Advertisements
Advertisements
प्रश्न
Evaluate: `int_((-π)/2)^(π/2) (sin|x| + cos|x|)dx`
उत्तर
We have, `int_((-π)/2)^(π/2) (sin|x| + cos|x|)dx`
Let f(x) = sin|x| + cos|x|
Then, f(x) = f(–x)
Since, f(x) is an even function
So, I = `int_((-π)/2)^(π/2) (sin|x| + cos|x|)dx`
= `2int_0^(π/2) (sinx + cosx)dx`
= `2[-cosx + sinx]_0^(π/2)`
= `2[-cos π/2 + sin π/2 + cos0 - sin0]`
= 2[0 + 1 + 1 – 0]
= 2(2)
= 4
APPEARS IN
संबंधित प्रश्न
Evaluate : `int "e"^(3"x")/("e"^(3"x") + 1)` dx
Evaluate : ∫ log (1 + x2) dx
Using properties of definite integrals, evaluate
`int_0^(π/2) sqrt(sin x )/ (sqrtsin x + sqrtcos x)dx`
Evaluate: `int_0^pi ("x"sin "x")/(1+ 3cos^2 "x") d"x"`.
Find : `int_ (2"x"+1)/(("x"^2+1)("x"^2+4))d"x"`.
Evaluate the following integrals : `int_2^5 sqrt(x)/(sqrt(x) + sqrt(7 - x))*dx`
`int_0^2 e^x dx` = ______.
`int_2^7 sqrt(x)/(sqrt(x) + sqrt(9 - x)) dx` = ______.
`int_(-7)^7 x^3/(x^2 + 7) "d"x` = ______
`int_0^1 ((x^2 - 2)/(x^2 + 1))`dx = ?
`int_0^{pi/2} log(tanx)dx` = ______
`int_0^1 (1 - x)^5`dx = ______.
If `int_0^"a" sqrt("a - x"/x) "dx" = "K"/2`, then K = ______.
`int_0^{pi/2} (cos2x)/(cosx + sinx)dx` = ______
If `f(a + b - x) = f(x)`, then `int_0^b x f(x) dx` is equal to
`int_(-5)^5 x^7/(x^4 + 10) dx` = ______.
Evaluate: `int_0^(2π) (1)/(1 + e^(sin x)`dx
Evaluate: `int_(-1)^3 |x^3 - x|dx`
`int_0^π(xsinx)/(1 + cos^2x)dx` equals ______.
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 ______.
`int_0^(π/4) x. sec^2 x dx` = ______.
Assertion (A): `int_2^8 sqrt(10 - x)/(sqrt(x) + sqrt(10 - x))dx` = 3.
Reason (R): `int_a^b f(x) dx = int_a^b f(a + b - x) dx`.
Evaluate the following limit :
`lim_("x"->3)[sqrt("x"+6)/"x"]`
Evaluate the following definite integral:
`int_4^9 1/sqrt"x" "dx"`
Solve the following.
`int_1^3 x^2 logx dx`
Evaluate `int_0^3root3(x+4)/(root3(x+4)+root3(7-x)) dx`
Evaluate the following integral:
`int_0^1x (1 - x)^5 dx`
Evaluate the following integral:
`int_0^1 x(1 - x)^5 dx`
