integration using sin/cos: int 0 to pi/2 cosx sqrt(sinx) dx

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crzymath

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Oct 27, 2008
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i did these two definite integral problems on my test and i got them wrong.

a) integral {0 to pi/2} (cos x) sqrt(sinx) dx
my answer: -1

b) integral {pi/4 to pi/2} (csc[sup:3mxosfwy]2[/sup:3mxosfwy]x+1) dx
my answer: -5pi/4

c) integral {1-2} (x-1)sqrt(2-x) dx
my answer: sqrt 2

i got all these wrong and i dont know why-can someone please use one of these problems to show me how to do it correctly? thanks!
 
Re: integration using sin/cos

can u tell me what the answers should be? it would help me if i had a key to these problems. ill post my work after i do it correctly :)
 
Re: integration using sin/cos

ok i redid one of the problems. hopefully its rite :D

integral {pi/4 to pi/2} (csc[sup:q7ax7yb1]2[/sup:q7ax7yb1]x+10) dx

= -cot x + x ] pi/4 to pi/2
next i used fundamental theorem of calculus
f(b)-f(a)
=(-cot pi/2 + pi/2) - (-cot pi/4 + pi/4)
=(0 + pi/2) - (-1 + pi/4)
=pi/2 + 1 - pi/4
=3pi/2 - pi/4
= 6pi/4 -pi/4
= 5pi/4 <---im pretty sure this is rite since i got the same answer on the test only it was negative and i got half a point off. hehe.

the second problem ive had a tougher time working with:

integral {1 to 2} (x-1)sqrt(2-x) dx
=integral {1 to 2} 2x[sup:q7ax7yb1]3/2[/sup:q7ax7yb1]-x[sup:q7ax7yb1]3/2[/sup:q7ax7yb1]-2[sup:q7ax7yb1]1/2[/sup:q7ax7yb1]+x[sup:q7ax7yb1]3/2[/sup:q7ax7yb1] <--- multiplied everything together
=4/5x[sup:q7ax7yb1]5/2[/sup:q7ax7yb1]-2/5x[sup:q7ax7yb1]5/2[/sup:q7ax7yb1]-4/3x[sup:q7ax7yb1]3/2[/sup:q7ax7yb1]+2/5x[sup:q7ax7yb1]5/2[/sup:q7ax7yb1] ]1 to 2
next i use fundamental theorem of calclus
f(b)-f(a)
...before i go on-can u tell me if the work ive done so far is correct? please correct me if im wrong. thanks alot~ :)
 
integral {1 to 2} (x-1)sqrt(2-x) dx

\(\displaystyle \int_1^2(x-1)\sqrt{2-x}dx\)

Substitute

\(\displaystyle u \, = \, \sqrt{2-x}\)

\(\displaystyle 2u \, du \, = \, - \, dx\)

\(\displaystyle \int_1^0(1-u^2)u(-2u \,du)\)

\(\displaystyle = \, -2\int_1^0(u^2 \, - \, u^4) \,du\)

\(\displaystyle = \, 2\cdot [\frac{1}{3} \, - \, \frac{1}{5}]\)
 
Re: integration using sin/cos

crzymath said:
ok i redid one of the problems. hopefully its rite :D

integral {pi/4 to pi/2} (csc[sup:rewm9a6z]2[/sup:rewm9a6z]x+10) dx

= -cot x + x ] pi/4 to pi/2
next i used fundamental theorem of calculus
f(b)-f(a)
=(-cot pi/2 + pi/2) - (-cot pi/4 + pi/4)
=(0 + pi/2) - (-1 + pi/4)
=pi/2 + 1 - pi/4
=3pi/2 - pi/4
= 6pi/4 -pi/4
= 5pi/4 <<<< Correct

<---im pretty sure this is rite since i got the same answer on the test only it was negative and i got half a point off. hehe.

the second problem ive had a tougher time working with:

integral {1 to 2} (x-1)sqrt(2-x) dx
=integral {1 to 2} 2x[sup:rewm9a6z]3/2[/sup:rewm9a6z]-x[sup:rewm9a6z]3/2[/sup:rewm9a6z]-2[sup:rewm9a6z]1/2[/sup:rewm9a6z]+x[sup:rewm9a6z]3/2[/sup:rewm9a6z] <--- multiplied everything together<<Cannot do that!!
=4/5x[sup:rewm9a6z]5/2[/sup:rewm9a6z]-2/5x[sup:rewm9a6z]5/2[/sup:rewm9a6z]-4/3x[sup:rewm9a6z]3/2[/sup:rewm9a6z]+2/5x[sup:rewm9a6z]5/2[/sup:rewm9a6z] ]1 to 2
next i use fundamental theorem of calclus
f(b)-f(a)
...before i go on-can u tell me if the work ive done so far is correct? please correct me if im wrong. thanks alot~ :)
Click to expand...
 
a) \(\displaystyle \int_{0}^{\frac{\pi}{2}}cos(x)\sqrt{sin(x)} dx\)
my answer: -1
Click to expand...

This one is actually quite easy. Just let \(\displaystyle u=sin(x), \;\ du=cos(x)dx\)

Don't forget to change your limits of integration. That is where most go wrong.

Then we have: \(\displaystyle \int_{0}^{1}\sqrt{u}du=\frac{2}{3}u^{\frac{3}{2}}|_{0}^{1}=\boxed{\frac{2}{3}}\)
 
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