Question

integration root over 1 + sin 2x DX barabar what​

Answer 1

Solution :

Let, f(x) = $\sqrt{1+sin2x}$

= $\sqrt{sin^{2}x+cos^{2}x+2sinxcosx}$

= $\sqrt{sin^{2}x+2sinxcosx+cos^{2}x}$

= $\sqrt{(sinx+cosx)^{2}}$

= sinx + cosx,

where sin2x = 2 sinx cosx

Now, $\int f(x)dx$

= $\int (sinx+cosx)dx$

= $\int sinxdx + \int cosxdx$

= - cosx + sinx + C,

where C = constant of integration

Thus the required integral is

= - cosx + sinx + C