Advertisements
Advertisements
प्रश्न
Evaluate the following:
`int_0^(1/2) ("d"x)/((1 + x^2)sqrt(1 - x^2))` (Hint: Let x = sin θ)
उत्तर
Let I = `int_0^(1/2) ("d"x)/((1 + x^2)sqrt(1 - x^2))`
Put x = sin θ
∴ dx = cos θ dθ
Changing the limits, we get
When x = 0
∴ sin θ = θ
∴ θ = 0
When x = `1/2`
∴ sin θ = `1/2`
∴ θ = `pi/6`
∴ I = `int_0^(pi/6) (cos theta "d"theta)/((1 + sin^2theta)sqrt(1 - sin^2theta))`
= `int_0^(pi/6) (cos theta "d"theta)/((1 + sin^2theta) costheta)`
= `int_0^(pi/6) 1/(1 + sin^2theta) "d"theta`
Now, dividing the numerator and denominator by cos2θ, we get
= `int_0^(pi/6) (1/cos^2theta)/(1/(cos^2theta) + (sin^2theta)/(cos^2theta)) "d"theta`
= `int_0^(pi/6) (sec^2theta)/(sec^2theta + tan^2theta) "d"theta`
= `int_0^(pi/6) (sec^2theta)/(1 + tan^2theta + tan^2theta) "d"theta`
= `int_0^(pi/6) (sec^2theta)/(2tan^2theta + 1) "d"theta`
Put tan θ = t
∴ sec2θ dθ = t
Changing the limits, we get
When θ = 0
∴ t = tan 0 = 0
When θ = `pi/6`
∴ t = `tan pi/6 = 1/sqrt(3)`
∴ I = `int_0^(1/sqrt(3)) "dt"/(2"t"^2 + 1)`
= `1/2 int_0^(1/sqrt(3)) "dt"/("t"^2 + 1/2)`
= `1/2 int_0^(1/sqrt(3)) "dt"/("t"^2 + (1/sqrt(2))^2)`
= `1/2 xx 1/(1/sqrt(12)) [tan^-1 "t"/(1/sqrt(12))]_0^(1/sqrt(3))`
= `1/sqrt(2) tan^-1 [sqrt(2)"t"]_0^(1/sqrt(3)`
= `1/sqrt(2) [tan^-1 sqrt(2)/sqrt(3) - tan^-1 0]`
= `1/sqrt(2) tan^-1 sqrt(2/3)`
APPEARS IN
संबंधित प्रश्न
Evaluate the following definite integrals as limit of sums.
`int_2^3 x^2 dx`
Evaluate the following definite integrals as limit of sums.
`int_1^4 (x^2 - x) dx`
Evaluate the definite integral:
`int_(pi/2)^pi e^x ((1-sinx)/(1-cos x)) dx`
Evaluate the definite integral:
`int_0^(pi/4) (sinx cos x)/(cos^4 x + sin^4 x)`dx
Evaluate the definite integral:
`int_0^(pi/2) (cos^2 x dx)/(cos^2 x + 4 sin^2 x)`
Evaluate the definite integral:
`int_1^4 [|x - 1|+ |x - 2| + |x -3|]dx`
Prove the following:
`int_(-1)^1 x^17 cos^4 xdx = 0`
Prove the following:
`int_0^1sin^(-1) xdx = pi/2 - 1`
Choose the correct answers The value of `int_0^1 tan^(-1) (2x -1)/(1+x - x^2)` dx is
(A) 1
(B) 0
(C) –1
(D) `pi/4`
if `int_0^k 1/(2+ 8x^2) dx = pi/16` then the value of k is ________.
(A) `1/2`
(B) `1/3`
(C) `1/4`
(D) `1/5`
\[\int\frac{1}{x} \left( \log x \right)^2 dx\]
Evaluate the following integral:
Evaluate the following integrals as limit of sums:
Evaluate `int_1^4 ( 1+ x +e^(2x)) dx` as limit of sums.
If f and g are continuous functions in [0, 1] satisfying f(x) = f(a – x) and g(x) + g(a – x) = a, then `int_0^"a" "f"(x) * "g"(x)"d"x` is equal to ______.
Evaluate the following:
`int_0^1 (x"d"x)/sqrt(1 + x^2)`
Evaluate the following:
`int_(pi/3)^(pi/2) sqrt(1 + cosx)/(1 - cos x)^(5/2) "d"x`
The value of `int_(-pi)^pi sin^3x cos^2x "d"x` is ______.
If f" = C, C ≠ 0, where C is a constant, then the value of `lim_(x -> 0) (f(x) - 2f (2x) + 3f (3x))/x^2` is
The value of `lim_(n→∞)1/n sum_(r = 0)^(2n-1) n^2/(n^2 + 4r^2)` is ______.
