Advertisements
Advertisements
Question
Evaluate each of the following integral:
Solution
\[ = \left.\frac{- \cos2x}{2}\right|_0^\frac{\pi}{4} \]
\[ = - \frac{1}{2}\left( \cos\frac{\pi}{2} - \cos0 \right)\]
\[ = - \frac{1}{2} \times \left( 0 - 1 \right)\]
\[ = \frac{1}{2}\]
APPEARS IN
RELATED QUESTIONS
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
If f is an integrable function, show that
If \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]
\[\int\limits_1^4 \left( x^2 + x \right) dx\]
Find : `∫_a^b logx/x` dx
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
Evaluate the following:
`int_0^oo "e"^(- x/2) x^5 "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 (2x + 3) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_0^1 x^2 "d"x`
Choose the correct alternative:
The value of `int_(- pi/2)^(pi/2) cos x "d"x` is
