हिंदी

Π / 2 ∫ 0 Sin θ √ 1 + Cos θ D θ - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\limits_0^{\pi/2} \frac{\sin \theta}{\sqrt{1 + \cos \theta}} d\theta\]
Advertisements

उत्तर

\[Let\ I = \int_0^\frac{\pi}{2} \frac{\sin \theta}{\sqrt{1 + \cos \theta}} d \theta . \]

\[Let\ \cos\ \theta = t . Then, - \sin\ \theta\ d\theta\ = dt\]

\[When\ \theta = 0, t = 1\ and\ \theta = \frac{\pi}{2}, t = 0\]

\[ \therefore I = \int_0^\frac{\pi}{2} \frac{\sin \theta}{\sqrt{1 + \cos \theta}} d \theta\]

\[ = \int_1^0 \frac{- dt}{\sqrt{1 + t}}\]

\[ = \int_0^1 \frac{dt}{\sqrt{1 + t}}\]

\[ = 2 \left[ \sqrt{1 + t} \right]_0^1 \]

\[ = 2\left( \sqrt{2} - 1 \right)\]

shaalaa.com
Definite Integrals
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 20: Definite Integrals - Exercise 20.2 [पृष्ठ ३९]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 20 Definite Integrals
Exercise 20.2 | Q 13 | पृष्ठ ३९

संबंधित प्रश्न

\[\int\limits_{- 2}^3 \frac{1}{x + 7} dx\]

\[\int\limits_0^{\pi/2} \cos^2 x\ dx\]

\[\int\limits_0^{\pi/6} \cos x \cos 2x\ dx\]

\[\int\limits_1^e \frac{\log x}{x} dx\]

\[\int\limits_0^\pi \frac{1}{3 + 2 \sin x + \cos x} dx\]

\[\int\limits_0^{\pi/2} x^2 \sin\ x\ dx\]

\[\int\limits_{- 1}^1 5 x^4 \sqrt{x^5 + 1} dx\]

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

\[\int_0^\frac{\pi}{2} \frac{\tan x}{1 + m^2 \tan^2 x}dx\]

\[\int_0^\frac{\pi}{2} \sqrt{\cos x - \cos^3 x}\left( \sec^2 x - 1 \right) \cos^2 xdx\]

\[\int\limits_0^5 \frac{\sqrt[4]{x + 4}}{\sqrt[4]{x + 4} + \sqrt[4]{9 - x}} dx\]

\[\int\limits_0^{\pi/2} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{\tan x}} dx\]

\[\int\limits_0^{\pi/2} \frac{\sin^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} dx\]

\[\int\limits_0^\pi x \sin^3 x\ dx\]

\[\int\limits_0^2 \left( x^2 + 1 \right) dx\]

\[\int\limits_1^2 x^2 dx\]

\[\int\limits_2^3 x^2 dx\]

\[\int\limits_0^1 \frac{1}{x^2 + 1} dx\]

Evaluate each of the following integral:

\[\int_0^\frac{\pi}{4} \tan\ xdx\]

 


\[\int\limits_2^3 \frac{1}{x}dx\]

If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.

 

 


If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]


\[\int\limits_0^2 \left[ x \right] dx .\]

The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is 

 


\[\int\limits_0^{\pi/2} \sin\ 2x\ \log\ \tan x\ dx\]  is equal to 

The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is

 


Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .


\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]


\[\int\limits_0^1 \left| 2x - 1 \right| dx\]


\[\int\limits_0^\pi x \sin x \cos^4 x dx\]


\[\int\limits_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx\]


\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]


Evaluate the following using properties of definite integral:

`int_(- pi/4)^(pi/4) x^3 cos^3 x  "d"x`


Evaluate the following using properties of definite integral:

`int_(-1)^1 log ((2 - x)/(2 + x))  "d"x`


Evaluate the following integrals as the limit of the sum:

`int_1^3 x  "d"x`


Choose the correct alternative:

The value of `int_(- pi/2)^(pi/2) cos  x  "d"x` is


Choose the correct alternative:

If n > 0, then Γ(n) is


If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.


Verify the following:

`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×