Can these integrals be solved by u-substitution using trig identities?
up vote
0
down vote
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$int frac{operatorname{sin} x}{2sin xcos x+5}dx$
$int frac{operatorname{sin}xcos x}{operatorname{sin}x+cos x}dx$
$int frac{dx}{operatorname{sin} x+cos x}$
I'd like to know if there's a way to solve some of these integrals by manipulating it and then doing u substitution, I tried but I'm not very good at trig identities. I got the second one like this:
$int frac{operatorname{sin}2x(operatorname{sin}x-cos x)}{cos 2x}dx$ but didn't know how to go from there.
calculus integration trigonometry indefinite-integrals trigonometric-integrals
edited Nov 25 at 17:32
Martin Sleziak
44.6k7115269
asked Nov 24 at 15:57
Andres Oropeza
355
add a comment |
up vote
0
down vote
favorite
$int frac{operatorname{sin} x}{2sin xcos x+5}dx$
$int frac{operatorname{sin}xcos x}{operatorname{sin}x+cos x}dx$
$int frac{dx}{operatorname{sin} x+cos x}$
I'd like to know if there's a way to solve some of these integrals by manipulating it and then doing u substitution, I tried but I'm not very good at trig identities. I got the second one like this:
$int frac{operatorname{sin}2x(operatorname{sin}x-cos x)}{cos 2x}dx$ but didn't know how to go from there.
calculus integration trigonometry indefinite-integrals trigonometric-integrals
edited Nov 25 at 17:32
Martin Sleziak
44.6k7115269
asked Nov 24 at 15:57
Andres Oropeza
355
Single question please
– lab bhattacharjee
Nov 24 at 16:01
Integrals of these types are quite often easily solved using the Weierstrass Substitution (en.wikipedia.org/wiki/Tangent_half-angle_substitution).
– DavidG
Dec 9 at 10:46
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$int frac{operatorname{sin} x}{2sin xcos x+5}dx$
$int frac{operatorname{sin}xcos x}{operatorname{sin}x+cos x}dx$
$int frac{dx}{operatorname{sin} x+cos x}$
I'd like to know if there's a way to solve some of these integrals by manipulating it and then doing u substitution, I tried but I'm not very good at trig identities. I got the second one like this:
$int frac{operatorname{sin}2x(operatorname{sin}x-cos x)}{cos 2x}dx$ but didn't know how to go from there.
calculus integration trigonometry indefinite-integrals trigonometric-integrals
edited Nov 25 at 17:32
Martin Sleziak
44.6k7115269
asked Nov 24 at 15:57
Andres Oropeza
355
$int frac{operatorname{sin} x}{2sin xcos x+5}dx$
$int frac{operatorname{sin}xcos x}{operatorname{sin}x+cos x}dx$
$int frac{dx}{operatorname{sin} x+cos x}$
I'd like to know if there's a way to solve some of these integrals by manipulating it and then doing u substitution, I tried but I'm not very good at trig identities. I got the second one like this:
$int frac{operatorname{sin}2x(operatorname{sin}x-cos x)}{cos 2x}dx$ but didn't know how to go from there.
calculus integration trigonometry indefinite-integrals trigonometric-integrals
calculus integration trigonometry indefinite-integrals trigonometric-integrals
edited Nov 25 at 17:32
Martin Sleziak
44.6k7115269
asked Nov 24 at 15:57
Andres Oropeza
355
edited Nov 25 at 17:32
Martin Sleziak
44.6k7115269
asked Nov 24 at 15:57
Andres Oropeza
355
edited Nov 25 at 17:32
Martin Sleziak
44.6k7115269
edited Nov 25 at 17:32
Martin Sleziak
44.6k7115269
edited Nov 25 at 17:32
Martin Sleziak
44.6k7115269
44.6k7115269
asked Nov 24 at 15:57
Andres Oropeza
355
asked Nov 24 at 15:57
Andres Oropeza
355
asked Nov 24 at 15:57
Andres Oropeza
355
355
Single question please
– lab bhattacharjee
Nov 24 at 16:01
Integrals of these types are quite often easily solved using the Weierstrass Substitution (en.wikipedia.org/wiki/Tangent_half-angle_substitution).
