Advertisements
Advertisements
Question
Evaluate: `int_(pi/6)^(pi/3) (dx)/(1 + sqrt(tanx)`
Advertisements
Solution
Let I = `int_(pi/6)^(pi/3) (dx)/(1 + sqrt(tanx)`
= `int_(pi/6)^(pi/3) sqrt(cosx)/(sqrt(sinx) + sqrt(cos x)) dx` ......(i)
Using `int_a^b f(x) dx = int_a^b f(a + b - x) dx`
I = `int_(pi/6)^(pi/3) sqrt(cos(pi/6 + pi/3 - x))/(sqrt(sin(pi/6 + pi/3 - x)) + sqrt(cos(pi/6 + pi/3 - x)))`
I = `int_(pi/6)^(pi/3) sqrt(sinx)/(sqrt(cosx) + sqrt(sinx)) dx` ......(ii)
Adding (i) and (ii), we get
2I = `int_(pi/6)^(pi/3) sqrt(cosx)/(sqrt(sinx) + sqrt(cosx)) dx + int_(pi/6)^(pi/3) sqrt(sinx)/(sqrt(cosx) + sqrt(sinx)) dx`
2I = `int_(pi/6)^(pi/3) dx`
= `[x]_(pi/6)^(pi/3)`
= `pi/3 - pi/6`
= `pi/6`
Hence, I = `int_(pi/6)^(pi/3) (dx)/(1 + sqrt(tanx)) = pi/12`
APPEARS IN
RELATED QUESTIONS
By using the properties of the definite integral, evaluate the integral:
`int_0^(pi/2) (cos^5 xdx)/(sin^5 x + cos^5 x)`
By using the properties of the definite integral, evaluate the integral:
`int_0^4 |x - 1| dx`
Evaluate : `int 1/("x" [("log x")^2 + 4]) "dx"`
`int_2^7 sqrt(x)/(sqrt(x) + sqrt(9 - x)) dx` = ______.
`int_0^1 "e"^(2x) "d"x` = ______
`int_0^(pi/4) (sec^2 x)/((1 + tan x)(2 + tan x))`dx = ?
`int_-9^9 x^3/(4 - x^2)` dx = ______
`int_0^{pi/2} xsinx dx` = ______
`int_0^{pi/4} (sin2x)/(sin^4x + cos^4x)dx` = ____________
`int_0^1 log(1/x - 1) "dx"` = ______.
`int_0^pi x*sin x*cos^4x "d"x` = ______.
`int_0^9 1/(1 + sqrtx)` dx = ______
`int_(-pi/4)^(pi/4) 1/(1 - sinx) "d"x` = ______.
`int_((-pi)/4)^(pi/4) "dx"/(1 + cos2x)` is equal to ______.
`int_(-5)^5 x^7/(x^4 + 10) dx` = ______.
Evaluate: `int_1^3 sqrt(x)/(sqrt(x) + sqrt(4) - x) dx`
`int_4^9 1/sqrt(x)dx` = ______.
The value of the integral `int_(-1)^1log_e(sqrt(1 - x) + sqrt(1 + x))dx` is equal to ______.
Let a be a positive real number such that `int_0^ae^(x-[x])dx` = 10e – 9 where [x] is the greatest integer less than or equal to x. Then, a is equal to ______.
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 ______.
Evaluate: `int_0^π 1/(5 + 4 cos x)dx`
With the usual notation `int_1^2 ([x^2] - [x]^2)dx` is equal to ______.
If `int_0^(2π) cos^2 x dx = k int_0^(π/2) cos^2 x dx`, then the value of k is ______.
Evaluate the following definite integral:
`int_4^9 1/sqrt"x" "dx"`
Evaluate `int_0^3root3(x+4)/(root3(x+4)+root3(7-x)) dx`
`int_-9^9 x^3/(4-x^2) dx` =______
Evaluate the following integral:
`int_0^1 x(1-x)^5 dx`
Solve.
`int_0^1e^(x^2)x^3dx`
Evaluate the following integral:
`int_-9^9x^3/(4-x^2)dx`
