Consider the integral `int_0^(2pi)(dx)/(5-2cosx)` making the substitution `tan(x/2)=t ,` we have `I=int_0^(2pi)(dx)/(5-2cosx)` `=int_0^0(2dt)/((1+t^2)[5-2(1-t^2)/(1+t^2)])=0` The result is obviously wrong, since the integrand is positive and consequently the integral of this function cannot be equal to zero. Find the mistake.

Consider the integral I=int_(0)^(2pi)(dx)/(5-2cosx) Making the substitution "tan"1/2x=t , we have I=int_(0)^(2pi)(dx)/(5-2cosx)=int_(0)^(0)(2dt)/((1+t^(2))[5-2(1-t^(2))//(1+t^(2))])=0 The result is obviously wrong, since the integrand is positive and consequently the integral of this function cannot be equal to zero. Find the mistake.

Consider the integral I=int_(0)^(2pi)(dx)/(5-2cosx) Making the substitution "tan"1/2x=t , we have I=int_(0)^(2pi)(dx)/(5-2cosx)=int_(0)^(0)(2dt)/((1+t^(2))[5-2(1-t^(2))//(1+t^(2))])=0 The result is obviously wrong, since the integrand is positive and consequently the integral of this function cannot be equal to zero. Find the mistake.

Consider the integral int_(0)^(2 pi)(dx)/(5-2cos x) making the substitution tan((x)/(2))=t, we have I=int_(0)^(2 pi)(dx)/(5-2cos x)=int_(0)^(0)(2dt)/((1+t^(2))[5-2(1-t^(2))/(1+t^(2))])=0 The result functive and consequently the integral of this functive and consequently the integral of this functake.

Evaluate the integral int_(0)^(pi//2) (dx)/(4+5cosx )

Find the Definite Integrals : int_0^(pi/2) dx /(2 + cosx)

Evaluate the following: int_0^(pi/2)dx/(1+cosx)

Evaluate the integral int_(0)^(2pi) (1+cosx)^(5) (1-cosx)^(3) dx

Evaluate: int_0^(pi//2)1/(3+2cosx)dx

Evaluate: int_0^(pi//2)cosx\ dx