Integrating ugly trig combination

[JR63]

Consider the integral 2Sin^3(x) Cos (x) (1-Sin^2(x))^18 dx.
1.Evaluate this integral by hand (with the appropriate substitution, the integral is asy to evaluate). Then, enter an antiderivative into Mathematica as my[x].

I started with let u=Sin(x), du=Cos(x)dx,
-> integral 2 u^3 (1-u^2)^18 du

I put it in a calculus solved program, and their solution gave me more confussion;
they used sin^2+Cos^2=1 indentity, then had Cos^19, then split it back up into
-> integral Sin^3 (x) Cos (x) (1-Sin^2 (x))^9dx -> then u-sub
-> integral u^3(1-u^2)^9 du

Please help, "mathematica" is bad enough without having these problems too! :?
It's causing me to pull my hair out.

John

 

[Deleted member 4993]

Consider the integral 2Sin^3(x) Cos (x) (1-Sin^2(x))^18 dx.
1.Evaluate this integral by hand (with the appropriate substitution, the integral is asy to evaluate). Then, enter an antiderivative into Mathematica as my[x].

I started with let u=Sin(x), du=Cos(x)dx,
-> integral 2 u^3 (1-u^2)^18 du

I put it in a calculus solved program, and their solution gave me more confussion;
they used sin^2+Cos^2=1 indentity, then had Cos^19, then split it back up into
-> integral Sin^3 (x) Cos (x) (1-Sin^2 (x))^9dx -> then u-sub
-> integral u^3(1-u^2)^9 du

Please help, "mathematica" is bad enough without having these problems too! :?
It's causing me to pull my hair out.

John

2Sin[sup:9h9lqcx6]3[/sup:9h9lqcx6](x) Cos (x) (1-Sin[sup:9h9lqcx6]2[/sup:9h9lqcx6](x))[sup:9h9lqcx6]18[/sup:9h9lqcx6] dx

= 2(1-cos[sup:9h9lqcx6]2[/sup:9h9lqcx6]x) sin(x) cos(x) cos[sup:9h9lqcx6]36[/sup:9h9lqcx6]x dx

=-2(1-u[sup:9h9lqcx6]2[/sup:9h9lqcx6]) u[sup:9h9lqcx6]37[/sup:9h9lqcx6] du

it is easy from here...

 

[pka]

Go to this website: W|A http://www.wolframalpha.com/In the input box type this exactly “integrate 2 u^3 (1-u^2)^18 du” without the quotes.
Click the equals icon at the right-hand side of the input box.
That will give you the answer. You can even ask to see the steps.

Try putting the original question in.

 

[galactus]

\(\displaystyle 2\int sin^{3}(x)cos(x)(1-sin^{2}(x))^{18}dx\)

If we make the sub \(\displaystyle u=sin^{2}(x), \;\ du=2sin(x)cos(x)dx\)

Then, we have:

\(\displaystyle \int u(1-u)^{18}du\)

Expand using a binomial expansion and do it term by term would be one way to go. It is long, but the terms are easy to integrate.

 

[BigGlenntheHeavy]

\(\displaystyle \int2sin^{3}(x)cos^{37}(x)dx \ = \ \frac{-(19sin^{2}(x)+1)(cos^{38}(x))}{380} \ +C\)

\(\displaystyle Now \ \int2sin^{3}(x)cos^{37}(x)dx \ = \ \int2sin^{2}(x)sin(x)cos(x)cos^{36}(x)dx\)

\(\displaystyle \int2sin^{2}(x)sin(x)cos(x)cos^{36}(x)dx \ = \ \int2sin^{2}(x)sin(x)cos(x)[1-sin^{2}(x)]^{18}dx\)

\(\displaystyle Let \ u \ = \ 1-sin^{2}(x), \ then \ -du \ = \ 2sin(x)cos(x)dx, \ sin^{2}(x) \ = \ 1-u.\)

\(\displaystyle Ergo; \ -\int(1-u)u^{18}du \ = \ \int(u^{19}-u^{18})du \ = \ \frac{u^{20}}{20}-\frac{u^{19}}{19} \ + \ C\)

\(\displaystyle Therefore: \ \frac{[1-sin^{2}(x)]^{20}}{20}-\frac{[1-sin^{2}(x)]^{19}}{19} \ + \ C\)

\(\displaystyle This \ equals \ \frac{19[1-sin^{2}(x)]^{20}-20[1-sin^{2}(x)]^{19}}{380} \ + \ C\)

\(\displaystyle Which \ in \ turn \ equals \ \frac{[1-sin^{2}(x)]^{19}[19(1-sin^{2}(x))-20]}{380} \ + \ C\)


\(\displaystyle And \ a \ little \ more \ algebra \ yields: \ -\frac{(19sin^{2}(x)+1)(cos^{38}(x))}{380} \ + \ C. \ QED\)

\(\displaystyle Note: \ If \ you \ let \ u \ = \ sin(x), \ then \ du \ = \ cos(x)dx \ and \ you \ get \ 2 \ \int[u^{3}(1-u^{2})^{18}]du\)

Now if you are going to find the antiderivative of that indefinite integral, I hope you have a couple of reams of paper and a lot of time on your hands.

 

[pka]

This is from W|A.

 

[galactus]

the integral is easy to evaluate

)

That's a good one

 

[Deleted member 4993]

This is from W|A.

I am sure it will boil down to:

=1/20 * cos[sup:2hjiparl]40[/sup:2hjiparl](x) - 1/19 * cos[sup:2hjiparl]38[/sup:2hjiparl](x) + C