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 :
`∫_(-pi)^pi (cos ax−sin bx)^2 dx`
Evaluate :
`∫_0^π(4x sin x)/(1+cos^2 x) dx`
Evaluate: `intsinsqrtx/sqrtxdx`
Evaluate the integral by using substitution.
`int_0^(pi/2) (sin x)/(1+ cos^2 x) dx`
Evaluate the integral by using substitution.
`int_(-1)^1 dx/(x^2 + 2x + 5)`
`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 of the following integral:
Evaluate of the following integral:
Evaluate:
Evaluate:
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 the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate :
Evaluate : \[\int\limits_{- 2}^1 \left| x^3 - x \right|dx\] .
Find : \[\int e^{2x} \sin \left( 3x + 1 \right) dx\] .
Evaluate: \[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x}dx\] .
Evaluate: `int_-π^π (1 - "x"^2) sin "x" cos^2 "x" 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` = ______.
Evaluate the following:
`int ("e"^(6logx) - "e"^(5logx))/("e"^(4logx) - "e"^(3logx)) "d"x`
Evaluate the following:
`int "dt"/sqrt(3"t" - 2"t"^2)`
Find: `int (dx)/sqrt(3 - 2x - x^2)`
If `int x^5 cos (x^6)dx = k sin (x^6) + C`, find the value of k.
