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 : `int1/(3+5cosx)dx`
Evaluate `int_(-1)^2|x^3-x|dx`
find `∫_2^4 x/(x^2 + 1)dx`
Evaluate :
`∫_0^π(4x sin x)/(1+cos^2 x) dx`
Evaluate the integral by using substitution.
`int_0^(pi/2) sqrt(sin phi) cos^5 phidphi`
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`
If `f(x) = int_0^pi t sin t dt`, then f' (x) is ______.
`int 1/(1 + cos x)` dx = _____
A) `tan(x/2) + c`
B) `2 tan (x/2) + c`
C) -`cot (x/2) + c`
D) -2 `cot (x/2)` + c
Evaluate `int_0^(pi/4) (sinx + cosx)/(16 + 9sin2x) dx`
Evaluate of the following integral:
(i) \[\int x^4 dx\]
Evaluate of the following integral:
Evaluate of the following integral:
Evaluate of the following integral:
Evaluate:
Evaluate:
Evaluate:
Evaluate:
Evaluate the following definite integral:
Evaluate the following integral:
\[\int\limits_0^2 \left| x^2 - 3x + 2 \right| dx\]
Evaluate the following integral:
Evaluate each of the following integral:
Evaluate the following integral:
Evaluate :
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_-1^2 (|"x"|)/"x"d"x"`.
Evaluate: `int_1^5{|"x"-1|+|"x"-2|+|"x"-3|}d"x"`.
`int_(pi/5)^((3pi)/10) [(tan x)/(tan x + cot x)]`dx = ?
`int_0^1 x(1 - x)^5 "dx" =` ______.
`int_0^3 1/sqrt(3x - x^2)"d"x` = ______.
`int_0^(pi4) sec^4x "d"x` = ______.
`int_0^1 sin^-1 ((2x)/(1 + x^2))"d"x` = ______.
Evaluate: `int_0^(π/2) sin 2x tan^-1 (sin x) dx`.
