Advertisements
Advertisements
Question
Evaluate:
Advertisements
Solution
\[\int\sqrt{\frac{1 - \cos 2x}{2}}dx\]
`=∫ \sqrt {{2 sin_2 x }/{2}} dx` `[∴ 1 - cos 2x = 2 sin ^2 x]`
\[ = \int\text{sin x dx}\]
\[ = - \cos x + C\]
APPEARS IN
RELATED QUESTIONS
Evaluate: `int1/(xlogxlog(logx))dx`
Evaluate : `int1/(3+5cosx)dx`
Evaluate `∫_0^(3/2)|x cosπx|dx`
Evaluate `int_(-1)^2|x^3-x|dx`
Evaluate :
`∫_(-pi)^pi (cos ax−sin bx)^2 dx`
find `∫_2^4 x/(x^2 + 1)dx`
Evaluate :
`∫_0^π(4x sin x)/(1+cos^2 x) dx`
If `int_0^a1/(4+x^2)dx=pi/8` , find the value of a.
Evaluate :
`int_e^(e^2) dx/(xlogx)`
Evaluate the integral by using substitution.
`int_0^(pi/2) sqrt(sin phi) cos^5 phidphi`
Evaluate the integral by using substitution.
`int_0^2 dx/(x + 4 - x^2)`
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:
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 the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Find : \[\int\frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}dx\] .
Evaluate: `int_ e^x ((2+sin2x))/cos^2 x dx`
Evaluate: `int_1^5{|"x"-1|+|"x"-2|+|"x"-3|}d"x"`.
`int_(pi/5)^((3pi)/10) [(tan x)/(tan x + cot x)]`dx = ?
If `I_n = int_0^(pi/4) tan^n theta "d"theta " then " I_8 + I_6` equals ______.
`int_0^3 1/sqrt(3x - x^2)"d"x` = ______.
`int_0^1 x^2e^x dx` = ______.
The value of `int_0^1 (x^4(1 - x)^4)/(1 + x^2) dx` is
