मराठी

Π / 2 ∫ 0 Log ( 3 + 5 Cos X 3 + 5 Sin X ) D X . - Mathematics

Advertisements
Advertisements

प्रश्न

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

 

बेरीज

उत्तर

\[Let\, I = \int_0^\frac{\pi}{2} \log\left( \frac{3 + 5\cos x}{3 + 5\sin x} \right) d x ................(1)\]
\[ = \int_0^\frac{\pi}{2} log\left[ \frac{3 + 5\cos\left( \frac{\pi}{2} - x \right)}{3 + 5\sin\left( \frac{\pi}{2} - x \right)} \right] dx\]
\[ = \int_0^\frac{\pi}{2} log\left( \frac{3 + 5\sin x}{3 + 5\cos x} \right) dx .................(2)\]
\[\text{Adding (1) and (2)}\]
\[2I = \int_0^\frac{\pi}{2} \left[ \log\left( \frac{3 + 5\cos x}{3 + 5\sin x} \right) + log\left( \frac{3 + 5\sin x}{3 + 5\cos x} \right) \right] d x\]
\[ = \int_0^\frac{\pi}{2} \log\left( \frac{3 + 5\cos x}{3 + 5\sin x} \times \frac{3 + 5\sin x}{3 + 5\cos x} \right) dx\]
\[ = \int_0^\frac{\pi}{2} \log1 dx = 0\]
\[Hence\ I = 0\]

shaalaa.com
Definite Integrals
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Definite Integrals - Very Short Answers [पृष्ठ ११५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
Very Short Answers | Q 15 | पृष्ठ ११५

संबंधित प्रश्‍न

\[\int\limits_0^{1/2} \frac{1}{\sqrt{1 - x^2}} dx\]

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

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

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

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

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

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

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

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

 


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


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

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

 


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

If f is an integrable function, show that

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


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

\[\int\limits_0^3 \frac{1}{x^2 + 9} dx .\]

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

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

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

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

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

\[\int\limits_1^2 \log_e \left[ x \right] dx .\]

\[\int\limits_0^\pi \frac{1}{a + b \cos x} dx =\]

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

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

 


If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals


The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is 


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


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


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


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


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


Evaluate the following:

f(x) = `{{:("c"x",", 0 < x < 1),(0",",  "otherwise"):}` Find 'c" if `int_0^1 "f"(x)  "d"x` = 2


Evaluate the following integrals as the limit of the sum:

`int_0^1 (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:

If n > 0, then Γ(n) is


Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`


Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×