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How to evaluate the definite integral $int_0^a/2 x^2(a^2-x^2)^-3/2dx $
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$$int_0^a/2x^2(a^2-x^2)^-3/2dx $$
I've been having trouble with this question for a while now. When I attempted the question I substituted $x=asintheta$ and $(acostheta)^3=(a^2-x^2)^-3/2$.
When I solved with the substitutions what I get at the end is $tan^2theta$ in the integral sign. So I integrated $sec^2theta-1$ and got an answer which is different to the real answer. And when I checked an online integral calculator the top $x^2$ was split into $x^2-a^2+a^2$ to make two fractions. which didn't make sense to me at all because somewhere along that line the $x^2$ itself was completely removed. The online calculator is www.integral-calculator.com
I learned trigonometric substitution last night and I am not very good at it. So a detailed explanation would be very much appreciated.
integration trigonometry substitution
edited Aug 3 at 10:34
Dylan
11.4k31026
asked Aug 2 at 3:49
Maiz Halyym
61
add a comment |Â
up vote
0
down vote
favorite
$$int_0^a/2x^2(a^2-x^2)^-3/2dx $$
I've been having trouble with this question for a while now. When I attempted the question I substituted $x=asintheta$ and $(acostheta)^3=(a^2-x^2)^-3/2$.
When I solved with the substitutions what I get at the end is $tan^2theta$ in the integral sign. So I integrated $sec^2theta-1$ and got an answer which is different to the real answer. And when I checked an online integral calculator the top $x^2$ was split into $x^2-a^2+a^2$ to make two fractions. which didn't make sense to me at all because somewhere along that line the $x^2$ itself was completely removed. The online calculator is www.integral-calculator.com
I learned trigonometric substitution last night and I am not very good at it. So a detailed explanation would be very much appreciated.
integration trigonometry substitution
edited Aug 3 at 10:34
Dylan
11.4k31026
asked Aug 2 at 3:49
Maiz Halyym
61
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$$int_0^a/2x^2(a^2-x^2)^-3/2dx $$
I've been having trouble with this question for a while now. When I attempted the question I substituted $x=asintheta$ and $(acostheta)^3=(a^2-x^2)^-3/2$.
When I solved with the substitutions what I get at the end is $tan^2theta$ in the integral sign. So I integrated $sec^2theta-1$ and got an answer which is different to the real answer. And when I checked an online integral calculator the top $x^2$ was split into $x^2-a^2+a^2$ to make two fractions. which didn't make sense to me at all because somewhere along that line the $x^2$ itself was completely removed. The online calculator is www.integral-calculator.com
I learned trigonometric substitution last night and I am not very good at it. So a detailed explanation would be very much appreciated.
integration trigonometry substitution
edited Aug 3 at 10:34
Dylan
11.4k31026
asked Aug 2 at 3:49
Maiz Halyym
61
$$int_0^a/2x^2(a^2-x^2)^-3/2dx $$
I've been having trouble with this question for a while now. When I attempted the question I substituted $x=asintheta$ and $(acostheta)^3=(a^2-x^2)^-3/2$.
When I solved with the substitutions what I get at the end is $tan^2theta$ in the integral sign. So I integrated $sec^2theta-1$ and got an answer which is different to the real answer. And when I checked an online integral calculator the top $x^2$ was split into $x^2-a^2+a^2$ to make two fractions. which didn't make sense to me at all because somewhere along that line the $x^2$ itself was completely removed. The online calculator is www.integral-calculator.com
I learned trigonometric substitution last night and I am not very good at it. So a detailed explanation would be very much appreciated.
integration trigonometry substitution
edited Aug 3 at 10:34
Dylan
11.4k31026
asked Aug 2 at 3:49
Maiz Halyym
61
edited Aug 3 at 10:34
Dylan
11.4k31026
edited Aug 3 at 10:34
Dylan
11.4k31026
edited Aug 3 at 10:34
Dylan
11.4k31026
11.4k31026
asked Aug 2 at 3:49
Maiz Halyym
61
asked Aug 2 at 3:49
Maiz Halyym
61
asked Aug 2 at 3:49
Maiz Halyym
61
61
add a comment |Â
add a comment |Â
2 Answers
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2
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Hint: Using substitution $x=asin t$ then
$$int_0^fraca2dfracx^2sqrta^2-x^2^3dx=int_0^pi/6tan^2t dt$$
answered Aug 2 at 4:05
user 108128
18.8k41544
That's exactly what i did, I then expressed tan^2t as sec^2t-1 and got the answer X/(a^2-x^2)-arcsin(x/a)+c. but when i tried to solve it for a/2 and 0, it didn't work out.
– Maiz Halyym
Aug 2 at 4:09
I couldn't read your notes. It's better to write them with mathjax. anyway, $$int tan^2t+1-1 dt=tan t-tBig|_0^pi/6$$ it's a number!
