Let f(x)=cosxsqrt(1+sinx).
A. Let F(x)=the integral of f(x)dx and F(0)=5/3, find F(pi/2).
I got 3.386 with U-substitution
B. If G(x)=the integral of f(x)sinxdx and G(0)=1/3, find G(x).
I got a reply to this but I did not understand the explanation. How would I solve this with U-substitution.
2 answers
Let G=integral cosxsinxsqrt(1+sinx)dx let u=1+sinx du/dx=cosx dx=du/cosx substituting dx cancel cos x in the integral so It become:integral(u-1)sqrtudu(since sinx=u-1) therefore open bracket gives:u^1/2(u-1)du=u^3/2-u^1/2du integrate each gives:2/5u^5/2-2/3u^3/2 substitute u gives 2/5(1+sinx)^5/2-2/3(1+sinx)^3/2+c
Sorry about my half-angle formula hint. Got me nowhere. Instead, for F(x), let
u^2 = 1+sinx
2u du = cosx dx
and now you have
∫2u^2 du = 2/3 u^3 = 2/3 (1+sinx)^(3/2) + c
Now you can find c, since you know F(0)
u^2 = 1+sinx
2u du = cosx dx
and now you have
∫2u^2 du = 2/3 u^3 = 2/3 (1+sinx)^(3/2) + c
Now you can find c, since you know F(0)
2 answers