हिंदी

Π / 2 ∫ π / 3 √ 1 + Cos X ( 1 − Cos X ) 5 / 2 D X

Advertisements
Advertisements

प्रश्न

\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]

योग
Advertisements

उत्तर

\[\int_\frac{\pi}{3}^\frac{\pi}{2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^\frac{5}{2}} d x\]

\[ = \int_\frac{\pi}{3}^\frac{\pi}{2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^\frac{5}{2}} \times \frac{\sqrt{1 - \cos x}}{\sqrt{1 - \cos x}} d x\]

\[ = \int_\frac{\pi}{3}^\frac{\pi}{2} \frac{\sin x}{\left( 1 - \cos x \right)^3}dx\]

\[ = - \frac{1}{2} \left[ \left( 1 - \cos x \right)^{- 2} \right]_\frac{\pi}{3}^\frac{\pi}{2} \]

\[ = - \frac{1}{2}\left[ 1 - 4 \right]\]

\[ = \frac{3}{2}\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Definite Integrals - Revision Exercise [पृष्ठ १२१]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Revision Exercise | Q 20 | पृष्ठ १२१

संबंधित प्रश्न

\[\int\limits_4^9 \frac{1}{\sqrt{x}} dx\]

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

\[\int\limits_0^{\pi/4} \sec x dx\]

\[\int\limits_{- \pi/4}^{\pi/4} \frac{1}{1 + \sin x} dx\]

\[\int\limits_0^{\pi/2} \left( a^2 \cos^2 x + b^2 \sin^2 x \right) dx\]

\[\int\limits_0^{\pi/2} \sqrt{1 + \cos x}\ dx\]

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

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

\[\int_0^1 x\log\left( 1 + 2x \right)dx\]

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

\[\int\limits_0^{\pi/2} \frac{\sin x \cos x}{1 + \sin^4 x} dx\]

\[\int_0^\frac{\pi}{4} \frac{\sin x + \cos x}{3 + \sin2x}dx\]

\[\int\limits_0^1 \frac{1 - x^2}{x^4 + x^2 + 1} dx\]

\[\int\limits_0^{\pi/6} \cos^{- 3} 2 \theta \sin 2\ \theta\ d\ \theta\]

\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{3/2}} dx\]

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

\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]


\[\int\limits_0^\pi \log\left( 1 - \cos x \right) dx\]

If f (x) is a continuous function defined on [0, 2a]. Then, prove that

\[\int\limits_0^{2a} f\left( x \right) dx = \int\limits_0^a \left\{ f\left( x \right) + f\left( 2a - x \right) \right\} dx\]

 


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

\[\int\limits_0^2 \sqrt{4 - x^2} dx\]

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

 

 


\[\int\limits_0^{15} \left[ x \right] dx .\]

Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]


\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan x} dx\]  is equal to

The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is

 


Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .


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


\[\int\limits_0^{\pi/4} e^x \sin x dx\]


\[\int\limits_0^{2\pi} \cos^7 x dx\]


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


\[\int\limits_0^4 x dx\]


\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]


Using second fundamental theorem, evaluate the following:

`int_1^2 (x - 1)/x^2  "d"x`


Evaluate the following using properties of definite integral:

`int_(- pi/2)^(pi/2) sin^2theta  "d"theta`


Evaluate the following:

Γ(4)


Find `int sqrt(10 - 4x + 4x^2)  "d"x`


Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×