मराठी

∫ 2 X Sec 3 ( X 2 + 3 ) Tan ( X 2 + 3 ) D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int2x    \sec^3 \left( x^2 + 3 \right) \tan \left( x^2 + 3 \right) dx\]
बेरीज
Advertisements

उत्तर

\[\int2x \sec^3 \left( x^2 + 3 \right) \cdot \tan \left( x^2 + 3 \right) dx\]
\[ = \int \sec^2 \left( x^2 + 3 \right) \cdot \sec \left( x^2 + 3 \right) \cdot \tan \left( x^2 + 3 \right) \cdot \text{2x dx}\]
\[\text{Let }\sec \left( x^2 + 3 \right) = t\]
\[ \Rightarrow \sec \left( x^2 + 3 \right) \cdot \tan \left( x^2 + 3 \right) \cdot 2x = \frac{dt}{dx}\]
\[ \Rightarrow \sec \left( x^2 + 3 \right) \cdot \tan \left( x^2 + 3 \right) \cdot \text{2x dx} = dt\]
\[Now, \int \sec^2 \left( x^2 + 3 \right) \cdot \sec \left( x^2 + 3 \right) \cdot \tan \left( x^2 + 3 \right) \cdot \text{2x dx}\]
\[ = \int t^2 dt\]
\[ = \frac{t^3}{3} + C\]
\[ = \frac{\sec^3 \left( x^2 + 3 \right)}{3} + C\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Exercise 19.09 [पृष्ठ ५८]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.09 | Q 39 | पृष्ठ ५८

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

`int{sqrtx(ax^2+bx+c)}dx`

\[\int\frac{5 \cos^3 x + 6 \sin^3 x}{2 \sin^2 x \cos^2 x} dx\]

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

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

\[\int\text{sin mx }\text{cos nx dx m }\neq n\]

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

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

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

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

\[\int\frac{x}{\sqrt{x^2 + 6x + 10}} \text{ dx }\]

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

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

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

\[\int \sin^{- 1} \left( 3x - 4 x^3 \right) \text{ dx }\]

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

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

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

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

 


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

\[\int\frac{5 x^2 + 20x + 6}{x^3 + 2 x^2 + x} dx\]

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

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

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

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

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

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

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

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

If `int(2x^(1/2))/(x^2)  dx = k  .  2^(1/x) + C`, then k is equal to ______.


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

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


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

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

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

\[\int\sqrt{a^2 - x^2}\text{  dx }\]

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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×