English

Π / 2 ∫ − π / 2 Log ( a − Sin θ a + Sin θ ) D θ - Mathematics

Advertisements
Advertisements

Question

\[\int\limits_{- \pi/2}^{\pi/2} \log\left( \frac{a - \sin \theta}{a + \sin \theta} \right) d\theta\]
Advertisements

Solution

\[Let\, I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \log\left( \frac{a - \sin\theta}{a + \sin\theta} \right) d \theta\]

\[Here\, f\left( \theta \right) = \log\left( \frac{a - \sin\theta}{a + \sin\theta} \right)\]

\[Consider\, f\left( - \theta \right) = \log\left[ \frac{a - \sin\left( - \theta \right)}{a + \sin\left( - \theta \right)} \right] = - \log\left( \frac{a - \sin\theta}{a + \sin\theta} \right) = - f\left( \theta \right)\]

\[i . e . , f\left( \theta \right) \text{is odd function} . \]

\[\text{Therefore}, I = 0\]

shaalaa.com
Definite Integrals
  Is there an error in this question or solution?
Chapter 20: Definite Integrals - Very Short Answers [Page 115]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 20 Definite Integrals
Very Short Answers | Q 18 | Page 115

RELATED QUESTIONS

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

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

 


Evaluate the following definite integrals:

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

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

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

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

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

\[\int_0^\frac{\pi}{2} \frac{\cos^2 x}{1 + 3 \sin^2 x}dx\]

\[\int\limits_1^4 f\left( x \right) dx, where\ f\left( x \right) = \begin{cases}4x + 3 & , & \text{if }1 \leq x \leq 2 \\3x + 5 & , & \text{if }2 \leq x \leq 4\end{cases}\]

 


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

 


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

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

 


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

 


Evaluate the following integral:

\[\int_{- 1}^1 \left| xcos\pi x \right|dx\]

 


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

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

\[\int\limits_1^4 \left( 3 x^2 + 2x \right) dx\]

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

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

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

 


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

Evaluate each of the following integral:

\[\int_0^\frac{\pi}{2} e^x \left( \sin x - \cos x \right)dx\]

 


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

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

The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\]  is 


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

 


\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]


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


\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]


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


\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]


\[\int\limits_0^{15} \left[ x^2 \right] dx\]


\[\int\limits_0^\pi \cos 2x \log \sin x dx\]


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


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


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


Using second fundamental theorem, evaluate the following:

`int_0^(pi/2) sqrt(1 + cos x)  "d"x`


Choose the correct alternative:

Γ(n) is


Find `int sqrt(10 - 4x + 4x^2)  "d"x`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×