`int(cos2x-1)/(cos2x+1)\ dx=`
(a)`tanx-x+C`
(b) `x+tanx+C`
(c) `x-tanx+C`
(d) ` x-cotx+C`

int(1+cos4x)/(cotx-tanx)\ dx

int(1+cos4x)/(cotx-tanx)dx

int1/(cos^2x(1-tanx)^2)\ dx

int1/(cos^2x(1-tanx)^2)\ dx

int(cos4x-1)/(cotx-tanx)dx is equal to

int(cos 4x-1)/(cot x-tanx)dx is equal to

int(2tanx+3)/(sin^2x+2cos^2x)dx

int(dx)/(sin^2xcos^2x) equals(A) tanx+cotx+C (B) tanx-cotx+C (C) tanxcotx+C (D) tanx-cot2x+C

Integral of the form int[xf\'(x)+f(x)]dx can be evaluated by using integration by parts in first integral and leaving second integral as it is.Now answer the question: int(x+sinx)/(1+cosx)dx= (A) xtanx/2+c (B) x/2tanx/2+c (C) x/2tanx+c (D) none of these

int\ (sin^6x + cos^6x)/(sin^2 x*cos^2x)\ dx (i) tanx+ cotx+3x+C (ii) tanx+ cotx-3x+C (iii) tanx-cotx-3x+C (iv) tanx-cotx+3x+C