– DavidG
Dec 9 at 10:46
add a comment |
Single question please
– lab bhattacharjee
Nov 24 at 16:01
Integrals of these types are quite often easily solved using the Weierstrass Substitution (en.wikipedia.org/wiki/Tangent_half-angle_substitution).
– DavidG
Dec 9 at 10:46
Single question please
– lab bhattacharjee
Nov 24 at 16:01
Single question please
– lab bhattacharjee
Nov 24 at 16:01
Integrals of these types are quite often easily solved using the Weierstrass Substitution (en.wikipedia.org/wiki/Tangent_half-angle_substitution).
– DavidG
Dec 9 at 10:46
Integrals of these types are quite often easily solved using the Weierstrass Substitution (en.wikipedia.org/wiki/Tangent_half-angle_substitution).
– DavidG
Dec 9 at 10:46
add a comment |
3 Answers
3
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up vote
1
down vote
accepted
Assuming $operatorname{sen}$ is synonymous with $sin$, the third integral is $$inttfrac{1}{sqrt{2}}csc (x+tfrac{pi}{4})dx=-tfrac{1}{sqrt{2}}ln |csc (x+tfrac{pi}{4})+cot (x+tfrac{pi}{4})|+C=-tfrac{1}{sqrt{2}}ln |cot (tfrac{x}{2}+tfrac{pi}{8})|+C,$$while the first integral is very messy (although integrating it between certain limits might be neater). Just to expand on lab bhattacharjee's treatment of the second integral,$$intfrac{1}{sqrt{8}}frac{sin (2y-tfrac{pi}{2})}{sin y}dy=-intfrac{1}{sqrt{8}}frac{cos 2y}{sin y}dy=intfrac{2sin y-csc y}{sqrt{8}}dy\=frac{-2cos y+ln|csc y+cot y|}{sqrt{8}}+C=frac{-2cos (x+tfrac{pi}{4})+ln|cot (tfrac{x}{2}+tfrac{pi}{8})|}{sqrt{8}}+C.$$
answered Nov 24 at 16:21
J.G.
21.5k21933
add a comment |
up vote
2
down vote
Hint:
$$dfrac{sin xcos x}{sin x+cos x}=dfrac{sin2x}{2sqrt2sin(x+pi/4)}$$
Set $x+pi/4=y$
$cos2y=1-2sin^2y$
answered Nov 24 at 16:05
lab bhattacharjee
222k15155273
add a comment |
up vote
2
down vote
For the first one, write numerator as $$dfrac{2sin x}{2sin xcos x+5}=dfrac{sin x-cos x}{2(2sin xcos x+5)}+dfrac{sin x+cos x}{2(2sin xcos x+5)}$$
For the first part, write $$int(sin x-cos x)dx=u, u^2=1+2sin xcos xiffsin xcos x=?$$
For the second, $$int(sin x+cos x)dx=v, v^2=1-2sin xcos xiffsin xcos x=?$$$
answered Nov 24 at 17:49
lab bhattacharjee
222k15155273
Could you expand a Little bit?
