मराठी

∫ √ Cot θ D θ - Mathematics

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प्रश्न

\[\int\sqrt{\cot \text{θ} d  } \text{ θ}\]
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उत्तर

\[\text{ We have,} \]

\[I = \int\sqrt{\cot \theta} d   \text{ θ}\]

\[\text{ Putting  cot} \text{ θ} = t^2 \]

\[ \Rightarrow - {cosec}^2 \text{ θ dθ}= 2t \text{ dt }\]

\[ \Rightarrow d\theta = - \frac{2t \text{ dt }}{\cos e c^2 \text{ θ }}\]

\[ \Rightarrow d\theta = \frac{- 2t \text{ dt}}{1 + co t^2 \text{ θ}}\]

\[ \Rightarrow d\theta = \frac{- 2t \text{ dt}}{1 + t^4}\]

\[ \therefore I = \int t\left( \frac{- 2t \text{ dt }}{1 + t^4} \right)\]

\[ = - \int\left( \frac{2 t^2}{1 + t^4} \right)dt\]

\[ = - \int\left( \frac{t^2 + 1 + t^2 - 1}{t^4 + 1} \right)dt\]

\[ = - \int\left( \frac{t^2 + 1}{t^4 + 1} \right)dt - \int\frac{\left( t^2 - 1 \right)dt}{t^4 + 1}\]

` \text{Dividing numerator and denominator by} \text{  t}^2 `

\[I = - \int\left( \frac{1 + \frac{1}{t^2}}{t^2 + \frac{1}{t^2}} \right)dt - \int\left( \frac{1 - \frac{1}{t^2}}{t^2 + \frac{1}{t^2}} \right)dt\]

\[ = - \int\frac{\left( 1 + \frac{1}{t^2} \right)dt}{t^2 + \frac{1}{t^2} - 2 + 2} - \int\frac{\left( 1 - \frac{1}{t^2} \right)dt}{t^2 + \frac{1}{t^2} + 2 - 2}\]

\[ = - \int\frac{\left( 1 + \frac{1}{t^2} \right)dt}{\left( t - \frac{1}{t} \right)^2 + \left( \sqrt{2} \right)^2} - \int\frac{\left( 1 - \frac{1}{t^2} \right)dt}{\left( t + \frac{1}{t} \right)^2 - \left( \sqrt{2} \right)^2}\]

\[\text{ Putting   t} - \frac{1}{t} = p\]

\[ \Rightarrow \left( 1 + \frac{1}{t^2} \right)dt = dp\]

\[\text{ Putting}\ t + \frac{1}{t} = q\]

\[ \Rightarrow \left( 1 - \frac{1}{t^2} \right)dt = dq\]

\[I = - \int \frac{dp}{p^2 + \left( \sqrt{2} \right)^2} - \int\frac{dq}{q^2 - \left( \sqrt{2} \right)^2}\]

\[ = - \frac{1}{\sqrt{2}} \tan^{- 1} \left( \frac{p}{\sqrt{2}} \right) - \frac{1}{2\sqrt{2}}\text{ log} \left| \frac{q - \sqrt{2}}{q + \sqrt{2}} \right| + C\]

\[ = - \frac{1}{\sqrt{2}} \tan^{- 1} \left( \frac{t - \frac{1}{t}}{\sqrt{2}} \right) - \frac{1}{2\sqrt{2}}\text{ log }\left| \frac{t + \frac{1}{t} - \sqrt{2}}{1 + \frac{1}{t} + \sqrt{2}} \right| + C\]

\[ = - \frac{1}{\sqrt{2}} \text{ tan}^{- 1} \left( \frac{t^2 - 1}{\sqrt{2} t} \right) - \frac{1}{2\sqrt{2}}\text{ log} \left| \frac{t^2 + 1 - \sqrt{2}t}{t^2 + 1 + \sqrt{2}t} \right| + C\]

\[ = - \frac{1}{\sqrt{2}} \tan^{- 1} \left( \frac{\cot \theta - 1}{2\sqrt{\cot \theta}} \right) - \frac{1}{2\sqrt{2}}\text{ log } \left| \frac{\cot \theta + 1 - \sqrt{2 \cot \theta}}{\cot \theta + 1 + \sqrt{2 \cot \theta}} \right| + C\]

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पाठ 19: Indefinite Integrals - Exercise 19.31 [पृष्ठ १९०]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.31 | Q 2 | पृष्ठ १९०

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