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int0^(pi/6)cos^4 3thetasin^2 6theta dthe...

`int_0^(pi/6)cos^4 3thetasin^2 6theta dtheta`

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Evaluate the following integrals: int_0^(pi//6)cos^(-3)2thetasin2thetad theta

Evaluate the following integrals: int_(0)^( pi/6)cos^(-3)2 theta sin2 theta d theta

int_0^(pi/8) cos^(3) 4 theta d theta =

int_(0)^(pi//8) cos^(3)4 theta d theta=

int_(0)^(pi//8) cos^(3)4 theta d theta=

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

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

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

If int_0^a(dx)/(sqrt(x+a)+sqrt(x))=int_0^(pi/8)(2tantheta)/(sin2theta)dtheta, then value of ' a ' is a equal to (a >0) 3/4 (b) pi/4 (c) (3pi)/4 (d) 9/(16)