Evaluate $int_{pi/6}^frac{pi}{2}cscleft(tright)cotleft(tright)dt $
So I consider the integral:
$$int_{pi/6}^frac{pi}{2}cscleft(tright)cotleft(tright)space text{dt}$$
In my notes I have: $$=-cscleft(tright)+C$$
Then I take the limit of $csc(t)$:
$$lim_{trightarrowfrac{pi}{6}}-csc(t)=-2$$
$$lim_{trightarrowfrac{pi}{2}}-csc(t)=-1$$
Then I subtract $F(b)-F(a)$,
$-1-(-2)=1$ but I don't know how I arrived at $-csc(t)+C$
calculus
edited Dec 10 '18 at 3:46
coreyman317
686219
asked Dec 10 '18 at 2:14
Eric Brown
716
add a comment |
So I consider the integral:
$$int_{pi/6}^frac{pi}{2}cscleft(tright)cotleft(tright)space text{dt}$$
In my notes I have: $$=-cscleft(tright)+C$$
Then I take the limit of $csc(t)$:
$$lim_{trightarrowfrac{pi}{6}}-csc(t)=-2$$
$$lim_{trightarrowfrac{pi}{2}}-csc(t)=-1$$
Then I subtract $F(b)-F(a)$,
$-1-(-2)=1$ but I don't know how I arrived at $-csc(t)+C$
calculus
edited Dec 10 '18 at 3:46
coreyman317
686219
asked Dec 10 '18 at 2:14
Eric Brown
716
5
You cannot split them. They are not equal at all.
– xbh
Dec 10 '18 at 2:22
$csc(t)cot(t)=frac{1}{sin(t)}cdotfrac{cos(t)}{sin(t)}=frac{cos(t)}{sin^2(t)}$
– coreyman317
Dec 10 '18 at 2:30
I think you are confusing the definite and indefinite integrals
– Aniruddh Venkatesan
Dec 10 '18 at 2:40
add a comment |
So I consider the integral:
$$int_{pi/6}^frac{pi}{2}cscleft(tright)cotleft(tright)space text{dt}$$
In my notes I have: $$=-cscleft(tright)+C$$
Then I take the limit of $csc(t)$:
$$lim_{trightarrowfrac{pi}{6}}-csc(t)=-2$$
$$lim_{trightarrowfrac{pi}{2}}-csc(t)=-1$$
Then I subtract $F(b)-F(a)$,
$-1-(-2)=1$ but I don't know how I arrived at $-csc(t)+C$
calculus
edited Dec 10 '18 at 3:46
coreyman317
686219
asked Dec 10 '18 at 2:14
Eric Brown
716
So I consider the integral:
$$int_{pi/6}^frac{pi}{2}cscleft(tright)cotleft(tright)space text{dt}$$
In my notes I have: $$=-cscleft(tright)+C$$
Then I take the limit of $csc(t)$:
$$lim_{trightarrowfrac{pi}{6}}-csc(t)=-2$$
$$lim_{trightarrowfrac{pi}{2}}-csc(t)=-1$$
Then I subtract $F(b)-F(a)$,
$-1-(-2)=1$ but I don't know how I arrived at $-csc(t)+C$
calculus
calculus
edited Dec 10 '18 at 3:46
coreyman317
686219
asked Dec 10 '18 at 2:14
Eric Brown
716
edited Dec 10 '18 at 3:46
coreyman317
686219
asked Dec 10 '18 at 2:14
Eric Brown
716
edited Dec 10 '18 at 3:46
coreyman317
686219
edited Dec 10 '18 at 3:46
coreyman317
686219
edited Dec 10 '18 at 3:46
coreyman317
686219
686219
asked Dec 10 '18 at 2:14
Eric Brown
716
asked Dec 10 '18 at 2:14
Eric Brown
716
asked Dec 10 '18 at 2:14
Eric Brown
716
716
5
You cannot split them. They are not equal at all.
– xbh
Dec 10 '18 at 2:22
$csc(t)cot(t)=frac{1}{sin(t)}cdotfrac{cos(t)}{sin(t)}=frac{cos(t)}{sin^2(t)}$
– coreyman317
Dec 10 '18 at 2:30
I think you are confusing the definite and indefinite integrals
– Aniruddh Venkatesan
Dec 10 '18 at 2:40
add a comment |
5
You cannot split them. They are not equal at all.
– xbh
Dec 10 '18 at 2:22
$csc(t)cot(t)=frac{1}{sin(t)}cdotfrac{cos(t)}{sin(t)}=frac{cos(t)}{sin^2(t)}$
– coreyman317
Dec 10 '18 at 2:30
I think you are confusing the definite and indefinite integrals
– Aniruddh Venkatesan
Dec 10 '18 at 2:40
5
5
You cannot split them. They are not equal at all.
– xbh
Dec 10 '18 at 2:22
You cannot split them. They are not equal at all.
