Advertisements
Advertisements
प्रश्न
Evaluate:
Advertisements
उत्तर
\[\int\left( \frac{\cos 2x + 2 \sin^2 x}{\sin^2 x} \right)dx\]
\[ = \int\left( \frac{1 - 2 \sin^2 x + 2 \sin^2 x}{\sin^2 x} \right)dx \left[ \because \cos 2x = 1 - 2 \sin^2 x \right]\]
\[ = \int {cosec}^2\text{ x dx}\]
\[ = - \cot x + C\]
APPEARS IN
संबंधित प्रश्न
Evaluate :`int_0^(pi/2)1/(1+cosx)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`
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 xsqrt(x+2)` (Put x + 2 = `t^2`)
Evaluate the integral by using substitution.
`int_1^2 (1/x- 1/(2x^2))e^(2x) dx`
`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:
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 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 the following integral:
Find : \[\int\frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}dx\] .
Evaluate: \[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x}dx\] .
`int_0^1 x(1 - x)^5 "dx" =` ______.
`int_0^1 sin^-1 ((2x)/(1 + x^2))"d"x` = ______.
