मराठी

∫ 1 1 + Cos 2 X D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[\int\frac{dx}{1 + \cos \left( 2x \right)} \left[ \therefore 1 + \cos\theta = 2 \cos^2 \left( \frac{\theta}{2} \right) \right]\]

\[ = \int\frac{dx}{2 \cos^2 x}\]

\[ = \frac{1}{2}\int \sec^2 x dx\]

\[ = \frac{1}{2}\tan x + C\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Exercise 19.02 [पृष्ठ १५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.02 | Q 33 | पृष्ठ १५

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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

 


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

\[\int\frac{3x + 5}{\sqrt{7x + 9}} dx\]

\[\  ∫    x   \text{ e}^{x^2} dx\]

\[\int\frac{x^2}{\sqrt{3x + 4}} dx\]

\[\int \cot^5 x  \text{ dx }\]

\[\int\frac{1}{\sin^4 x \cos^2 x} dx\]

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

\[\int\frac{1}{\sqrt{3 x^2 + 5x + 7}} dx\]

\[\int\frac{1}{\sqrt{\left( x - \alpha \right)\left( \beta - x \right)}} dx, \left( \beta > \alpha \right)\]

\[\int\frac{\sec^2 x}{\sqrt{4 + \tan^2 x}} dx\]

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

\[\int\frac{2x + 1}{\sqrt{x^2 + 4x + 3}} \text{ dx }\]

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

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

\[\int\frac{\log x}{x^n}\text{  dx }\]

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

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

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

\[\int\frac{x^2 + 6x - 8}{x^3 - 4x} dx\]

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

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

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

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

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

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

\[\int\frac{\sin^6 x}{\cos^8 x} dx =\]

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

\[\int\frac{x^3}{x + 1}dx\] is equal to

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


\[\int\sqrt{\frac{a + x}{x}}dx\]
 

\[\int\sqrt{1 + 2x - 3 x^2}\text{  dx } \]

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

\[\int\left( 2x + 3 \right) \sqrt{4 x^2 + 5x + 6} \text{ dx}\]

\[ \int\left( 1 + x^2 \right) \ \cos 2x \ dx\]


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

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

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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×