Advertisements
Advertisements
Question
Advertisements
Solution
\[Let I = \int_0^\frac{\pi}{2} \left( 2 \log \cos x - \log\sin2x \right) d x\]
\[ = \int_0^\frac{\pi}{2} \left[ 2 \log \cos x - \log\left( 2\sin x \cos x \right) \right] d x\]
\[ = \int_0^\frac{\pi}{2} \left( 2\log\cos x - \log2 - \log\sin x - \log\cos x \right)dx\]
\[ = \int_0^\frac{\pi}{2} \left( \log\cos x - \log2 - \log\sin x \right)dx\]
\[ = \int_0^\frac{\pi}{2} \log\cos x dx - \int_0^\frac{\pi}{2} \log2 dx - \int_0^\frac{\pi}{2} \log\sin x dx\]
\[ = \int_0^\frac{\pi}{2} \log\cos x dx - \int_0^\frac{\pi}{2} \log2 dx - \int_0^\frac{\pi}{2} \log\sin\left( \frac{\pi}{2} - x \right) dx ..........................\left[\text{Using }\int_0^a f\left( x \right) dx = \int_0^a f\left( a - x \right) dx \right]\]
\[ = \int_0^\frac{\pi}{2} \log\cos x dx - \int_0^\frac{\pi}{2} \log2 dx - \int_0^\frac{\pi}{2} \log\cos x dx\]
\[ = - \log2 \left[ x \right]_0^\frac{\pi}{2} \]
\[ = - \frac{\pi}{2} \log2\]
APPEARS IN
RELATED QUESTIONS
Prove that:
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] 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
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
Evaluate the following:
`Γ (9/2)`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
What is the result of a definite integral?
