Advertisements
Advertisements
प्रश्न
Integrate the following with respect to x :
sin5x cos3x
Advertisements
उत्तर
`int sin^5x cos^3x "d"x = int sin^5x cos^2x * cos x "d"x`
= `int sin^5x (1 - sin^2x) * cos x "d"x`
= `int (sin^5x - sin^7x) * cos x "d"x`
= `int sin^5x cos x "d'x - nt sin^7x cos x "d"x`
Put u = sin x
du = cos x dx
`int sin^5x cos^3x "d"x = int "u"^5 "du" - int"u"^7 "du"`
= `"u"^6/6 - "u"^8/8 + "c"`
`int sin^5x cos^3x "d"x = (sinx)^6/6 - (sinx)^8/8 + "c"`
APPEARS IN
संबंधित प्रश्न
Evaluate : `∫_0^(pi/2) (sinx.cosx)/(1 + sin^4x)`.dx
Evaluate : `int _0^1 ("x" . ("sin"^-1 "x")^2)/sqrt (1 - "x"^2)` dx
Integrate the following functions with respect to x :
`(sin4x)/sinx`
Integrate the following functions with respect to x :
`(3x + 4) sqrt(3x + 7)`
Integrate the following functions with respect to x :
`(3x - 9)/((x - 1)(x + 2)(x^2 + 1))`
Integrate the following with respect to x :
`(sin^-1 x)/sqrt(1 - x^2)`
Integrate the following with respect to x :
`1/(x log x log (log x))`
Integrate the following with respect to x:
27x2e3x
Integrate the following with respect to x:
`tan^-1 ((8x)/(1 - 16x^2))`
Integrate the following with respect to x:
`"e"^(-x) cos 2x`
Integrate the following with respect to x:\
`logx/(1 + log)^2`
Find the integrals of the following:
`1/(6x - 7 - x^2)`
Find the integrals of the following:
`1/sqrt((2 + x)^2 - 1)`
Integrate the following with respect to x:
`(2x + 3)/sqrt(x^2 + 4x + 1)`
Integrate the following functions with respect to x:
`sqrt(x^2 - 2x - 3)`
Choose the correct alternative:
If `int 3^(1/x)/x^2 "d"x = "k"(3^(1/x)) + "c"`, then the value of k is
Choose the correct alternative:
`int ("e"^x(x^2 tan^-1x + tan^-1x + 1))/(x^2 + 1) "d"x` is
Choose the correct alternative:
`int x^2 cos x "d"x` is
Choose the correct alternative:
`int "e"^(sqrt(x)) "d"x` is
