हिंदी

∫ Sec 2 X Cos 2 2 X Dx - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[\int\left( \sec^2 x \cdot \cos^2 2x \right)dx\]
\[ = \int \sec^2 x \times \left( 2 \cos^2 x - 1 \right)^2 dx\]
\[ = \int \sec^2 x \left[ 4 \cos^4 x - 4 \cos^2 x + 1 \right]dx\]
\[ \Rightarrow \int\left( 4 \cos^2 x - 4 + \sec^2 x \right)dx\]
\[ = 4\int \cos^2 x \text{ dx } + \int \sec^2 x \text{ dx }- 4\int dx\]
\[ \Rightarrow 4\int\left( \frac{1 + \cos 2x}{2} \right)dx + \int \sec^2 x - 4\int dx\]
\[ \Rightarrow 2 \left[ x + \frac{\sin 2x}{2} \right] + \tan x - 4x + C\]
\[ \Rightarrow \sin 2x + \tan x - 2x + C\]

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 9 | पृष्ठ २०३

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

\[\int \left( \tan x + \cot x \right)^2 dx\]

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

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

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

`  ∫  sin 4x cos  7x  dx  `

Integrate the following integrals:

\[\int\text { sin  x  cos  2x     sin 3x   dx}\]

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

` ∫ {"cosec"   x }/ { log  tan   x/2 ` dx 

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

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

\[\int\frac{\sin 2x}{\sin \left( x - \frac{\pi}{6} \right) \sin \left( x + \frac{\pi}{6} \right)} dx\]

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

\[\int\sqrt {e^x- 1}  \text{dx}\] 

Evaluate the following integrals:

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

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

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

\[\int\frac{3 x^5}{1 + x^{12}} dx\]

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

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

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

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

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

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

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

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

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

\[\int e^x \left[ \sec x + \log \left( \sec x + \tan x \right) \right] dx\]

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

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

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

 


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

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

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

Write a value of

\[\int e^{3 \text{ log x}} x^4\text{ dx}\]

If \[\int\frac{\cos 8x + 1}{\tan 2x - \cot 2x} dx\]


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

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


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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×