Advertisements
Advertisements
प्रश्न
Integrate the following with respect to x.
`sqrt(1 - sin 2x)`
Advertisements
उत्तर
`sqrt(1 - sin 2x) = sqrt(1 - 2sinx cos x)`
= `sqrt(sin^2x + cos^2x - 2sinx cosx)`
= `sqrt((cosx - sin x)^2`
= `cos x - sin x`
So `int sqrt(1 - sin 2x) "d"x = int cos x "d"x - int sin x + "c"`
= `sin x - (- cos x) + "c"`
= `sin x+ cosx + "c"`
`sqrt(sin^2x + cos^2x - 2sinx cosx) = sqrt((sinx - cosx)^2)`
= `sin x - cos x`
So `int sqrt(1 - sin^2x) "d"x = int sin x "d"x - int cos x "d"x + "c"`
= `- cos x - sinx + "c"`
= `- (cos x + sin x) + "c"`
APPEARS IN
संबंधित प्रश्न
Integrate the following with respect to x.
If f'(x) = 8x3 – 2x and f(2) = 8, then find f(x)
Integrate the following with respect to x.
`(x^4 - x^2 + 2)/(x - 1)`
Integrate the following with respect to x.
`(3x + 2)/((x - 2)(x - 3))`
Integrate the following with respect to x.
`"e"^(xlog"a") + "e"^("a"log"a") - "e"^("n"logx)`
Integrate the following with respect to x.
`[1 - 1/2]"e"^((x + 1/x))`
Integrate the following with respect to x.
2 cos x – 3 sin x + 4 sec2x – 5 cosec2x
Integrate the following with respect to x
`1/(x log x)`
Choose the correct alternative:
`int_2^4 ("d"x)/x` is
Choose the correct alternative:
`int_0^4 (sqrt(x) + 1/sqrt(x)) "d"x` is
Evaluate the following integral:
`int_0^1 sqrt(x(x - 1)) "d"x`
