`(cosx)/((1-sinx)(2-sinx))` [Hint : Put sin x = t]

Evaluate int(cosx)/((1-sinx)(2+sinx))dx

The value of the integral int(cos^3x+cos^5x)/(sin^2x+sin^4x)dxi s (a). sinx-6tan^(-1)(sinx)+C (b). sinx-2(sinx)^(-1)+C (c). sinx-2(sinx)^(-1)-6tan^(-1)(sinx)+C (d). sinx-2(sinx)^(-1)+5tan^(-1)(sinx)+C

Let f(x) = |(2cos^2x, sin2x, -sinx), (sin2x, 2 sin^2x, cosx), (sinx, -cosx,0)| , then the value of int_0^(pi//2){f(x) + f'(x)} dx is

The function f(x)=cos^(-1)((2[|sinx|+|cosx|])/(sin^2x+2sinx+11/4)) is defined if x belongs to (where [.] represents the greatest integer function)

If A = ((cosx,sinx),(-sinx, cosx)) and A (adj A) = lambda ((1,0),(0,1)) then lambda is

Find int(cos2x)/((cosx+sinx)^(2))dx .

If y= (sinx + cosx)/(sinx - cosx) then (dy)/(dx) at x = pi/2 is …………………….. .

Maximum value of function f(x)=(sin^(-1)(sinx)^(2)-sin^(-1)(sinx) is:

A function f is continuous for all x (and not everywhere zero) such that f^2(x)=int_0^xf(t)(cost)/(2+sint)dtdot Then f(x) is 1/2 1n((x+cosx)/2); x!=0 1/2 1n(3/(x+cosx)); x!=0 1/2 1n((2+sinx)/2); x!=npi,n in I (cosx+sinx)/(2+sinx); x!=npi+(3pi)/4,n in I

If y=(sinx)^(tanx),t h e n(dy)/(dx)= (a) (sinx)^(tanx)(1+sec^2xlogsinx) (b) tanx(sinx)^(tanx-1)cosx (c) (sinx)^(tanx) (d) sec^2xlogsinx tanx(sinx)^(tanx-1)