Advertisements
Advertisements
प्रश्न
Integrate the function:
`cos x/sqrt(4 - sin^2 x)`
Advertisements
उत्तर
Let I = `int (cos x)/sqrt(4 - sin^2 x)`dx
Substituting sin x = 2t
cos x dx = 2 dt
Hence, `I = int (2 dt)/sqrt(4 - 4t^2)`
`= int (2 dt)/(2sqrt(1 - t^2))`
`= int 1/sqrt(1 - t^2) dt = sin^-1 t + C`
`= sin^-1 ((sin x)/2) + C ....[(because sin x = 2 t), (=> t = (sin x)/2)]`
APPEARS IN
संबंधित प्रश्न
Evaluate : `∫(sin^6x+cos^6x)/(sin^2x.cos^2x)dx`
If `f(x) =∫_0^xt sin t dt` , then write the value of f ' (x).
Find an anti derivative (or integral) of the following function by the method of inspection.
(axe + b)2
Find the following integrals:
`int (4e^(3x) + 1)`
Find the following integrals:
`intx^2 (1 - 1/x^2)dx`
Find the following integrals:
`int(2x^2 + e^x)dx`
Find the following integrals:
`int (x^3 + 3x + 4)/sqrtx dx`
Find the following integrals:
`intsec x (sec x + tan x) dx`
Find the following integrals:
`int(sec^2x)/(cosec^2x) dx`
Integrate the function:
`1/(x - x^3)`
Integrate the function:
`1/(sqrt(x+a) + sqrt(x+b))`
Integrate the function:
`x^3/(sqrt(1-x^8)`
Integrate the function:
`cos^3 xe^(log sinx)`
Integrate the function:
f' (ax + b) [f (ax + b)]n
Integrate the functions `(sin^(-1) sqrtx - cos^(-1) sqrtx)/ (sin^(-1) sqrtx + cos^(-1) sqrtx) , x in [0,1]`
Integrate the function:
`(2+ sin 2x)/(1+ cos 2x) e^x`
Integrate the function:
`(x^2 + x + 1)/((x + 1)^2 (x + 2))`
Evaluate `int(x^3+5x^2 + 4x + 1)/x^2 dx`
Find : \[\int\frac{\left( x^2 + 1 \right)\left( x^2 + 4 \right)}{\left( x^2 + 3 \right)\left( x^2 - 5 \right)}dx\] .
Evaluate: `int (1 - cos x)/(cos x(1 + cos x)) dx`
The anti derivative of `(sqrt(x) + 1/sqrt(x))` is equals:
`sqrt((10x^9 + 10^x log e^10)/(x^10 + 10^x)) dx` equals
`int (dx)/sqrt(9x - 4x^2)` equals
`int (dx)/(x(x^2 + 1))` equals
What is integration called because it reverses the operation of finding derivatives?
In \[\int f(x)\,dx\], what is \(x\) called?
What is integration?
Which statement about indefinite integrals is correct?
Evaluate \[\int \cos x\,dx\]
Evaluate \[\int e^x\,dx\]
Evaluate \[\int \frac{1}{x}\,dx\]
Evaluate \[\int a^x\,dx\]
