हिंदी

∫ √ 1 − Cos X 1 + Cos X D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

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

` ∫   \sqrt {{2 sin^2  x/2} / {2 cos^2   x/2 }}      dx`  `[ ∵ 1 - cos x = 2 sin^2      x/2    & 1 + cos x = 2 cos ^2   x/2]`

\[ = \int\tan\frac{x}{2} dx\]

\[ =\text{ - 2 }\text{ln }\left| \cos\frac{x}{2} \right| + C\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - Exercise 19.08 [पृष्ठ ४७]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.08 | Q 4 | पृष्ठ ४७

संबंधित प्रश्न

Evaluate : `int_0^3dx/(9+x^2)`


Integrate the following w.r.t. x `(x^3-3x+1)/sqrt(1-x^2)`


`∫   x    \sqrt{x + 2}     dx ` 

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

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

Evaluate the following integrals: 

`int "sec x"/"sec 2x" "dx"`

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

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

\[\int\frac{2 \cos x - 3 \sin x}{6 \cos x + 4 \sin x} dx\]

\[\int\frac{10 x^9 + {10}^x \log_e 10}{{10}^x + x^{10}} dx\]

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

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

Evaluate the following integrals:

\[\int\frac{1}{\left( x^2 + 2x + 10 \right)^2}dx\]

 


Evaluate the following integrals: 

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

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

Evaluate the following integrals:

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

Evaluate the following integrals:

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

\[\int\frac{a x^2 + bx + c}{\left( x - a \right) \left( x - b \right) \left( x - c \right)} dx,\text{ where a, b, c are distinct}\]

Evaluate the following integral :-

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

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

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

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

Evaluate the following integral:

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

Evaluate:\[\int\frac{\sin \sqrt{x}}{\sqrt{x}} \text{ dx }\]


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


Evaluate:\[\int\frac{\log x}{x} \text{ dx }\]


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


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


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


Write the value of\[\int\sec x \left( \sec x + \tan x \right)\text{  dx }\]


Evaluate: \[\int\frac{x + \cos6x}{3 x^2 + \sin6x}\text{ dx }\]


Evaluate:  \[\int\frac{2}{1 - \cos2x}\text{ dx }\]


Evaluate : \[\int\frac{1}{x(1 + \log x)} \text{ dx}\]


Evaluate: `int_  (x + sin x)/(1 + cos x )  dx`


Evaluate the following:

`int x/(x^4 - 1) "d"x`


Evaluate the following:

`int sqrt(x)/(sqrt("a"^3 - x^3)) "d"x`


Evaluate the following:

`int ("d"x)/(xsqrt(x^4 - 1))`  (Hint: Put x2 = sec θ)


Evaluate the following:

`int_1^2 ("d"x)/sqrt((x - 1)(2 - x))`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×