English

∫ Sec 2 X √ 4 + Tan 2 X D X - Mathematics

Advertisements
Advertisements

Question

\[\int\frac{\sec^2 x}{\sqrt{4 + \tan^2 x}} dx\]
Sum

Solution

 

` ∫   { sec^2 x  dx}/{\sqrt{4 + tan^2 x}} `


\`text{ let } tan  x }= t `
\[ \Rightarrow \sec^2 x dx = dt\]
Now, ` ∫   { sec^2 x  dx}/{\sqrt{4 + tan^2 x}} `
\[ = \int\frac{dt}{\sqrt{2^2 + t^2}}\]
\[ = \text{ log } \left| t + \sqrt{4 + t^2} \right| + C\]
\[ = \text{ log }\left| \text{ tan x }+ \sqrt{4 + \tan^2 x} \right| + C\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - Exercise 19.18 [Page 98]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.18 | Q 2 | Page 98

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

 
\[\int\frac{\cos x}{1 - \cos x} \text{dx }or \int\frac{\cot x}{\text{cosec         } {x }- \cot x} dx\]

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

`∫     cos ^4  2x   dx `


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

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

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

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

` ∫  sec^6   x  tan    x   dx `

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

\[\int\frac{1}{4 x^2 + 12x + 5} dx\]

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

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

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

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

\[\int\frac{1}{5 - 4 \sin x} \text{ dx }\]

`int 1/(sin x - sqrt3 cos x) dx`

\[\int\frac{1}{4 + 3 \tan x} dx\]

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

 
` ∫  x tan ^2 x dx 

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

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

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

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

\[\int\frac{\sqrt{16 + \left( \log x \right)^2}}{x} \text{ dx}\]

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

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

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

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

\[\int\frac{x}{4 + x^4} \text{ dx }\] is equal to

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

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

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


\[\int \tan^5 x\ dx\]

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


\[\int {cosec}^4 2x\ dx\]


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

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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×