`I=int_1^(2)(x^(2)+sin x+9)dx`

int_ (0) ^ (1) sin (x ^ (2) + 2x + 1) dx-int_ (1) ^ (2) sin x ^ (2) dx is

If I_(1)=int_(0)^(pi//2)"x.sin x dx" and I_(2)=int_(0)^(pi//2)"x.cos x dx" , then

int(1-sin^(2)x)/(sin^(2)x)dx

int1/(3+2 sin x)dx

If I_(1)=int_(0)^( pi/2)(sin x)^(sqrt(3)+1)dx,I_(2)=int_(0)^(pi/2)(sin x)^(sqrt(3)-1))dx then (I_(1))/(I_(2))=

I = int_ (1) ^ (2) | x sin pi x | dx

If I_(1)=int_(1-x)^(k) x sin{x(1-x)}dx and I_(2)=int_(1-x)^(k) sin{x(1-x)}dx , then

If (a,b) be the orthocentre of the triangle whose vertices are (1,2),(2,3) and (3,1) ,and I_(1)=int_(a)^(b)x sin(4x-x^(2))dx,I_(2)=int_(a)^(b)sin(4x-x^(2))dx ,then 36(I_(1))/(I_(2)) is equal to:

If int _(a )^(b) |sin x |dx =8 and int _(0)^(a+b) |cos x| dx=9 then the value of (1)/(sqrt2pi) |int _(a)^(b) x sin x dx | is: