Advertisements
Advertisements
प्रश्न
Find `int_0^(pi/4) sqrt(1 + sin 2x) "d"x`
उत्तर
We have I = `int_0^(pi/4) sqrt(1 + sin 2x) "d"x`
= `int_0^(pi/4) sqrt((sinx + cosx)^2) "d"x`
= `int_0^(pi/4) (sinx + cosx) "d"x`
= `(-cosx + sinx)_0^(pi/4)`
I = 1.
APPEARS IN
संबंधित प्रश्न
By using the properties of the definite integral, evaluate the integral:
`int_0^(pi/2) sin^(3/2)x/(sin^(3/2)x + cos^(3/2) x) dx`
By using the properties of the definite integral, evaluate the integral:
`int_2^8 |x - 5| dx`
By using the properties of the definite integral, evaluate the integral:
`int_0^(pi/2) (2log sin x - log sin 2x)dx`
Evaluate `int e^x [(cosx - sin x)/sin^2 x]dx`
Evaluate : ∫ log (1 + x2) dx
Evaluate: `int_0^pi ("x"sin "x")/(1+ 3cos^2 "x") d"x"`.
`int_2^7 sqrt(x)/(sqrt(x) + sqrt(9 - x)) dx` = ______.
`int_0^{pi/2}((3sqrtsecx)/(3sqrtsecx + 3sqrt(cosecx)))dx` = ______
`int_"a"^"b" sqrtx/(sqrtx + sqrt("a" + "b" - x)) "dx"` = ______.
`int_0^{pi/4} (sin2x)/(sin^4x + cos^4x)dx` = ____________
`int_0^(pi/2) 1/(1 + cosx) "d"x` = ______.
Which of the following is true?
`int_(-1)^1 (x^3 + |x| + 1)/(x^2 + 2|x| + 1) "d"x` is equal to ______.
Evaluate: `int_1^3 sqrt(x)/(sqrt(x) + sqrt(4) - x) dx`
If `intxf(x)dx = (f(x))/2` then f(x) = ex.
Let f be continuous periodic function with period 3, such that `int_0^3f(x)dx` = 1. Then the value of `int_-4^8f(2x)dx` is ______.
If f(x) = `{{:(x^2",", "where" 0 ≤ x < 1),(sqrt(x)",", "when" 1 ≤ x < 2):}`, then `int_0^2f(x)dx` equals ______.
`int_0^(pi/4) (sec^2x)/((1 + tanx)(2 + tanx))dx` equals ______.
If `int_0^(π/2) log cos x dx = π/2 log(1/2)`, then `int_0^(π/2) log sec dx` = ______.
`int_((-π)/2)^(π/2) log((2 - sinx)/(2 + sinx))` is equal to ______.
`int_-1^1 (17x^5 - x^4 + 29x^3 - 31x + 1)/(x^2 + 1) dx` is equal to ______.
Evaluate : `int_-1^1 log ((2 - x)/(2 + x))dx`.
`int_-9^9 x^3/(4-x^2) dx` =______
Evaluate:
`int_0^1 |2x + 1|dx`
Evaluate the following integral:
`int_0^1 x(1 - x)^5 dx`
Evaluate the following integral:
`int_0^1 x (1 - x)^5 dx`
Solve the following.
`int_0^1e^(x^2)x^3dx`
Evaluate:
`int_0^6 |x + 3|dx`
