A trigonometric function question for calculus [closed]
Compute the definite integral $$int_0^1 left(2-x^2right)^{3/2}dx$$
Please help me, i suppose $x=sqrt{2} sin theta$ but I just couldn't get the answer.
Thank you
definite-integrals trigonometric-integrals
edited Nov 29 at 16:25
gt6989b
33k22452
asked Nov 29 at 16:18
Kizaru
253
closed as off-topic by amWhy, Brahadeesh, Alexander Gruber♦ Nov 30 at 3:30
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Brahadeesh, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
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show 2 more comments
Compute the definite integral $$int_0^1 left(2-x^2right)^{3/2}dx$$
Please help me, i suppose $x=sqrt{2} sin theta$ but I just couldn't get the answer.
Thank you
definite-integrals trigonometric-integrals
edited Nov 29 at 16:25
gt6989b
33k22452
asked Nov 29 at 16:18
Kizaru
253
closed as off-topic by amWhy, Brahadeesh, Alexander Gruber♦ Nov 30 at 3:30
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Brahadeesh, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
What happened when using your substitution?
– coffeemath
Nov 29 at 16:21
Using that substitution, what does $dx$ need to be? $dx = sqrt 2 cos theta$
– amWhy
Nov 29 at 16:23
If you don't tell us where you get stuck, you won't get the most helpful of answers.
– amWhy
Nov 29 at 16:33
imgur.com/a/wZSjmZS this is my first try and couldn't figure out where going wrong .i used another way and got the answer. I wanted to know why this is wrong
– Kizaru
Nov 29 at 16:45
Note: I used Wallis in the process
– Kizaru
Nov 29 at 16:47
|
show 2 more comments
Compute the definite integral $$int_0^1 left(2-x^2right)^{3/2}dx$$
Please help me, i suppose $x=sqrt{2} sin theta$ but I just couldn't get the answer.
Thank you
definite-integrals trigonometric-integrals
edited Nov 29 at 16:25
gt6989b
33k22452
asked Nov 29 at 16:18
Kizaru
253
Compute the definite integral $$int_0^1 left(2-x^2right)^{3/2}dx$$
Please help me, i suppose $x=sqrt{2} sin theta$ but I just couldn't get the answer.
Thank you
definite-integrals trigonometric-integrals
definite-integrals trigonometric-integrals
edited Nov 29 at 16:25
gt6989b
33k22452
asked Nov 29 at 16:18
Kizaru
253
edited Nov 29 at 16:25
gt6989b
33k22452
asked Nov 29 at 16:18
Kizaru
253
edited Nov 29 at 16:25
gt6989b
33k22452
edited Nov 29 at 16:25
gt6989b
33k22452
edited Nov 29 at 16:25
gt6989b
33k22452
33k22452
asked Nov 29 at 16:18
Kizaru
253
asked Nov 29 at 16:18
Kizaru
253
asked Nov 29 at 16:18
Kizaru
253
253
closed as off-topic by amWhy, Brahadeesh, Alexander Gruber♦ Nov 30 at 3:30
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Brahadeesh, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Brahadeesh, Alexander Gruber♦ Nov 30 at 3:30
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Brahadeesh, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
What happened when using your substitution?
– coffeemath
Nov 29 at 16:21
Using that substitution, what does $dx$ need to be? $dx = sqrt 2 cos theta$
– amWhy
Nov 29 at 16:23
If you don't tell us where you get stuck, you won't get the most helpful of answers.
– amWhy
Nov 29 at 16:33
imgur.com/a/wZSjmZS this is my first try and couldn't figure out where going wrong .i used another way and got the answer. I wanted to know why this is wrong
– Kizaru
Nov 29 at 16:45
Note: I used Wallis in the process
– Kizaru
Nov 29 at 16:47
|
show 2 more comments
What happened when using your substitution?
– coffeemath
Nov 29 at 16:21
Using that substitution, what does $dx$ need to be? $dx = sqrt 2 cos theta$
– amWhy
Nov 29 at 16:23
If you don't tell us where you get stuck, you won't get the most helpful of answers.
– amWhy
Nov 29 at 16:33
imgur.com/a/wZSjmZS this is my first try and couldn't figure out where going wrong .i used another way and got the answer. I wanted to know why this is wrong
– Kizaru
Nov 29 at 16:45
Note: I used Wallis in the process
– Kizaru
Nov 29 at 16:47
What happened when using your substitution?
– coffeemath
Nov 29 at 16:21
What happened when using your substitution?
