मराठी

∫ 1 Cos X ( 5 − 4 Sin X ) D X - Mathematics

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प्रश्न

\[\int\frac{1}{\cos x \left( 5 - 4 \sin x \right)} dx\]
बेरीज
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उत्तर

We have,
\[I = \int\frac{dx}{\cos x \left( 5 - 4 \sin x \right)}\]
\[ = \int\frac{\cos x dx}{\cos^2 x \left( 5 - 4 \sin x \right)}\]
\[ = \int\frac{\cos x dx}{\left( 1 - \sin^2 x \right) \left( 5 - 4 \sin x \right)}\]
\[ = \int\frac{\cos x dx}{\left( 1 - \sin x \right) \left( 1 + \sin x \right) \left( 5 - 4 \sin x \right)}\]
\[\text{Putting }\sin x = t\]
\[ \Rightarrow \cos x dx = dt\]
\[ \therefore I = \int\frac{dt}{\left( 1 - t \right) \left( 1 + t \right) \left( 5 - 4t \right)}\]
\[\text{Let }\frac{1}{\left( 1 - t \right) \left( 1 + t \right) \left( 5 - 4t \right)} = \frac{A}{1 - t} + \frac{B}{1 + t} + \frac{C}{5 - 4t}\]
\[ \Rightarrow \frac{1}{\left( 1 - t \right) \left( 1 + t \right) \left( 5 - 4t \right)} = \frac{A\left( 1 + t \right) \left( 5 - 4t \right) + B\left( 1 - t \right) \left( 5 - 4t \right) + C\left( 1 - t \right) \left( 1 + t \right)}{\left( 1 - t \right) \left( 1 + t \right) \left( 5 - 4t \right)}\]
\[ \Rightarrow 1 = A\left( 1 + t \right) \left( 5 - 4t \right) + B\left( 1 - t \right) \left( 5 - 4t \right) + C\left( 1 - t \right) \left( 1 + t \right)\]
\[\text{Putting 1 + t = 0}\]
\[ \Rightarrow t = - 1\]
\[1 = B\left( 2 \right) \left( 5 + 4 \right)\]
\[B = \frac{1}{18}\]
\[\text{Putting 1 - t = 0}\]
\[ \Rightarrow t = 1\]
\[1 = A \left( 2 \right) \left( 5 - 4 \right) + B \times 0 + C \times 0\]
\[A = \frac{1}{2}\]
\[\text{Putting 5 - 4t = 0}\]
\[ \Rightarrow 4t = 5\]
\[ \Rightarrow t = \frac{5}{4}\]
\[1 = C \left( 1 - \frac{5}{4} \right) \left( 1 + \frac{5}{4} \right)\]
\[ \Rightarrow 1 = C \left( - \frac{1}{4} \right) \left( \frac{9}{4} \right)\]
\[ \Rightarrow C = - \frac{16}{9}\]
\[ \therefore I = \frac{1}{2}\int\frac{dt}{1 - t} + \frac{1}{18}\int\frac{dt}{1 + t} - \frac{16}{9}\int\frac{dt}{5 - 4t}\]
\[ = \frac{1}{2} \frac{\log \left| 1 - t \right|}{- 1} + \frac{1}{18} \log \left| 1 + t \right| - \frac{16}{9} \times \frac{\log \left| 5 - 4t \right|}{- 4} + C\]
\[ = \frac{1}{18} \log \left| 1 + t \right| - \frac{1}{2} \log \left| 1 - t \right| + \frac{4}{9}\log \left| 5 - 4t \right| + C\]
\[ = \frac{1}{18} \log \left| 1 + \sin x \right| - \frac{1}{2} \log \left| 1 - \sin x \right| + \frac{4}{9} \log \left| 5 - 4 \sin x \right| + C\]

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पाठ 19: Indefinite Integrals - Exercise 19.30 [पृष्ठ १७७]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.30 | Q 59 | पृष्ठ १७७

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