Integration of a trigonometic function

[chwala]

Homework Statement

find $\int_0^{0.5π} \frac {sin^2 x}{sin x + cos x}dx$

Relevant Equations

integration

my thinking was to have everything changed to a function that has cosine only...

$\int_0^{0.5π} \frac {1-cos^2x}{sin x + cos x}dx$

$\int_0^{0.5π} \frac {(1-cos x)(1+cos x)}{(1-cos^2x)^{0.5} + cos x}dx$ ...

first of all is this integration possible? if so then let me know if i am on the right track...chain rule?

 

[pasmith]

Hint: <br /> \begin{align*}<br /> \cos(x - \tfrac{\pi}4) &amp;= \frac1{\sqrt 2}(\cos x + \sin x) \\<br /> \sin(x + \tfrac{\pi}4) &amp;= \frac1{\sqrt 2}(\cos x + \sin x)<br /> \end{align*}<br />

 

[chwala]

Or am I supposed to use the relation of expressing sum of trig functions as a product...for the denominator...

 

[chwala]

Hint: <br /> \begin{align*}<br /> \cos(x - \tfrac{\pi}4) &amp;= \frac1{\sqrt 2}(\cos x + \sin x) \\<br /> \sin(x + \tfrac{\pi}4) &amp;= \frac1{\sqrt 2}(\cos x + \sin x)<br /> \end{align*}<br />

Never seen this identity...how did you arrive at it?I would like to see how it was derived or reference

 

[chwala]

i am getting,
$\frac {4}{π√2}\int_0^{0.5π} \frac {sin(u-0.25π)}{sin u}du$
=$\frac {4}{π√2}\int_0^{0.5π} \frac {sinu-cos u}{sin u}du$
==$\frac {4}{π√2}\int_0^{0.5π} [1-cot u ]du=\frac {4}{π√2}\left.[u-ln sin u]\right|_0^{0.5π}$
=$\frac {4}{π√2}\left.[(x+0.25π)-ln sin (x+0.25π)]\right|_0^{0.5π}$
correct?

 

[pasmith]

Never seen this identity...how did you arrive at it?I would like to see how it was derived or reference

Use the identities <br /> \begin{align*}<br /> \cos(x \pm y) &amp;= \cos x \cos y \mp \sin x \sin y \\<br /> \sin(x \pm y) &amp;= \sin x \cos y \pm \cos x \sin y<br /> \end{align*}<br /> and \cos \frac{\pi}4 = \sin \frac{\pi}4 = \frac1{\sqrt 2}.

You can use @etotheipi's suggestion (which they have now deleted) of using the identities <br /> \cos x = \sin (\tfrac{\pi}2 - x), \qquad \sin x = \cos(\tfrac{\pi}2 - x)<br /> to show that \int_0^{\pi/2} \frac{\sin^2 x}{\cos x + \sin x}\,dx =<br /> \frac12 \int_0^{\pi/2} \frac{1}{\sin x + \cos x}\,dx and then use the identity I posted to simplify the denominator.

 

[chwala]

Use the identities <br /> \begin{align*}<br /> \cos(x \pm y) &amp;= \cos x \cos y \mp \sin x \sin y \\<br /> \sin(x \pm y) &amp;= \sin x \cos y \pm \cos x \sin y<br /> \end{align*}<br /> and \cos \frac{\pi}4 = \sin \frac{\pi}4 = \frac1{\sqrt 2}.

You can use @etotheipi's suggestion (which they have now deleted) of using the identities <br /> \cos x = \sin (\tfrac{\pi}2 - x), \qquad \sin x = \cos(\tfrac{\pi}2 - x)<br /> to show that \int_0^{\pi/2} \frac{\sin^2 x}{\cos x + \sin x}\,dx =<br /> \frac12 \int_0^{\pi/2} \frac{1}{\sin x + \cos x}\,dx and then use the identity I posted to simplify the denominator.

am still not getting it...how does the numerator change from $sin^2 x$ to $1$
what i know probably is using $cos 2x= 1-2sin^2x$, this way its clear to me...
ok i have seen it,..you want me to use $sin x=\frac {1}{√2}$→$sin^2x=0.5$

 

[etotheipi]

You can use @etotheipi's suggestion (which they have now deleted) of using the identities <br /> \cos x = \sin (\tfrac{\pi}2 - x), \qquad \sin x = \cos(\tfrac{\pi}2 - x)<br /> to show that \int_0^{\pi/2} \frac{\sin^2 x}{\cos x + \sin x}\,dx =<br /> \frac12 \int_0^{\pi/2} \frac{1}{\sin x + \cos x}\,dx and then use the identity I posted to simplify the denominator.

Yeah sorry about that, I looked at the integral and went "yeah, I know how to do that", wrote a little comment but then realized that the resulting integral isn't too easy either. But then I read your post #2, and with those identities you can indeed then solve the $1/(\sin{x} + \cos{x})$ integral, so I guess the suggestion was fine all along

 

[chwala]

i am now getting,
$\frac {1}{2√2}\int_0^{0.5π} \frac {1}{cos (x-0.25π)}dx$
=$\frac {1}{2√2}\left.ln |sec (x-0.25π)+ tan (x-0.25π)|\right|_0^{0.5π}$

 

[etotheipi]

N.B. one can also use the Weierstrass $t=\tan{x/2}$ on the integral in #6, i.e. with$$\frac{dt}{dx} = \frac{1}{2} \sec^2{(x/2)} = \frac{1}{2}(1+t^2), \quad \sin{x} = \frac{2t}{1+t^2},\quad \cos{x} = \frac{1-t^2}{1+t^2}$$

 

[chwala]

is post $9$ correct?

 

[chwala]

 

[chwala]

yep, i nailed it bingo! Africa power...

 

[chwala]

Hint: <br /> \begin{align*}<br /> \cos(x - \tfrac{\pi}4) &amp;= \frac1{\sqrt 2}(\cos x + \sin x) \\<br /> \sin(x + \tfrac{\pi}4) &amp;= \frac1{\sqrt 2}(\cos x + \sin x)<br /> \end{align*}<br />

aaarrgh i have always known this, i guess my brain was off at the time...

 

[chwala]

Never seen this identity...how did you arrive at it?I would like to see how it was derived or reference

$cos (m+p)=cosmcosp-sinmsinp$
if $p=45^0$...then we end up getting your hint indicated in your post $2$.
I know this very well, cheers