How to evaluate $int_pi/6^pi/3fracsqrt[3]sin xsqrt[3]sin x+sqrt[3]cos x$?
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$$int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos x$$
I have tried it using transformation but its getting lengthy..it must be short question. please help
integration definite-integrals
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$$int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos x$$
I have tried it using transformation but its getting lengthy..it must be short question. please help
integration definite-integrals
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up vote
0
down vote
favorite
up vote
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down vote
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$$int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos x$$
I have tried it using transformation but its getting lengthy..it must be short question. please help
integration definite-integrals
$$int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos x$$
I have tried it using transformation but its getting lengthy..it must be short question. please help
integration definite-integrals
integration definite-integrals
edited Nov 5 '16 at 10:46
StubbornAtom
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4,11711136
asked Nov 5 '16 at 10:20
Rajnesh Singh
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2 Answers
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Hint use $f (x)=f (a+b-x ) $ where a,b are lower ,upper limits . Then everything is straightforward
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Using the hint given by Archis Welankar:
$$int_a^b f(x) dx=int_a^bf(a+b-x)dx;\
int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx=int_pi/6^pi/3 frac sqrt[3]sin (pi/2-x) sqrt[3]sin (pi/2-x) + sqrt[3]cos (pi/2-x)dx Rightarrow \
underbraceint_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx_A=underbraceint_pi/6^pi/3 frac sqrt[3]cos x sqrt[3]cos x + sqrt[3]sin xdx_B Rightarrow \
A-B=0.$$
Also:
$$int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx=int_pi/6^pi/3 frac sqrt[3]sin x+sqrt[3]cos x-sqrt[3]cos x sqrt[3]sin x + sqrt[3]cos xdx=\
int_pi/6^pi/3 1-frac sqrt[3]cos x sqrt[3]cos x + sqrt[3]sin xdx Rightarrow \
A+B=fracpi6.$$
Hence:
$$A=fracpi12.$$
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint use $f (x)=f (a+b-x ) $ where a,b are lower ,upper limits . Then everything is straightforward
add a comment |Â
up vote
1
down vote
accepted
Hint use $f (x)=f (a+b-x ) $ where a,b are lower ,upper limits . Then everything is straightforward
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint use $f (x)=f (a+b-x ) $ where a,b are lower ,upper limits . Then everything is straightforward
Hint use $f (x)=f (a+b-x ) $ where a,b are lower ,upper limits . Then everything is straightforward
answered Nov 5 '16 at 10:23
Archis Welankar
11.9k41239
11.9k41239
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Using the hint given by Archis Welankar:
$$int_a^b f(x) dx=int_a^bf(a+b-x)dx;\
int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx=int_pi/6^pi/3 frac sqrt[3]sin (pi/2-x) sqrt[3]sin (pi/2-x) + sqrt[3]cos (pi/2-x)dx Rightarrow \
underbraceint_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx_A=underbraceint_pi/6^pi/3 frac sqrt[3]cos x sqrt[3]cos x + sqrt[3]sin xdx_B Rightarrow \
A-B=0.$$
Also:
$$int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx=int_pi/6^pi/3 frac sqrt[3]sin x+sqrt[3]cos x-sqrt[3]cos x sqrt[3]sin x + sqrt[3]cos xdx=\
int_pi/6^pi/3 1-frac sqrt[3]cos x sqrt[3]cos x + sqrt[3]sin xdx Rightarrow \
A+B=fracpi6.$$
Hence:
$$A=fracpi12.$$
add a comment |Â
up vote
0
down vote
Using the hint given by Archis Welankar:
$$int_a^b f(x) dx=int_a^bf(a+b-x)dx;\
int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx=int_pi/6^pi/3 frac sqrt[3]sin (pi/2-x) sqrt[3]sin (pi/2-x) + sqrt[3]cos (pi/2-x)dx Rightarrow \
underbraceint_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx_A=underbraceint_pi/6^pi/3 frac sqrt[3]cos x sqrt[3]cos x + sqrt[3]sin xdx_B Rightarrow \
A-B=0.$$
Also:
$$int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx=int_pi/6^pi/3 frac sqrt[3]sin x+sqrt[3]cos x-sqrt[3]cos x sqrt[3]sin x + sqrt[3]cos xdx=\
int_pi/6^pi/3 1-frac sqrt[3]cos x sqrt[3]cos x + sqrt[3]sin xdx Rightarrow \
A+B=fracpi6.$$
Hence:
$$A=fracpi12.$$
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Using the hint given by Archis Welankar:
$$int_a^b f(x) dx=int_a^bf(a+b-x)dx;\
int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx=int_pi/6^pi/3 frac sqrt[3]sin (pi/2-x) sqrt[3]sin (pi/2-x) + sqrt[3]cos (pi/2-x)dx Rightarrow \
underbraceint_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx_A=underbraceint_pi/6^pi/3 frac sqrt[3]cos x sqrt[3]cos x + sqrt[3]sin xdx_B Rightarrow \
A-B=0.$$
Also:
$$int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx=int_pi/6^pi/3 frac sqrt[3]sin x+sqrt[3]cos x-sqrt[3]cos x sqrt[3]sin x + sqrt[3]cos xdx=\
int_pi/6^pi/3 1-frac sqrt[3]cos x sqrt[3]cos x + sqrt[3]sin xdx Rightarrow \
A+B=fracpi6.$$
Hence:
$$A=fracpi12.$$
Using the hint given by Archis Welankar:
$$int_a^b f(x) dx=int_a^bf(a+b-x)dx;\
int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx=int_pi/6^pi/3 frac sqrt[3]sin (pi/2-x) sqrt[3]sin (pi/2-x) + sqrt[3]cos (pi/2-x)dx Rightarrow \
underbraceint_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx_A=underbraceint_pi/6^pi/3 frac sqrt[3]cos x sqrt[3]cos x + sqrt[3]sin xdx_B Rightarrow \
A-B=0.$$
Also:
$$int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx=int_pi/6^pi/3 frac sqrt[3]sin x+sqrt[3]cos x-sqrt[3]cos x sqrt[3]sin x + sqrt[3]cos xdx=\
int_pi/6^pi/3 1-frac sqrt[3]cos x sqrt[3]cos x + sqrt[3]sin xdx Rightarrow \
A+B=fracpi6.$$
Hence:
$$A=fracpi12.$$
answered Sep 9 at 12:38
farruhota
15.6k2734
15.6k2734
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