Advertisements
Advertisements
प्रश्न
Evaluate:
Advertisements
उत्तर
\[\int\sqrt{\frac{1 + \cos 2x}{2}}dx\]
\[ \int\sqrt{\frac{\text{2 cos}^2 x}{2}}dx \left[ \therefore 1 + \cos2A = 2 \cos^2 A \right]\]
\[ = \int\ \text{cos x dx}\]
\[ = \sin x + C\]
APPEARS IN
संबंधित प्रश्न
Evaluate: `int (1+logx)/(x(2+logx)(3+logx))dx`
Evaluate :`int_0^(pi/2)1/(1+cosx)dx`
Evaluate : `int_0^4(|x|+|x-2|+|x-4|)dx`
Evaluate `int_(-1)^2|x^3-x|dx`
Evaluate :
`∫_0^π(4x sin x)/(1+cos^2 x) dx`
Evaluate :
`int_e^(e^2) dx/(xlogx)`
Evaluate: `intsinsqrtx/sqrtxdx`
Evaluate the integral by using substitution.
`int_0^1 sin^(-1) ((2x)/(1+ x^2)) dx`
Evaluate the integral by using substitution.
`int_0^2 dx/(x + 4 - x^2)`
Evaluate the integral by using substitution.
`int_(-1)^1 dx/(x^2 + 2x + 5)`
Evaluate the integral by using substitution.
`int_1^2 (1/x- 1/(2x^2))e^(2x) dx`
The value of the integral `int_(1/3)^4 ((x- x^3)^(1/3))/x^4` dx is ______.
Evaluate `int_0^(pi/4) (sinx + cosx)/(16 + 9sin2x) dx`
Evaluate of the following integral:
Evaluate of the following integral:
Evaluate:
Evaluate:
Evaluate the following integral:
Evaluate the following integral:
Evaluate 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:
Evaluate : \[\int\limits_{- 2}^1 \left| x^3 - x \right|dx\] .
Find : \[\int e^{2x} \sin \left( 3x + 1 \right) dx\] .
Find : \[\int\frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}dx\] .
Evaluate: `int_ e^x ((2+sin2x))/cos^2 x dx`
Evaluate: `int_-π^π (1 - "x"^2) sin "x" cos^2 "x" d"x"`.
Find: `int_ (3"x"+ 5)sqrt(5 + 4"x"-2"x"^2)d"x"`.
`int_(pi/5)^((3pi)/10) [(tan x)/(tan x + cot x)]`dx = ?
`int_0^(pi4) sec^4x "d"x` = ______.
Find: `int (dx)/sqrt(3 - 2x - x^2)`
Which limits are used to evaluate the resulting expression in Method: Resubstitution?
For the integral \[\int_{0}^{1}\frac{\tan^{-1}x}{1+x^2}\,dx\], which substitution is suitable?
