Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\text{ Let I }= \int e^x \left( \cot x - {cosec}^2 x \right)dx\]
\[\text{ here f(x) } = \text{ cot x put e}^x f(x) = t\]
\[ f'(x) = - {cosec}^2 x\]
\[\text{ let e}^x \cot x = t\]
\[\text{ Diff both sides w . r . t x }\]
\[ e^x \cot x + e^x \left( - {cosec}^2 x \right) = \frac{dt}{dx}\]
\[ \Rightarrow e^x \left( \cot x - {cosec}^2 x \right)dx = dt\]
\[ \therefore \int e^x \left( \cot x - {cosec}^2 x \right)dx = \int dt\]
\[ = t + C\]
\[ = e^x \cot x + C\]
APPEARS IN
संबंधित प्रश्न
\[\int\frac{x^2 + 5x + 2}{x + 2} dx\]
` ∫ \sqrt{tan x} sec^4 x dx `
\[\int\frac{x}{\sqrt{8 + x - x^2}} dx\]
\[\int\frac{x^3}{\sqrt{1 + x^2}}dx = a \left( 1 + x^2 \right)^\frac{3}{2} + b\sqrt{1 + x^2} + C\], then
\[\int\sqrt{\frac{1 - x}{x}} \text{ dx}\]
\[ \int\left( 1 + x^2 \right) \ \cos 2x \ dx\]
Find : \[\int\frac{dx}{\sqrt{3 - 2x - x^2}}\] .
