We can derive different reduction formulas by using integration by parts.
If `int tan^(6)x dx=(1)/(5) tan^(5)x+A tan^(3)x+tanx+c`, then A is equal to-

We can derive different reduction formulas by using integration by parts. If int sec^(6)x dx=(1)/(5) tan^(3)x+A tan^(3)x+tanx+c , then A is equal to-

We can derive different reduction formulas by using integration by parts. If int cosec^(n)x dx=(-cosec^(n-2)x cotx)/(n-1)+A int cosec^(n-2)x dx , then A is equal to-

Method of integration by parts : int tan^(-1)x dx=....

int tan 2x tan 3 x tan 5x dx

Reduction formulas can be used to compute integrals of higher power of sinx,cosx,tanx etc. inttan^(6)xdx=(1)/(5)tan^(5)x+Atan^(3)x+tanx-x+C then

Reduction formulas can be used to compute integrals of higher power of sinx,cosx,tanx etc. intsec^(6)xdx=(1)/(5)tan^(5)x+Atan^(3)x+tanx+C then

Method of integration by parts : int (x tan^(-1)x)/((1+x^(2))^((3)/(2)))dx=.....

int (tanx sec^(2)x)/((1-tan^(2)x))dx.

int[1+tanx.tan(x+alpha)]dx is equal to