \Rightarrow \int\text{ cosec}^2 x \left( 1 - 2 \sin^2 x \right)^2 dx\]
\[ \Rightarrow \int \text{ cosec}^2 x \left( 1 + 4 \sin^4 x - 4 \sin^2 x \right)dx\]
\[ \Rightarrow \int\left( {cosec}^2 x + 4 \sin^2 x - 4 \right)dx\]
\[ \Rightarrow \int {cosec}^2 x \text{ dx} + 4\int\left( \frac{1 - \cos 2x}{2} \right)dx - 4\int dx\]
\[ \Rightarrow - \cot x + 2 \left[ x - \frac{\sin 2x}{2} \right] - 4x + C\]
\[ \Rightarrow - \cot x + 2x - \sin 2x - 4x + C\]
\[ \Rightarrow - \cot x - \sin 2x - 2x + C\]
APPEARS IN
संबंधित प्रश्न
If f' (x) = a sin x + b cos x and f' (0) = 4, f(0) = 3, f
` ∫ {sec x "cosec " x}/{log ( tan x) }` dx
` ∫ tan x sec^4 x dx `
` = ∫1/{sin^3 x cos^ 2x} dx`
` ∫ {x-3} /{ x^2 + 2x - 4 } dx `
\[\int\frac{x}{\sqrt{8 + x - x^2}} dx\]
If \[\int\frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx\]
The primitive of the function \[f\left( x \right) = \left( 1 - \frac{1}{x^2} \right) a^{x + \frac{1}{x}} , a > 0\text{ is}\]
\[\int\sin x \sin 2x \text{ sin 3x dx }\]
