`Iflambda=int_0^(pi/2)costhetaf(sintheta+cos^2theta)dtheta` `a n dI_2=int_0^(pi//2)sin2thetaf(sintheta+cos^2theta)dtheta,t h e n` `I_1=-2I_2` (b) `I_1=I_2` `2I_1=I_2` (d) `I_1=-I_2`

I=int_(0)^( pi/4)(sin2 theta)/(sin^(2)theta+cos^(4)theta)d theta

Let the be an integrable function defined on [0,a] if I_(1)=int_(pi//2)^(pi//2) cos theta f(sin theta cos ^(2) theta)d theta and I_(2)=int_(0)^(pi//2)sin 2 theta f(sin theta cos ^(2) theta)d theta then

The value of I(n)=int_(0)^( pi)(sin^(2)n theta)/(sin^(2)theta)d theta is (AA n in N)

Integrate the following I=int_0^((pi)/2) d theta

If I_(n)=int_(0)^(2 pi)(cos(n theta))/(cos theta)d(theta), where n in N then.

if I=int_(0)^(pi//2)(3sqrt(costheta))/((sqrt(sintheta)+sqrt(costheta))^(5))d theta , then I^(2) is equal to

If(cos theta+i sin theta)(cos2 theta+i sin2 theta).....(cos n theta+i sin n theta)=1 then the value of theta is:

If I_n=int_0^(pi/2) (sin(2n-1)x)/sinx)dx , and a_n=int_0^(pi/2) ((sinntheta)/sintheta)^2 d theta , then a_(n+1)-a_n= (A) I_n (B) 2I_n (C) I_(n+1) (D) 0

Let f(x) be an integrable function defined on [a,b], b gt a gt 0 . If I_(1)=int_(pi//6)^(pi//3) f(tan theta+cos theta)sec^(2) theta d theta and, I_(2)=int_(pi//6)^(pi//3) f(tan theta +cot theta)cosec^(2) theta d theta , then (I_(1))/(I_(2))=