मराठी

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

Advertisements
Advertisements

प्रश्न

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

 

बेरीज
Advertisements

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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - Very Short Answers [पृष्ठ ११५]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Very Short Answers | Q 15 | पृष्ठ ११५

संबंधित प्रश्‍न

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

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

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

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

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

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

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

\[\int\limits_4^9 \frac{\sqrt{x}}{\left( 30 - x^{3/2} \right)^2} dx\]

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

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

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

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

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

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

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

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

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

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

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

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

\[\int\limits_0^\pi \cos^5 x\ dx .\]

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

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

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


\[\int\limits_{- 1}^1 \left| 1 - x \right| dx\]  is equal to

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


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


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


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


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


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 x  "d"x`


Evaluate the following:

`int ((x^2 + 2))/(x + 1) "d"x`


If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then ______.


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


Which integral has Constant C present?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×