Advertisements
Advertisements
प्रश्न
Find: `intsqrt(1 - sin 2x) dx, pi/4 < x < pi/2`
Advertisements
उत्तर
`I = int sqrt(1 - sin 2x) dx`
`I = int sqrt(sin^2 x + cos^2 x - 2sin x cos x) dx`
`I = int (sin x - cos x )dx`
`I = int sin x dx - int cos x dx`
`I = -cos x - sin x + "C"`
APPEARS IN
संबंधित प्रश्न
Evaluate :`int_(pi/6)^(pi/3) dx/(1+sqrtcotx)`
Evaluate : `intsin(x-a)/sin(x+a)dx`
Find the integrals of the function:
sin2 (2x + 5)
Find the integrals of the function:
sin 3x cos 4x
Find the integrals of the function:
cos 2x cos 4x cos 6x
Find the integrals of the function:
sin3 (2x + 1)
Find the integrals of the function:
sin3 x cos3 x
Find the integrals of the function:
sin x sin 2x sin 3x
Find the integrals of the function:
`cos x/(1 + cos x)`
Find the integrals of the function:
`(cos 2x - cos 2 alpha)/(cos x - cos alpha)`
Find the integrals of the function:
`1/(sin xcos^3 x)`
Find the integrals of the function:
`(cos 2x)/(cos x + sin x)^2`
Find the integrals of the function:
`1/(cos(x - a) cos(x - b))`
Evaluate `int_0^(3/2) |x sin pix|dx`
Differentiate : \[\tan^{- 1} \left( \frac{1 + \cos x}{\sin x} \right)\] with respect to x .
Find: `int_ (cos"x")/((1 + sin "x") (2+ sin"x")) "dx"`
Find `int x^2tan^-1x"d"x`
`int "e"^x (cosx - sinx)"d"x` is equal to ______.
`int "dx"/(sin^2x cos^2x)` is equal to ______.
Evaluate the following:
`int sqrt(1 + sinx)"d"x`
Evaluate the following:
`int (sin^6x + cos^6x)/(sin^2x cos^2x) "d"x`
Evaluate the following:
`int (cosx - cos2x)/(1 - cosx) "d"x`
Evaluate the following:
`int "e"^(tan^-1x) ((1 + x + x^2)/(1 + x^2)) "d"x`
Evaluate the following:
`int sin^-1 sqrt(x/("a" + x)) "d"x` (Hint: Put x = a tan2θ)
`int (x + sinx)/(1 + cosx) "d"x` is equal to ______.
`int sinx/(3 + 4cos^2x) "d"x` = ______.
