Advertisements
Advertisements
Question
Advertisements
Solution
\[\int\limits_0^{\pi/2} \sin^2 x\ dx .\]
\[ = \int_0^\frac{\pi}{2} \frac{1 - \cos2x}{2} dx\]
\[ = \frac{1}{2} \int_0^\frac{\pi}{2} \left( 1 - \cos2x \right) dx\]
\[ = \frac{1}{2} \left[ x - \frac{\sin2x}{2} \right]_0^\frac{\pi}{2} \]
\[ = \frac{1}{2}\left( \frac{\pi}{2} - 0 \right)\]
\[ = \frac{\pi}{4}\]
APPEARS IN
RELATED QUESTIONS
Evaluate each of the following integral:
Evaluate :
\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
Find : `∫_a^b logx/x` dx
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 x "d"x`
Choose the correct alternative:
`Γ(3/2)`
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
