हिंदी

Π / 2 ∫ 0 √ 1 − Cos 2 X D X .

Advertisements
Advertisements

प्रश्न

\[\int\limits_0^{\pi/2} \sqrt{1 - \cos 2x}\ dx .\]
योग
Advertisements

उत्तर

\[\int_0^\frac{\pi}{2} \sqrt{1 - \cos2x}\ dx\]

\[ = \int_0^\frac{\pi}{2}\sqrt{2 \sin^2 x}\ dx\]

\[ = \int_0^\frac{\pi}{2} \sqrt{2} \sin x\ dx\]

\[ = - \sqrt{2} \left[ \cos x \right]_0^\frac{\pi}{2} \]

\[ = - \left( 0 - \sqrt{2} \right)\]

\[ = \sqrt{2}\]

\[\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Definite Integrals - Very Short Answers [पृष्ठ ११५]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Very Short Answers | Q 13 | पृष्ठ ११५

संबंधित प्रश्न

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

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

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

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

\[\int\limits_4^{12} x \left( x - 4 \right)^{1/3} dx\]

\[\int_0^2 2x\left[ x \right]dx\]

\[\int\limits_0^7 \frac{\sqrt[3]{x}}{\sqrt[3]{x} + \sqrt[3]{7} - x} dx\]

If  \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]

 


\[\int\limits_0^\pi \frac{x \tan x}{\sec x \ cosec x} dx\]

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

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

If f is an integrable function, show that

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


If f(x) is a continuous function defined on [−a, a], then prove that 

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

Prove that:

\[\int_0^\pi xf\left( \sin x \right)dx = \frac{\pi}{2} \int_0^\pi f\left( \sin x \right)dx\]

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

\[\int\limits_0^4 \left( x + e^{2x} \right) dx\]

\[\int\limits_1^4 \left( x^2 - x \right) dx\]

\[\int\limits_{- \pi/2}^{\pi/2} \sin^3 x\ dx .\]

\[\int\limits_0^{\pi/2} \log \left( \frac{3 + 5 \cos x}{3 + 5 \sin x} \right) dx .\]

 


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


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

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

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

If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals

 


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

Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .


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


\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]


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


\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]


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


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


\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]


Using second fundamental theorem, evaluate the following:

`int_0^1 "e"^(2x)  "d"x`


Evaluate the following:

`int_0^oo "e"^(-4x) x^4  "d"x`


Choose the correct alternative:

Using the factorial representation of the gamma function, which of the following is the solution for the gamma function Γ(n) when n = 8 is


Choose the correct alternative:

If n > 0, then Γ(n) is


`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.


Find: `int logx/(1 + log x)^2 dx`


The value of `int_2^3 x/(x^2 + 1)`dx is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×