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
If f' (x) = x − \[\frac{1}{x^2}\] and f (1) \[\frac{1}{2}, find f(x)\]
` = ∫1/{sin^3 x cos^ 2x} dx`
Evaluate the following integrals:
If \[\int\frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx\]
The primitive of the function \[f\left( x \right) = \left( 1 - \frac{1}{x^2} \right) a^{x + \frac{1}{x}} , a > 0\text{ is}\]
\[\int\sin x \sin 2x \text{ sin 3x dx }\]
