English

∫ Tan 4 X D X - Mathematics

Advertisements
Advertisements

Question

\[\int \tan^4 x\ dx\]
Sum
Advertisements

Solution

\[\text{ Let I } = \int \text{ tan}^4 \text{ x dx }\]
\[ = \int \tan^2 x \cdot \tan^2 \text{ x dx}\]
\[ = \int\left( \sec^2 x - 1 \right) \tan^2 \text{ x  dx}\]
\[ = \int \sec^2 x \cdot \tan^2\text{  x dx }- \int \tan^2 \text{ x  dx}\]
\[ = \int \tan^2 x \cdot \sec^2 x - \int\left( \sec^2 x - 1 \right) dx\]
\[\text{ Putting tan x } = \text{ t  in   the  Ist  integral}\]
\[ \Rightarrow \sec^2 \text{ x  dx } = dt\]
\[ \therefore I = \int t^2 \cdot dt - \int\left( \sec^2 x - 1 \right) dx\]
\[ = \frac{t^3}{3} - \tan x + x + C\]
\[ = \frac{\tan^3 x}{3} - \text{ tan x + x + C }..............\left( \because t = \tan x \right)\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - Revision Excercise [Page 203]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Revision Excercise | Q 29 | Page 203

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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


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)
 

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

\[\int\frac{\text{sin} \left( x - a \right)}{\text{sin}\left( x - b \right)} dx\]

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

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

` ∫  tan^3    x   sec^2  x   dx  `

` ∫      tan^5    x   dx `


` ∫    \sqrt{tan x}     sec^4  x   dx `


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

\[\int\frac{\cos x}{\sqrt{\sin^2 x - 2 \sin x - 3}} dx\]

\[\int\frac{1}{4 \cos^2 x + 9 \sin^2 x}\text{  dx }\]

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

\[\int x^3 \cos x^2 dx\]

\[\int {cosec}^3 x\ dx\]

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

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

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

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

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

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

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

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

\[\int\frac{x^9}{\left( 4 x^2 + 1 \right)^6}dx\]  is equal to 

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

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


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

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

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


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

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

\[\int\frac{\sin^2 x}{\cos^6 x} \text{ dx }\]

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

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

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

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

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

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

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

 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×