मराठी

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

Advertisements
Advertisements

प्रश्न

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

उत्तर

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

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Exercise 19.30 [पृष्ठ १७७]

APPEARS IN

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

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

\[\int\left( 2^x + \frac{5}{x} - \frac{1}{x^{1/3}} \right)dx\]

\[\int\sqrt{x}\left( 3 - 5x \right) dx\]

 


\[\int\frac{x^3 - 3 x^2 + 5x - 7 + x^2 a^x}{2 x^2} dx\]

\[\int\frac{x + 3}{\left( x + 1 \right)^4} dx\]

\[\int\frac{1}{1 - \sin\frac{x}{2}} dx\]

\[\int\frac{2x + 1}{\sqrt{3x + 2}} dx\]

`  =  ∫ root (3){ cos^2 x}  sin x   dx `


\[\int\frac{\sin 2x}{\left( a + b \cos 2x \right)^2} dx\]

\[\int\frac{e^\sqrt{x} \cos \left( e^\sqrt{x} \right)}{\sqrt{x}} dx\]

\[\int\frac{e^{m \tan^{- 1} x}}{1 + x^2} dx\]

\[\int \sin^5 x \text{ dx }\]

` ∫  {x-3} /{ x^2 + 2x - 4 } dx `


\[\int\frac{x^2 + 1}{x^2 - 5x + 6} dx\]

\[\int\frac{x + 1}{\sqrt{x^2 + 1}} dx\]

\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]

`int 1/(sin x - sqrt3 cos x) dx`

\[\int\frac{1}{1 - \cot x} dx\]

\[\int\frac{1}{4 + 3 \tan x} dx\]

\[\int2 x^3 e^{x^2} dx\]

\[\int \log_{10} x\ dx\]

\[\int \cos^3 \sqrt{x}\ dx\]

\[\int \sin^{- 1} \sqrt{\frac{x}{a + x}} \text{ dx }\]

\[\int e^x \left( \frac{x - 1}{2 x^2} \right) dx\]

\[\int e^x \left( \log x + \frac{1}{x^2} \right) dx\]

\[\int\frac{5x}{\left( x + 1 \right) \left( x^2 - 4 \right)} dx\]

\[\int\frac{1}{x\left[ 6 \left( \log x \right)^2 + 7 \log x + 2 \right]} dx\]

\[\int\frac{x^2 + 1}{\left( x - 2 \right)^2 \left( x + 3 \right)} dx\]

\[\int\frac{x^2}{\left( x^2 + 1 \right) \left( 3 x^2 + 4 \right)} dx\]

\[\int\frac{x^2 + 1}{x^4 + x^2 + 1} \text{  dx }\]

\[\int\frac{x^2 + 1}{x^4 + 7 x^2 + 1} 2 \text{ dx }\]

\[\int\frac{x}{\left( x - 3 \right) \sqrt{x + 1}} dx\]

\[\int\frac{\sin 2x}{\sin^4 x + \cos^4 x} \text{ dx }\]

\[\int\frac{\sin x}{\sqrt{\cos^2 x - 2 \cos x - 3}} \text{ dx }\]

\[\int\sqrt{\frac{1 - x}{x}} \text{ dx}\]


\[\int\frac{1}{1 + 2 \cos x} \text{ dx }\]

\[\int \sec^6 x\ dx\]

\[\int\frac{1}{x\sqrt{1 + x^3}} \text{ dx}\]

\[\int \left( \sin^{- 1} x \right)^3 dx\]

\[\int\frac{x^2}{\left( x - 1 \right)^3 \left( x + 1 \right)} \text{ dx}\]

\[\int\frac{x^2}{x^2 + 7x + 10}\text{ dx }\]

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×