Advertisements
Advertisements
Question
Find the integrals of the function:
sin−1 (cos x)
Advertisements
Solution
Let `I - int sin^-1 (cos x) dx`
`int sin^-1 [sin (pi/2 - x)] dx`
`= int (pi/2 - x) dx`
`= pi/2 int dx - int x dx`
`= (pix)/2 - x^2/2 +C`
APPEARS IN
RELATED QUESTIONS
Find the integrals of the function:
sin2 (2x + 5)
Find the integrals of the function:
sin3 (2x + 1)
Find the integrals of the function:
`cos x/(1 + cos x)`
Find the integrals of the function:
cos4 2x
Find the integrals of the function:
`(cos 2x - cos 2 alpha)/(cos x - cos alpha)`
Find the integrals of the function:
`(cos x - sinx)/(1+sin 2x)`
Find the integrals of the function:
`(cos 2x+ 2sin^2x)/(cos^2 x)`
Find the integrals of the function:
`1/(sin xcos^3 x)`
Find the integrals of the function:
`(cos 2x)/(cos x + sin x)^2`
`int (e^x(1 +x))/cos^2(e^x x) dx` equals ______.
Find `int dx/(x^2 + 4x + 8)`
Evaluate: `int_0^π (x sin x)/(1 + cos^2x) dx`.
Find `int((3 sin x - 2) cos x)/(13 - cos^2 x- 7 sin x) 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"`
Integrate the function `cos("x + a")/sin("x + b")` w.r.t. x.
Find: `int sec^2 x /sqrt(tan^2 x+4) dx.`
Find: `intsqrt(1 - sin 2x) dx, pi/4 < x < pi/2`
Evaluate `int tan^8 x sec^4 x"d"x`
Find `int "dx"/(2sin^2x + 5cos^2x)`
Find `int x^2tan^-1x"d"x`
`int (sin^6x)/(cos^8x) "d"x` = ______.
Evaluate the following:
`int ((1 + cosx))/(x + sinx) "d"x`
Evaluate the following:
`int ("d"x)/(1 + cos x)`
Evaluate the following:
`int tan^2x sec^4 x"d"x`
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 "e"^(tan^-1x) ((1 + x + x^2)/(1 + x^2)) "d"x`
`int (x + sinx)/(1 + cosx) "d"x` is equal to ______.
The value of the integral `int_(1/3)^1 (x - x^3)^(1/3)/x^4 dx` is
`int (cos^2x)/(sin x + cos x)^2 dx` is equal to