– user 108128
Aug 2 at 4:13
I'll have to learn how to use mathjax, thanks. And also what i meant was that's exactly the answer i got, but when i re substitute the original values and calculate, The final answer is wrong.
– Maiz Halyym
Aug 2 at 4:20
What's te final answer ?
– user 108128
Aug 2 at 4:21
What I did was, after re-substituting, I used the same limits as were originally given a/2 and 0. The answer should be 1/sqrt3 - pi/6. but i couldn't get this.
– Maiz Halyym
Aug 2 at 4:31
 |Â
show 4 more comments
up vote
0
down vote
Hint:
How about integration by parts
$$int xcdotdfrac x(a^2-x^2)^3/2dx=xintdfrac x(a^2-x^2)^3/2dx-intleft(dfracdxdxintdfrac x(a^2-x^2)^3/2dxright)dx$$
Now choose $sqrta^2-x^2=y$
answered Aug 2 at 9:40
lab bhattacharjee
214k14152263
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Hint: Using substitution $x=asin t$ then
$$int_0^fraca2dfracx^2sqrta^2-x^2^3dx=int_0^pi/6tan^2t dt$$
answered Aug 2 at 4:05
user 108128
18.8k41544
That's exactly what i did, I then expressed tan^2t as sec^2t-1 and got the answer X/(a^2-x^2)-arcsin(x/a)+c. but when i tried to solve it for a/2 and 0, it didn't work out.
– Maiz Halyym
Aug 2 at 4:09
I couldn't read your notes. It's better to write them with mathjax. anyway, $$int tan^2t+1-1 dt=tan t-tBig|_0^pi/6$$ it's a number!
– user 108128
Aug 2 at 4:13
I'll have to learn how to use mathjax, thanks. And also what i meant was that's exactly the answer i got, but when i re substitute the original values and calculate, The final answer is wrong.
– Maiz Halyym
Aug 2 at 4:20
What's te final answer ?
– user 108128
Aug 2 at 4:21
What I did was, after re-substituting, I used the same limits as were originally given a/2 and 0. The answer should be 1/sqrt3 - pi/6. but i couldn't get this.
– Maiz Halyym
Aug 2 at 4:31
 |Â
show 4 more comments
up vote
2
down vote
Hint: Using substitution $x=asin t$ then
$$int_0^fraca2dfracx^2sqrta^2-x^2^3dx=int_0^pi/6tan^2t dt$$
answered Aug 2 at 4:05
user 108128
18.8k41544
That's exactly what i did, I then expressed tan^2t as sec^2t-1 and got the answer X/(a^2-x^2)-arcsin(x/a)+c. but when i tried to solve it for a/2 and 0, it didn't work out.
– Maiz Halyym
Aug 2 at 4:09
I couldn't read your notes. It's better to write them with mathjax. anyway, $$int tan^2t+1-1 dt=tan t-tBig|_0^pi/6$$ it's a number!
– user 108128
Aug 2 at 4:13
I'll have to learn how to use mathjax, thanks. And also what i meant was that's exactly the answer i got, but when i re substitute the original values and calculate, The final answer is wrong.
– Maiz Halyym
Aug 2 at 4:20
What's te final answer ?
– user 108128
Aug 2 at 4:21
What I did was, after re-substituting, I used the same limits as were originally given a/2 and 0. The answer should be 1/sqrt3 - pi/6. but i couldn't get this.
– Maiz Halyym
Aug 2 at 4:31
 |Â
show 4 more comments
up vote
2
down vote
up vote
2
down vote
Hint: Using substitution $x=asin t$ then
$$int_0^fraca2dfracx^2sqrta^2-x^2^3dx=int_0^pi/6tan^2t dt$$
answered Aug 2 at 4:05
user 108128
18.8k41544
Hint: Using substitution $x=asin t$ then
$$int_0^fraca2dfracx^2sqrta^2-x^2^3dx=int_0^pi/6tan^2t dt$$
answered Aug 2 at 4:05
user 108128
18.8k41544
answered Aug 2 at 4:05
user 108128
18.8k41544
answered Aug 2 at 4:05
user 108128
18.8k41544
answered Aug 2 at 4:05
user 108128
18.8k41544
18.8k41544
That's exactly what i did, I then expressed tan^2t as sec^2t-1 and got the answer X/(a^2-x^2)-arcsin(x/a)+c. but when i tried to solve it for a/2 and 0, it didn't work out.
– Maiz Halyym
Aug 2 at 4:09
I couldn't read your notes. It's better to write them with mathjax. anyway, $$int tan^2t+1-1 dt=tan t-tBig|_0^pi/6$$ it's a number!
– user 108128
Aug 2 at 4:13
I'll have to learn how to use mathjax, thanks. And also what i meant was that's exactly the answer i got, but when i re substitute the original values and calculate, The final answer is wrong.
– Maiz Halyym
Aug 2 at 4:20
What's te final answer ?
– user 108128
Aug 2 at 4:21
What I did was, after re-substituting, I used the same limits as were originally given a/2 and 0. The answer should be 1/sqrt3 - pi/6. but i couldn't get this.
