Advertisements
Advertisements
Question
Evaluate the integral by using substitution.
`int_(-1)^1 dx/(x^2 + 2x + 5)`
Advertisements
Solution
`int_-1^1 dx/(x^2 + 2x + 5)`
`= int_-1^1 dx/(x^2 + 2x + 4 + 1)`
`int_-1^1 dx/((x + 2)^2 + (1)^2)`
`= 1/2 [tan^-1 ((x + 1)/2)]_-1^1`
`= 1/2 [tan^-1 (1) - tan^-1 0]`
`= 1/2 [pi/4 - 0]`
`= pi/8`
APPEARS IN
RELATED QUESTIONS
Evaluate : `int1/(3+5cosx)dx`
Evaluate `∫_0^(3/2)|x cosπx|dx`
Evaluate the integral by using substitution.
`int_0^2 dx/(x + 4 - x^2)`
The value of the integral `int_(1/3)^4 ((x- x^3)^(1/3))/x^4` dx is ______.
If `f(x) = int_0^pi t sin t dt`, then f' (x) is ______.
Evaluate of the following integral:
Evaluate of the following integral:
Evaluate of the following integral:
Evaluate:
Evaluate:
Evaluate:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate each of the following integral:
Evaluate each of the following integral:
Evaluate each of the following integral:
Evaluate each of the following integral:
Evaluate each of the following integral:
Evaluate the following integral:
Evaluate
\[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\]
Evaluate :
Evaluate : \[\int\limits_{- 2}^1 \left| x^3 - x \right|dx\] .
Find : \[\int\frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}dx\] .
Evaluate: \[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x}dx\] .
Evaluate: `int_ e^x ((2+sin2x))/cos^2 x dx`
Find: `int_ (3"x"+ 5)sqrt(5 + 4"x"-2"x"^2)d"x"`.
`int_0^1 x(1 - x)^5 "dx" =` ______.
`int_0^3 1/sqrt(3x - x^2)"d"x` = ______.
Find: `int (dx)/sqrt(3 - 2x - x^2)`
`int_0^1 x^2e^x dx` = ______.
Evaluate: `int x/(x^2 + 1)"d"x`
Evaluate:
`int (1 + cosx)/(sin^2x)dx`
