Thread:

1. Re-substituion problem

this is part of an integration, thats why i posted it here.
the needed substitution here is $u=sin(x)$

i end up with this sometime, which is correct:

$2\sqrt{sin(x)}-\frac{2*\sqrt{sin(x^5)}}{5}+C$

now my book says the result should be:

$\frac{2}{5}\sqrt{sin(x)}[4+cos^2(x)]+C$

and wofram alpha says it should be:

$\frac{1}{5}\sqrt{sin(x)}[cos(2x)+9]+C$

to be honest... i tried and i tried, i dont get it how i will get any of those 2 solutions. would be so nice if someone could explain it me reeeeal slow

thx in advance

2. I think there's a typo in there. Did you mean $\sin^5( x)$ instead of $\sin (x^5)$?

In that case:

$2\sqrt{\sin x}-\frac{2\sqrt{\sin^5 x}}{5}+C=2\cdot \frac{5}{5}\sqrt{\sin x}-\frac{2\sqrt{\sin x}\sqrt{\sin^4 x}}{5}+C=$ $\frac{2\sqrt{\sin x}}{5}(5-\sqrt{\sin^4 x})+C=\frac{2\sqrt{\sin x}}{5}(5-\sin^2 x)+C=$ $\frac{2\sqrt{\sin x}}{5}(4 + (1-\sin^2 x))+C=$ $\frac{2\sqrt{\sin x}}{5}(4+\cos^2 x)+C$

3. oh yes that was a typo... thank you very much, all clear now!

need to improve my algebra skills

4. Originally Posted by coobe
this is part of an integration, thats why i posted it here.
the needed substitution here is $u=sin(x)$

i end up with this sometime, which is correct:

$2\sqrt{sin(x)}-\frac{2*\sqrt{sin(x^5)}}{5}+C$ ${\color{red} <font color=$2\sqrt{sin(x)}-\frac{2*\sqrt{(sin(x))^5}}{5}+CMr F says: You mean }" alt="{\color{red} 2\sqrt{sin(x)}-\frac{2*\sqrt{(sin(x))^5}}{5}+CMr F says: You mean }" />.

now my book says the result should be:

$\frac{2}{5}\sqrt{sin(x)}[4+cos^2(x)]+C$

and wofram alpha says it should be:

$\frac{1}{5}\sqrt{sin(x)}[cos(2x)+9]+C$

to be honest... i tried and i tried, i dont get it how i will get any of those 2 solutions. would be so nice if someone could explain it me reeeeal slow

thx in advance
$2\sqrt{\sin(x)}-\frac{2 \sqrt{(\sin(x))^5}}{5}+C = 2\sqrt{\sin(x)}-\frac{2 \sqrt{(\sin(x))^4 \sin x}}{5}+C$

$= 2\sqrt{\sin(x)}-\frac{2 \sin^2 x\sqrt{\sin x}}{5}+C$.

Now take out a factor of $\frac{2 \sqrt{\sin x}}{5}$ and substitute $\sin^2 x = 1 - \cos^2 x$ to get your book's answer.

To show that Wolfram's answer is equivalent to the book's answer you need to recall that $\cos (2x) = 2 \cos^2 (x) - 1$ ....
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