Question

solve the integration​

Answer 1

Answer:

The function is given as,

y=∫π20√sinx√sinx+√cosxdxy = ∫0π2sinxsinx+cosxdx11

From the property of integration,

∫b0f(x)dx=∫b0f(b−x)dx∫0bf(x)dx=∫0bf(b−x)dx

y=∫π20√sin(π2−x)√sin(π2−x)+√cos(π2−x)dxy=∫π20√cos(x)√cos(x)+√sin(x)dxy= ∫0π2sin(π2−x)sin(π2−x)+cos(π2−x)dxy=∫0π2cos(x)cos(x)+sin(x)dx22

We have to add equations 11 and 22 to get the solution.

y+y=∫π20√sin(x)√sin(x)+√cos(x)dx+∫π20√cos(x)√cos(x)+√sin(x)dx2y=∫π20√sin(x)+