English

∫ Tan − 1 ( Sin 2 X 1 + Cos 2 X ) D X - Mathematics

Advertisements
Advertisements

Question

\[\int \tan^{- 1} \left( \frac{\sin 2x}{1 + \cos 2x} \right) dx\]
Sum
Advertisements

Solution

\[\int \tan^{- 1} \left[ \frac{\sin \left( 2x \right)}{1 + \cos2x} \right]dx\]

`=  ∫  tan ^-1  [ (2 sin  x cos  x) / ( 2 cos^2 x)] `dx ` [∴     sin 2x  = 2  sin x cos x  &   1 + cos 2x = 2    cos ^2 x ]`

\[ = \int \tan^{- 1}  \left[ \tan x \right]\]

\[ = \int \tan^{- 1}     \left[ \tan x \right]\]

` =  ∫   x dx `

\[ = \frac{x^2}{2} + C\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - Exercise 19.02 [Page 15]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.02 | Q 35 | Page 15

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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

 


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

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

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

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

\[\int\frac{3x + 5}{\sqrt{7x + 9}} dx\]

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

`∫     cos ^4  2x   dx `


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

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

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

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

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

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

\[\int \tan^3 \text{2x sec 2x dx}\]

` ∫  sec^6   x  tan    x   dx `

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

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

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

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

\[\int\frac{x^2 + 1}{x^2 - 5x + 6} dx\]

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


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

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

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

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

\[\int\frac{1}{\sin x + \sqrt{3} \cos x} \text{ dx  }\]

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

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

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

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


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

\[\int\frac{2 x^2 + 7x - 3}{x^2 \left( 2x + 1 \right)} dx\]

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

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

 


\[\int\frac{1}{\cos x + \sqrt{3} \sin x} \text{ dx } \] is equal to

\[\int\frac{1}{a + b \tan x} \text{ dx }\]

\[\int \tan^3 x\ \sec^4 x\ dx\]

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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×