English

∫ E X ( 1 + X ) Cos 2 ( X E X ) D X = - Mathematics

Advertisements
Advertisements

Question

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

Options

  • 2 loge cos (xex) + C

  • sec (xex) + C

  • tan (xex) + C

  •  tan (x + ex) + C

MCQ
Advertisements

Solution

tan (xex) + C

 

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

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - MCQ [Page 201]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
MCQ | Q 22 | Page 201

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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


If f' (x) = 8x3 − 2xf(2) = 8, find f(x)


Write the primitive or anti-derivative of
\[f\left( x \right) = \sqrt{x} + \frac{1}{\sqrt{x}} .\]

 


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

\[\int\frac{1}{\left( 7x - 5 \right)^3} + \frac{1}{\sqrt{5x - 4}} dx\]

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

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

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

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

` ∫  tan 2x tan 3x  tan 5x    dx  `

\[\int\left\{ 1 + \tan x \tan \left( x + \theta \right) \right\} dx\]

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

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

` ∫    x   {tan^{- 1} x^2}/{1 + x^4} dx`

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

\[\int\frac{1}{x \left( x^6 + 1 \right)} dx\]

\[\int\frac{1}{\sqrt{8 + 3x - x^2}} dx\]

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

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

\[\int\frac{x + 7}{3 x^2 + 25x + 28}\text{ dx}\]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Find :  \[\int\frac{e^x}{\left( 2 + e^x \right)\left( 4 + e^{2x} \right)}dx.\] 

 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×