Question #89694

Question

1520 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-14T00:46:31

#int cosxsqrt(1+cos2x)dx =(x +sinxcosx)/sqrt2 + C#

Use the trigonometric identity:

#cos^2x = (1+cos2x)/2#

so:

#cosxsqrt(1+cos2x) = cosx sqrt(2cos^2x) =sqrt2cos^2x = (1+cos2x)/sqrt2#

and:

#int cosxsqrt(1+cos2x)dx = int (1+cos2x)/sqrt2dx#

#int cosxsqrt(1+cos2x)dx = 1/sqrt2 int dx +1/(2sqrt2) int cos2xd(2x)#

#int cosxsqrt(1+cos2x)dx = x/sqrt2 +(sin2x)/(2sqrt2) +C#

#int cosxsqrt(1+cos2x)dx =(x +sinxcosx)/sqrt2 + C#