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

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

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

cos m theta-sin n theta=0

If f (n)= 1/pi int _(0) ^(pi//2) (sin ^(2) (n theta) d theta)/(sin ^(2) theta), n in N, then evulate (f (15)+ f (3))/( f (12) -f (10)).

y=sin^(2)theta+cos ec^(2)theta,theta!=n pi n in z then

Solve int((sin theta+sqrt(cos^(2)theta+2sin^(2)theta))^(n))/(sqrt(1+sin^(2)theta))cos theta d theta , where n ne -1 .

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

If I_(n) = int _(0)^(pi//4) tan^(n) theta " " d theta, "then "I_(8)+I_(6)=

If u_(n)=int_(0)^((pi)/(2))theta sin^(n)theta d theta and n>=1, then prove that u_(n)=((n-1)/(n))u_(n-2)+(1)/(n^(2))