हिंदी

∫ Cosec 2 X Cos 2 2x Dx - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int \text{cosec}^2 x \text{ cos}^2 \text{  2x  dx} \]
योग
Advertisements

उत्तर

\[\int \text{cosec}^2 x  .\text{ cos}^2 \text{  2x  dx} \]
\[ \Rightarrow \int\text{ cosec}^2 x \left( 1 - 2 \sin^2 x \right)^2 dx\]
\[ \Rightarrow \int \text{ cosec}^2 x \left( 1 + 4 \sin^4 x - 4 \sin^2 x \right)dx\]
\[ \Rightarrow \int\left( {cosec}^2 x + 4 \sin^2 x - 4 \right)dx\]
\[ \Rightarrow \int {cosec}^2 x \text{ dx} + 4\int\left( \frac{1 - \cos 2x}{2} \right)dx - 4\int dx\]
\[ \Rightarrow - \cot x + 2 \left[ x - \frac{\sin 2x}{2} \right] - 4x + C\]
\[ \Rightarrow - \cot x + 2x - \sin 2x - 4x + C\]
\[ \Rightarrow - \cot x - \sin 2x - 2x + C\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - Revision Excercise [पृष्ठ २०३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Revision Excercise | Q 10 | पृष्ठ २०३

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

\[\int\left( \sec^2  x + {cosec}^2  x \right)  dx\]

` ∫  1/ {1+ cos   3x}  ` dx


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

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

\[\int \sin^2\text{ b x dx}\]

Integrate the following integrals:

\[\int\text{sin 2x  sin 4x    sin 6x  dx} \]

\[\int\frac{1}{x (3 + \log x)} dx\]

\[\int\frac{1 + \cot x}{x + \log \sin x} dx\]

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


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

\[\int\frac{\sin^5 x}{\cos^4 x} \text{ dx }\]

` ∫  tan^3    x   sec^2  x   dx  `

\[\int\frac{1}{\sqrt{5 - 4x - 2 x^2}} dx\]

\[\int\frac{\cos 2x}{\sqrt{\sin^2 2x + 8}} dx\]

\[\int\frac{\cos x - \sin x}{\sqrt{8 - \sin2x}}dx\]

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

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

\[\int e^\sqrt{x} \text{ dx }\]

\[\int \sec^{- 1} \sqrt{x}\ dx\]

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

\[\int e^x \left( \cot x + \log \sin x \right) dx\]

\[\int\frac{\sqrt{16 + \left( \log x \right)^2}}{x} \text{ dx}\]

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

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

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

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

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

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

\[\int \sin^5 x\ dx\]

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

\[\int\frac{5x + 7}{\sqrt{\left( x - 5 \right) \left( x - 4 \right)}} \text{ dx }\]

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

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

\[\int x \sec^2 2x\ dx\]

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

\[\int x^2 \tan^{- 1} x\ dx\]

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

\[\int\frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x/2} \text{ dx}\]

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×