– coffeemath
Nov 29 at 16:21
Using that substitution, what does $dx$ need to be? $dx = sqrt 2 cos theta$
– amWhy
Nov 29 at 16:23
Using that substitution, what does $dx$ need to be? $dx = sqrt 2 cos theta$
– amWhy
Nov 29 at 16:23
If you don't tell us where you get stuck, you won't get the most helpful of answers.
– amWhy
Nov 29 at 16:33
If you don't tell us where you get stuck, you won't get the most helpful of answers.
– amWhy
Nov 29 at 16:33
imgur.com/a/wZSjmZS this is my first try and couldn't figure out where going wrong .i used another way and got the answer. I wanted to know why this is wrong
– Kizaru
Nov 29 at 16:45
imgur.com/a/wZSjmZS this is my first try and couldn't figure out where going wrong .i used another way and got the answer. I wanted to know why this is wrong
– Kizaru
Nov 29 at 16:45
Note: I used Wallis in the process
– Kizaru
Nov 29 at 16:47
Note: I used Wallis in the process
– Kizaru
Nov 29 at 16:47
|
show 2 more comments
1 Answer
1
active
oldest
votes
So if $x = sqrt{2} sin t$ then $2-x^2 = 2 cos^2 t$, and also $dx = sqrt{2} cos t dt$ and you get
$$
int_0^1 left(2-x^2right)^{3/2}dx
= 2^{3/2} sqrt{2} int_0^{pi/4} cos^4 t dt
= 4int_0^{pi/4} cos^4 t dt
$$
Can you complete this? (Hint: use the relation between $cos 2x$ and $cos^2 x$ twice)
(You can see here for how to do it if in doubt)
answered Nov 29 at 16:23
gt6989b
33k22452
That all falls out from the mere substitution, and if the asker was stuck on the dx conversion, that was covered. Chances are you stopped where it gets difficult.
– amWhy
Nov 29 at 16:27
1
@amWhy interesting, will update -- thought he made a mistake in substituting -- but it's a standard technique from here and on, tabulated integral really
– gt6989b
Nov 29 at 16:29
imgur.com/a/wZSjmZS. This is my original try
– Kizaru
Nov 29 at 16:31
@amWhy updated with the link to stackexchange post how to solve the trig integral
– gt6989b
Nov 29 at 16:32
I used Wallis in the process
– Kizaru
Nov 29 at 16:32
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
So if $x = sqrt{2} sin t$ then $2-x^2 = 2 cos^2 t$, and also $dx = sqrt{2} cos t dt$ and you get
$$
int_0^1 left(2-x^2right)^{3/2}dx
= 2^{3/2} sqrt{2} int_0^{pi/4} cos^4 t dt
= 4int_0^{pi/4} cos^4 t dt
$$
Can you complete this? (Hint: use the relation between $cos 2x$ and $cos^2 x$ twice)
(You can see here for how to do it if in doubt)
answered Nov 29 at 16:23
gt6989b
33k22452
That all falls out from the mere substitution, and if the asker was stuck on the dx conversion, that was covered. Chances are you stopped where it gets difficult.
– amWhy
Nov 29 at 16:27
1
@amWhy interesting, will update -- thought he made a mistake in substituting -- but it's a standard technique from here and on, tabulated integral really
– gt6989b
Nov 29 at 16:29
imgur.com/a/wZSjmZS. This is my original try
– Kizaru
Nov 29 at 16:31
@amWhy updated with the link to stackexchange post how to solve the trig integral
– gt6989b
Nov 29 at 16:32
I used Wallis in the process
– Kizaru
Nov 29 at 16:32
add a comment |
So if $x = sqrt{2} sin t$ then $2-x^2 = 2 cos^2 t$, and also $dx = sqrt{2} cos t dt$ and you get
$$
int_0^1 left(2-x^2right)^{3/2}dx
= 2^{3/2} sqrt{2} int_0^{pi/4} cos^4 t dt
= 4int_0^{pi/4} cos^4 t dt
$$
Can you complete this? (Hint: use the relation between $cos 2x$ and $cos^2 x$ twice)
(You can see here for how to do it if in doubt)
answered Nov 29 at 16:23
gt6989b
33k22452
That all falls out from the mere substitution, and if the asker was stuck on the dx conversion, that was covered. Chances are you stopped where it gets difficult.
