Question

Evaluate the integral with the aid of an appropriate u-sub:

Integral sign { cos^19 (x)}^1/2 sinx dx

Answer 1

tailor made for u-sub.
put \[u=cos(x)\]
\[du=-sin(x)dx\]
\[-\int u^{\frac{19}{2}}du\]

Answer 2

I see what I did wrong, I messed up the power. The whole cos^19 (x) messed me up. Thanks for your help, I appreciate it. : )

Answer 3

You should get \[-\int\limits_{}^{}\sin ^{1/2}{(u ^{19})} du\]
From here you may have to jump into some trig identity im not sure

Answer 4

actually there is no \[sin^{\frac{1}{2}}\] in it. just take the anti derivative of \[u^{\frac{9}{2}}\] which is \[\frac{2}{11} u^{\frac{11}{9}}\] and then replace u by cos(x) don't forget the "-" sign in front like i did.

Answer 5

oops i meant the antiderivative is \[-\frac{2}{11}u^{\frac{11}{2}}\]

Answer 6

the problem they wrote there is \[\int\limits_{}^{}\sin ^{1/2}[\cos ^{19}(x)]*\sin(x)*dx\]