मराठी

Π ∫ 0 1 5 + 3 Cos X D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\]
Advertisements

उत्तर

\[Let\ I = \int_0^\pi \frac{1}{5 + 3 \cos x} d\ x . Then, \]
\[I = \int_0^\pi \frac{1}{5 + 3\left( \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \right)} d x\]
\[ \Rightarrow I = \int_0^\pi \frac{1 + \tan^2 \frac{x}{2}}{5 + 5 \tan^2 \frac{x}{2} - 3 \tan^2 \frac{x}{2}} dx\]
\[ \Rightarrow I = \int_0^\pi \frac{\sec^2 \frac{x}{2}}{5 + 2 \tan^2 \frac{x}{2}} dx\]
\[Let\ \tan \frac{x}{2} = t . Then, \frac{1}{2} \sec^2 \frac{x}{2} dx = dt\]
\[When\ x = 0, t = 0\ and\ x = \pi, t = \infty \]
\[ \therefore I = \int_0^\infty \frac{dt}{5 + 2 t^2}\]
\[ \Rightarrow I = \frac{1}{2} \int_0^\infty \frac{dt}{\frac{5}{2} + t^2}\]
\[ \Rightarrow I = \frac{1}{2} \left[ \tan^{- 1} \frac{\sqrt{5}t}{\sqrt{2}} \right]_0^\infty \]
\[ \Rightarrow I = \frac{1}{2}\left( \frac{\pi}{2} - 0 \right)\]
\[ \Rightarrow I = \frac{\pi}{4}\]

shaalaa.com
Definite Integrals
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Definite Integrals - Exercise 20.2 [पृष्ठ ३९]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
Exercise 20.2 | Q 27 | पृष्ठ ३९

संबंधित प्रश्‍न

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

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

\[\int\limits_1^3 \frac{\log x}{\left( x + 1 \right)^2} dx\]

\[\int\limits_1^2 \frac{x + 3}{x \left( x + 2 \right)} dx\]

\[\int\limits_{- 1}^1 \frac{1}{x^2 + 2x + 5} dx\]

\[\int\limits_0^1 x \left( 1 - x \right)^5 dx\]

\[\int\limits_0^\pi \left( \sin^2 \frac{x}{2} - \cos^2 \frac{x}{2} \right) dx\]

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

\[\int\limits_0^{\pi/4} \left( \sqrt{\tan}x + \sqrt{\cot}x \right) dx\]

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

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

Evaluate the following integral:

\[\int\limits_{- 3}^3 \left| x + 1 \right| dx\]

\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{- \frac{\pi}{2}}{\sqrt{\cos x \sin^2 x}}dx\]

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

 


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

 


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

If f(2a − x) = −f(x), prove that

\[\int\limits_0^{2a} f\left( x \right) dx = 0 .\]

\[\int\limits_3^5 \left( 2 - x \right) dx\]

\[\int\limits_{- 2}^1 \frac{\left| x \right|}{x} dx .\]

\[\int\limits_0^{\pi/2} \log \tan x\ dx .\]

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

The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .


The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is

 


Evaluate : \[\int e^{2x} \cdot \sin \left( 3x + 1 \right) dx\] .


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


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


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


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


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


\[\int\limits_2^3 e^{- x} dx\]


Using second fundamental theorem, evaluate the following:

`int_0^1 x"e"^(x^2)  "d"x`


Evaluate the following:

`int_0^oo "e"^(-4x) x^4  "d"x`


Evaluate the following integrals as the limit of the sum:

`int_1^3 (2x + 3)  "d"x`


Choose the correct alternative:

`int_(-1)^1 x^3 "e"^(x^4)  "d"x` is


Choose the correct alternative:

`Γ(3/2)`


Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×