Advertisements
Advertisements
प्रश्न
`int (sin^6x)/(cos^8x) "d"x` = ______.
Advertisements
उत्तर
`int (sin^6x)/(cos^8x) "d"x` = `(tan^7x)/7 + "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:
cos 2x cos 4x cos 6x
Find the integrals of the function:
sin x sin 2x sin 3x
Find the integrals of the function:
cos4 2x
Find the integrals of the function:
`(sin^2 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:
tan3 2x sec 2x
Find the integrals of the function:
`(sin^3 x + cos^3 x)/(sin^2x cos^2 x)`
Find the integrals of the function:
`(cos 2x+ 2sin^2x)/(cos^2 x)`
Find the integrals of the function:
`(cos 2x)/(cos x + sin x)^2`
Find the integrals of the function:
sin−1 (cos x)
Find the integrals of the function:
`1/(cos(x - a) cos(x - b))`
Find `int (sin^2 x - cos^2x)/(sin x cos x) dx`
Evaluate: `int_0^π (x sin x)/(1 + cos^2x) dx`.
Evaluate `int_0^(3/2) |x sin pix|dx`
Find `int (2x)/((x^2 + 1)(x^4 + 4))`dx
Evaluate : \[\int\limits_0^\pi \frac{x \tan x}{\sec x \cdot cosec x}dx\] .
Find `int_ (sin2"x")/((sin^2 "x"+1)(sin^2"x"+3))d"x"`
Find:
`int"dx"/sqrt(5-4"x" - 2"x"^2)`
Integrate the function `cos("x + a")/sin("x + b")` w.r.t. x.
Find: `intsqrt(1 - sin 2x) dx, pi/4 < x < pi/2`
Find: `int sin^-1 (2x) dx.`
Find `int "dx"/(2sin^2x + 5cos^2x)`
Find `int x^2tan^-1x"d"x`
`int "dx"/(sin^2x cos^2x)` is equal to ______.
Evaluate the following:
`int ("d"x)/(1 + cos x)`
Evaluate the following:
`int (sin^6x + cos^6x)/(sin^2x cos^2x) "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` = ______.
The value of the integral `int_(1/3)^1 (x - x^3)^(1/3)/x^4 dx` is
