Question 344817
I'm assuming that the problem is:
{{{ int (sin (sqrt ( 2u ) ) / sqrt( 2u ), du) }}}
If not, then repost your question.<br>
With a good knowledge of derivatives and integration formulas and with some practice you will develop an "eye" for substitutions which can be made to make an integral easy to solve. In this case I see {{{sqrt(2u)}}} whose derivative is {{{1/sqrt(2u)}}} which is present in your integrand! And since sin is an easy integral to solve the problem becomes very easy.<br>
First we'll manipulate the integrand Algebraically:
{{{ int (sin (sqrt ( 2u ) )*(1/sqrt( 2u )), du) }}}
Now we start our substitution.
Let {{{v = sqrt(2u) = (2u)^(1/2)}}}
Then {{{dv/du = (1/2)*2u^(1/2-1)*2 = (2u)^(-1/2) = 1/(2u)^(1/2) = 1/sqrt(2u)}}}
and, in differential form:
{{{dv = (1/sqrt(2u))du}}}
We can substitute these into
{{{ int (sin (sqrt ( 2u ) )*(1/sqrt( 2u )), du) }}}
giving:
{{{ int (sin (v), dv) }}}
which integrates to:
{{{-cos(v) + C}}}
Substituting back in for v we get:
{{{-cos(sqrt(2u)) + C}}}