– xbh
Dec 10 '18 at 2:22
$csc(t)cot(t)=frac{1}{sin(t)}cdotfrac{cos(t)}{sin(t)}=frac{cos(t)}{sin^2(t)}$
– coreyman317
Dec 10 '18 at 2:30
$csc(t)cot(t)=frac{1}{sin(t)}cdotfrac{cos(t)}{sin(t)}=frac{cos(t)}{sin^2(t)}$
– coreyman317
Dec 10 '18 at 2:30
I think you are confusing the definite and indefinite integrals
– Aniruddh Venkatesan
Dec 10 '18 at 2:40
I think you are confusing the definite and indefinite integrals
– Aniruddh Venkatesan
Dec 10 '18 at 2:40
add a comment |
2 Answers
2
active
oldest
votes
You cannot split up your integral like you have tried in your original post. In order to evaluate this, notice that $$frac{d}{dx} (-csc x) = csc x cot x$$
Therefore, $intcsc x cot x = -csc x + C$. Then you use the Fundamental Theorem of Calculus, as you have noted.
answered Dec 10 '18 at 2:34
BSplitter
509215
add a comment |
If you write it terms of sine and cosine, you'll get $$frac{cos x}{sin^2x}$$ which you can do u-sub on ($u=sin x$), neatly cancelling the cosine in the numerator.
P.S. in general, you can't split products inside integrals. That only works for addition or subtraction, or multiplication by a constant.
answered Dec 10 '18 at 2:30
D.R.
1,453620
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
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votes
You cannot split up your integral like you have tried in your original post. In order to evaluate this, notice that $$frac{d}{dx} (-csc x) = csc x cot x$$
Therefore, $intcsc x cot x = -csc x + C$. Then you use the Fundamental Theorem of Calculus, as you have noted.
answered Dec 10 '18 at 2:34
BSplitter
509215
add a comment |
You cannot split up your integral like you have tried in your original post. In order to evaluate this, notice that $$frac{d}{dx} (-csc x) = csc x cot x$$
Therefore, $intcsc x cot x = -csc x + C$. Then you use the Fundamental Theorem of Calculus, as you have noted.
answered Dec 10 '18 at 2:34
BSplitter
509215
add a comment |
You cannot split up your integral like you have tried in your original post. In order to evaluate this, notice that $$frac{d}{dx} (-csc x) = csc x cot x$$
Therefore, $intcsc x cot x = -csc x + C$. Then you use the Fundamental Theorem of Calculus, as you have noted.
answered Dec 10 '18 at 2:34
BSplitter
509215
You cannot split up your integral like you have tried in your original post. In order to evaluate this, notice that $$frac{d}{dx} (-csc x) = csc x cot x$$
Therefore, $intcsc x cot x = -csc x + C$. Then you use the Fundamental Theorem of Calculus, as you have noted.
answered Dec 10 '18 at 2:34
BSplitter
509215
answered Dec 10 '18 at 2:34
BSplitter
509215
answered Dec 10 '18 at 2:34
BSplitter
509215
answered Dec 10 '18 at 2:34
BSplitter
509215
509215
add a comment |
add a comment |
If you write it terms of sine and cosine, you'll get $$frac{cos x}{sin^2x}$$ which you can do u-sub on ($u=sin x$), neatly cancelling the cosine in the numerator.
P.S. in general, you can't split products inside integrals. That only works for addition or subtraction, or multiplication by a constant.
answered Dec 10 '18 at 2:30
D.R.
1,453620
add a comment |
If you write it terms of sine and cosine, you'll get $$frac{cos x}{sin^2x}$$ which you can do u-sub on ($u=sin x$), neatly cancelling the cosine in the numerator.
P.S. in general, you can't split products inside integrals. That only works for addition or subtraction, or multiplication by a constant.
answered Dec 10 '18 at 2:30
D.R.
1,453620
add a comment |
If you write it terms of sine and cosine, you'll get $$frac{cos x}{sin^2x}$$ which you can do u-sub on ($u=sin x$), neatly cancelling the cosine in the numerator.
P.S. in general, you can't split products inside integrals. That only works for addition or subtraction, or multiplication by a constant.
answered Dec 10 '18 at 2:30
D.R.
1,453620
If you write it terms of sine and cosine, you'll get $$frac{cos x}{sin^2x}$$ which you can do u-sub on ($u=sin x$), neatly cancelling the cosine in the numerator.
P.S. in general, you can't split products inside integrals. That only works for addition or subtraction, or multiplication by a constant.
answered Dec 10 '18 at 2:30
D.R.
1,453620
answered Dec 10 '18 at 2:30
D.R.
1,453620
answered Dec 10 '18 at 2:30
D.R.
1,453620
answered Dec 10 '18 at 2:30
D.R.
1,453620
1,453620
add a comment |
add a comment |
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5
You cannot split them. They are not equal at all.
– xbh
Dec 10 '18 at 2:22
$csc(t)cot(t)=frac{1}{sin(t)}cdotfrac{cos(t)}{sin(t)}=frac{cos(t)}{sin^2(t)}$
– coreyman317
Dec 10 '18 at 2:30
I think you are confusing the definite and indefinite integrals
– Aniruddh Venkatesan
Dec 10 '18 at 2:40