– Maiz Halyym
Aug 2 at 4:31
 |Â
show 4 more comments
That's exactly what i did, I then expressed tan^2t as sec^2t-1 and got the answer X/(a^2-x^2)-arcsin(x/a)+c. but when i tried to solve it for a/2 and 0, it didn't work out.
– Maiz Halyym
Aug 2 at 4:09
I couldn't read your notes. It's better to write them with mathjax. anyway, $$int tan^2t+1-1 dt=tan t-tBig|_0^pi/6$$ it's a number!
– user 108128
Aug 2 at 4:13
I'll have to learn how to use mathjax, thanks. And also what i meant was that's exactly the answer i got, but when i re substitute the original values and calculate, The final answer is wrong.
– Maiz Halyym
Aug 2 at 4:20
What's te final answer ?
– user 108128
Aug 2 at 4:21
What I did was, after re-substituting, I used the same limits as were originally given a/2 and 0. The answer should be 1/sqrt3 - pi/6. but i couldn't get this.
– Maiz Halyym
Aug 2 at 4:31
That's exactly what i did, I then expressed tan^2t as sec^2t-1 and got the answer X/(a^2-x^2)-arcsin(x/a)+c. but when i tried to solve it for a/2 and 0, it didn't work out.
– Maiz Halyym
Aug 2 at 4:09
That's exactly what i did, I then expressed tan^2t as sec^2t-1 and got the answer X/(a^2-x^2)-arcsin(x/a)+c. but when i tried to solve it for a/2 and 0, it didn't work out.
– Maiz Halyym
Aug 2 at 4:09
I couldn't read your notes. It's better to write them with mathjax. anyway, $$int tan^2t+1-1 dt=tan t-tBig|_0^pi/6$$ it's a number!
– user 108128
Aug 2 at 4:13
I couldn't read your notes. It's better to write them with mathjax. anyway, $$int tan^2t+1-1 dt=tan t-tBig|_0^pi/6$$ it's a number!
– user 108128
Aug 2 at 4:13
I'll have to learn how to use mathjax, thanks. And also what i meant was that's exactly the answer i got, but when i re substitute the original values and calculate, The final answer is wrong.
– Maiz Halyym
Aug 2 at 4:20
I'll have to learn how to use mathjax, thanks. And also what i meant was that's exactly the answer i got, but when i re substitute the original values and calculate, The final answer is wrong.
– Maiz Halyym
Aug 2 at 4:20
What's te final answer ?
– user 108128
Aug 2 at 4:21
What's te final answer ?
– user 108128
Aug 2 at 4:21
What I did was, after re-substituting, I used the same limits as were originally given a/2 and 0. The answer should be 1/sqrt3 - pi/6. but i couldn't get this.
– Maiz Halyym
Aug 2 at 4:31
What I did was, after re-substituting, I used the same limits as were originally given a/2 and 0. The answer should be 1/sqrt3 - pi/6. but i couldn't get this.
– Maiz Halyym
Aug 2 at 4:31
 |Â
show 4 more comments
up vote
0
down vote
Hint:
How about integration by parts
$$int xcdotdfrac x(a^2-x^2)^3/2dx=xintdfrac x(a^2-x^2)^3/2dx-intleft(dfracdxdxintdfrac x(a^2-x^2)^3/2dxright)dx$$
Now choose $sqrta^2-x^2=y$
answered Aug 2 at 9:40
lab bhattacharjee
214k14152263
add a comment |Â
up vote
0
down vote
Hint:
How about integration by parts
$$int xcdotdfrac x(a^2-x^2)^3/2dx=xintdfrac x(a^2-x^2)^3/2dx-intleft(dfracdxdxintdfrac x(a^2-x^2)^3/2dxright)dx$$
Now choose $sqrta^2-x^2=y$
answered Aug 2 at 9:40
lab bhattacharjee
214k14152263
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint:
How about integration by parts
$$int xcdotdfrac x(a^2-x^2)^3/2dx=xintdfrac x(a^2-x^2)^3/2dx-intleft(dfracdxdxintdfrac x(a^2-x^2)^3/2dxright)dx$$
Now choose $sqrta^2-x^2=y$
answered Aug 2 at 9:40
lab bhattacharjee
214k14152263
Hint:
How about integration by parts
$$int xcdotdfrac x(a^2-x^2)^3/2dx=xintdfrac x(a^2-x^2)^3/2dx-intleft(dfracdxdxintdfrac x(a^2-x^2)^3/2dxright)dx$$
Now choose $sqrta^2-x^2=y$
answered Aug 2 at 9:40
lab bhattacharjee
214k14152263
answered Aug 2 at 9:40
lab bhattacharjee
214k14152263
answered Aug 2 at 9:40
lab bhattacharjee
214k14152263
answered Aug 2 at 9:40
lab bhattacharjee
214k14152263
214k14152263
add a comment |Â
add a comment |Â
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