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प्रश्न
` = ∫1/{sin^3 x cos^ 2x} dx`
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उत्तर
\[\int\frac{dx}{\sin^3 x . \cos x}\]
` "Dividing numerator and denominator by" sin^4 x`
\[ = \int\frac{\frac{1}{\sin^4 x}dx}{\frac{\sin^3 x . \cos x}{\sin^4 x}}\]
\[ = \int\frac{{cosec}^4 x dx}{\cot x}\]
\[ = \int\frac{{cosec}^2 x . {cosec}^2 x dx}{\cot x}\]
`= {( 1 + cot^2 x ) . "cosec"^2 x dx}/cot x`
\[Let \cot x = t\]
` ⇒ "-cosec"^2 x = dt / dx `
` ⇒ "cosec"^2 x dx = - dt `
\[Now, \int\frac{\left( 1 + \cot^2 x \right) . {cosec}^2 x}{\cot x}dx\]
\[ = \int\frac{\left( 1 + t^2 \right) . \left( - dt \right)}{t}\]
\[ = - \int\left( \frac{1}{t} + t \right)dt\]
\[ = - \log \left| t \right| - \frac{t^2}{2} + C\]
\[ = - \log \left| \cot x \right| - \frac{\cot^2 x}{2} + C\]
\[ = \log \left| \cot x \right|^{- 1} - \frac{\left( {cosec}^2 x - 1 \right)}{2} + C\]
\[ = \log \left| \frac{1}{\cot x} \right| - \frac{{cosec}^2 x}{2} + \frac{1}{2} + C\]
\[ = \log \left| \tan x \right| - \frac{1}{2 \sin^2 x} + C' \left[ \therefore C' = C + \frac{1}{2} \right]\]
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