INTEGRAL CHALLENGE TO TIITIANS

Just apply the trigonometrical identity ,

sin x = 2 tan ( x / 2 )1 + tan ² ( x / 2 )

And then the universal subsitution , tan ( x / 2 ) = z

I = 2 ∫ ( z ² + 1 ) dzz ⁴ + 2 z ³ + 6 z ² + 1

Now , it becomes a rational function in " z " which can always be integrated .

nice work , but seems to be lengthy

another trick is splitting
1+\sin x+\sin^2x=(1-\omega \sin x)(1-\omega^2 \sin x)

this trick was applied by bhatt sir in another forum

and now go for partial fractions ;)

another one :
find :

Err , are you sure about the question ? The answer comes in terms of Gauss ' s confluent hypergeometric series . The integral as such could be evaluated if n is given , but a general case doesn't give a nice answer as far as I know .

Basically that ' s exactly why I said that this integral cannot be generalised . You see , that for a fixed n , ( in the link given , n = 16 ) , we easily can evaluate the integral in spite of calculations . But as such you cannot " integrate " the function .