हिंदी

∫ X Sec X 2 Dx is Equal to - Mathematics

Advertisements
Advertisements

प्रश्न

` \int \text{ x} \text{ sec x}^2 \text{  dx  is  equal  to }`

 

विकल्प

  • \[\frac{1}{2}\] log (sec x2 + tan x2) + C

  • \[\frac{x^2}{2}\]  log (sec x2 + tan x2) + C

  • 2 log (sec x2 + tan x2) + C

  • none of these

MCQ

उत्तर

\[\frac{1}{2}\]  log (sec x2 + tan x2) + C

\[\text{ Let I }= \int x \sec x^2 dx\]
\[\text{ Putting x}^2 = t\]
\[ \Rightarrow 2x \text{ dx }= dt\]
\[ \Rightarrow x \text{ dx} = \frac{dt}{2}\]
\[ \therefore I = \frac{1}{2}\int\sec t \cdot dt\]
\[ = \frac{1}{2} \text{ log } \left| \sec t + \tan t \right| + C\]
\[ = \frac{1}{2} \text{ log }\left| \sec x^2 + \tan x^2 \right| + C \left( \because t = x^2 \right)\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - MCQ [पृष्ठ २००]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
MCQ | Q 3 | पृष्ठ २००

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

\[\int\frac{2 x^4 + 7 x^3 + 6 x^2}{x^2 + 2x} dx\]

` ∫  {cosec x} / {"cosec x "- cot x} ` dx      


If f' (x) = x − \[\frac{1}{x^2}\]  and  f (1)  \[\frac{1}{2},    find  f(x)\]

 


If f' (x) = a sin x + b cos x and f' (0) = 4, f(0) = 3, f

\[\left( \frac{\pi}{2} \right)\] = 5, find f(x)
 

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

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

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

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

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

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

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

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

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

\[\int \cos^7 x \text{ dx  } \]

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

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

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

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

\[\int\frac{2x + 3}{\sqrt{x^2 + 4x + 5}} \text{ dx }\]

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

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

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

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

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


\[\int x^3 \tan^{- 1}\text{  x dx }\]

\[\int x \sin^3 x\ dx\]

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

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

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

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

 


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

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

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

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

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

\[\int \sin^3 x \cos^4 x\ \text{ dx }\]

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

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

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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×