– amWhy
Nov 29 at 16:27
1
@amWhy interesting, will update -- thought he made a mistake in substituting -- but it's a standard technique from here and on, tabulated integral really
– gt6989b
Nov 29 at 16:29
imgur.com/a/wZSjmZS. This is my original try
– Kizaru
Nov 29 at 16:31
@amWhy updated with the link to stackexchange post how to solve the trig integral
– gt6989b
Nov 29 at 16:32
I used Wallis in the process
– Kizaru
Nov 29 at 16:32
add a comment |
So if $x = sqrt{2} sin t$ then $2-x^2 = 2 cos^2 t$, and also $dx = sqrt{2} cos t dt$ and you get
$$
int_0^1 left(2-x^2right)^{3/2}dx
= 2^{3/2} sqrt{2} int_0^{pi/4} cos^4 t dt
= 4int_0^{pi/4} cos^4 t dt
$$
Can you complete this? (Hint: use the relation between $cos 2x$ and $cos^2 x$ twice)
(You can see here for how to do it if in doubt)
answered Nov 29 at 16:23
gt6989b
33k22452
So if $x = sqrt{2} sin t$ then $2-x^2 = 2 cos^2 t$, and also $dx = sqrt{2} cos t dt$ and you get
$$
int_0^1 left(2-x^2right)^{3/2}dx
= 2^{3/2} sqrt{2} int_0^{pi/4} cos^4 t dt
= 4int_0^{pi/4} cos^4 t dt
$$
Can you complete this? (Hint: use the relation between $cos 2x$ and $cos^2 x$ twice)
(You can see here for how to do it if in doubt)
answered Nov 29 at 16:23
gt6989b
33k22452
edited Nov 29 at 16:33
answered Nov 29 at 16:23
gt6989b
33k22452
answered Nov 29 at 16:23
gt6989b
33k22452
answered Nov 29 at 16:23
gt6989b
33k22452
33k22452
That all falls out from the mere substitution, and if the asker was stuck on the dx conversion, that was covered. Chances are you stopped where it gets difficult.
– amWhy
Nov 29 at 16:27
1
@amWhy interesting, will update -- thought he made a mistake in substituting -- but it's a standard technique from here and on, tabulated integral really
– gt6989b
Nov 29 at 16:29
imgur.com/a/wZSjmZS. This is my original try
– Kizaru
Nov 29 at 16:31
@amWhy updated with the link to stackexchange post how to solve the trig integral
– gt6989b
Nov 29 at 16:32
I used Wallis in the process
– Kizaru
Nov 29 at 16:32
add a comment |
That all falls out from the mere substitution, and if the asker was stuck on the dx conversion, that was covered. Chances are you stopped where it gets difficult.
– amWhy
Nov 29 at 16:27
1
@amWhy interesting, will update -- thought he made a mistake in substituting -- but it's a standard technique from here and on, tabulated integral really
– gt6989b
Nov 29 at 16:29
imgur.com/a/wZSjmZS. This is my original try
– Kizaru
Nov 29 at 16:31
@amWhy updated with the link to stackexchange post how to solve the trig integral
– gt6989b
Nov 29 at 16:32
I used Wallis in the process
– Kizaru
Nov 29 at 16:32
That all falls out from the mere substitution, and if the asker was stuck on the dx conversion, that was covered. Chances are you stopped where it gets difficult.
– amWhy
Nov 29 at 16:27
That all falls out from the mere substitution, and if the asker was stuck on the dx conversion, that was covered. Chances are you stopped where it gets difficult.
– amWhy
Nov 29 at 16:27
1
1
@amWhy interesting, will update -- thought he made a mistake in substituting -- but it's a standard technique from here and on, tabulated integral really
– gt6989b
Nov 29 at 16:29
@amWhy interesting, will update -- thought he made a mistake in substituting -- but it's a standard technique from here and on, tabulated integral really
– gt6989b
Nov 29 at 16:29
imgur.com/a/wZSjmZS. This is my original try
– Kizaru
Nov 29 at 16:31
imgur.com/a/wZSjmZS. This is my original try
– Kizaru
Nov 29 at 16:31
@amWhy updated with the link to stackexchange post how to solve the trig integral
– gt6989b
Nov 29 at 16:32
@amWhy updated with the link to stackexchange post how to solve the trig integral
– gt6989b
Nov 29 at 16:32
I used Wallis in the process
– Kizaru
Nov 29 at 16:32
I used Wallis in the process
– Kizaru
Nov 29 at 16:32
add a comment |
What happened when using your substitution?
– coffeemath
Nov 29 at 16:21
Using that substitution, what does $dx$ need to be? $dx = sqrt 2 cos theta$
– amWhy
Nov 29 at 16:23
If you don't tell us where you get stuck, you won't get the most helpful of answers.
– amWhy
Nov 29 at 16:33
imgur.com/a/wZSjmZS this is my first try and couldn't figure out where going wrong .i used another way and got the answer. I wanted to know why this is wrong
– Kizaru
Nov 29 at 16:45
Note: I used Wallis in the process
– Kizaru
Nov 29 at 16:47