– Andres Oropeza
Nov 24 at 19:21
@AndresOropeza, Please find the updated answer
– lab bhattacharjee
Nov 25 at 16:34
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Assuming $operatorname{sen}$ is synonymous with $sin$, the third integral is $$inttfrac{1}{sqrt{2}}csc (x+tfrac{pi}{4})dx=-tfrac{1}{sqrt{2}}ln |csc (x+tfrac{pi}{4})+cot (x+tfrac{pi}{4})|+C=-tfrac{1}{sqrt{2}}ln |cot (tfrac{x}{2}+tfrac{pi}{8})|+C,$$while the first integral is very messy (although integrating it between certain limits might be neater). Just to expand on lab bhattacharjee's treatment of the second integral,$$intfrac{1}{sqrt{8}}frac{sin (2y-tfrac{pi}{2})}{sin y}dy=-intfrac{1}{sqrt{8}}frac{cos 2y}{sin y}dy=intfrac{2sin y-csc y}{sqrt{8}}dy\=frac{-2cos y+ln|csc y+cot y|}{sqrt{8}}+C=frac{-2cos (x+tfrac{pi}{4})+ln|cot (tfrac{x}{2}+tfrac{pi}{8})|}{sqrt{8}}+C.$$
answered Nov 24 at 16:21
J.G.
21.5k21933
add a comment |
up vote
1
down vote
accepted
Assuming $operatorname{sen}$ is synonymous with $sin$, the third integral is $$inttfrac{1}{sqrt{2}}csc (x+tfrac{pi}{4})dx=-tfrac{1}{sqrt{2}}ln |csc (x+tfrac{pi}{4})+cot (x+tfrac{pi}{4})|+C=-tfrac{1}{sqrt{2}}ln |cot (tfrac{x}{2}+tfrac{pi}{8})|+C,$$while the first integral is very messy (although integrating it between certain limits might be neater). Just to expand on lab bhattacharjee's treatment of the second integral,$$intfrac{1}{sqrt{8}}frac{sin (2y-tfrac{pi}{2})}{sin y}dy=-intfrac{1}{sqrt{8}}frac{cos 2y}{sin y}dy=intfrac{2sin y-csc y}{sqrt{8}}dy\=frac{-2cos y+ln|csc y+cot y|}{sqrt{8}}+C=frac{-2cos (x+tfrac{pi}{4})+ln|cot (tfrac{x}{2}+tfrac{pi}{8})|}{sqrt{8}}+C.$$
answered Nov 24 at 16:21
J.G.
21.5k21933
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Assuming $operatorname{sen}$ is synonymous with $sin$, the third integral is $$inttfrac{1}{sqrt{2}}csc (x+tfrac{pi}{4})dx=-tfrac{1}{sqrt{2}}ln |csc (x+tfrac{pi}{4})+cot (x+tfrac{pi}{4})|+C=-tfrac{1}{sqrt{2}}ln |cot (tfrac{x}{2}+tfrac{pi}{8})|+C,$$while the first integral is very messy (although integrating it between certain limits might be neater). Just to expand on lab bhattacharjee's treatment of the second integral,$$intfrac{1}{sqrt{8}}frac{sin (2y-tfrac{pi}{2})}{sin y}dy=-intfrac{1}{sqrt{8}}frac{cos 2y}{sin y}dy=intfrac{2sin y-csc y}{sqrt{8}}dy\=frac{-2cos y+ln|csc y+cot y|}{sqrt{8}}+C=frac{-2cos (x+tfrac{pi}{4})+ln|cot (tfrac{x}{2}+tfrac{pi}{8})|}{sqrt{8}}+C.$$
answered Nov 24 at 16:21
J.G.
21.5k21933
Assuming $operatorname{sen}$ is synonymous with $sin$, the third integral is $$inttfrac{1}{sqrt{2}}csc (x+tfrac{pi}{4})dx=-tfrac{1}{sqrt{2}}ln |csc (x+tfrac{pi}{4})+cot (x+tfrac{pi}{4})|+C=-tfrac{1}{sqrt{2}}ln |cot (tfrac{x}{2}+tfrac{pi}{8})|+C,$$while the first integral is very messy (although integrating it between certain limits might be neater). Just to expand on lab bhattacharjee's treatment of the second integral,$$intfrac{1}{sqrt{8}}frac{sin (2y-tfrac{pi}{2})}{sin y}dy=-intfrac{1}{sqrt{8}}frac{cos 2y}{sin y}dy=intfrac{2sin y-csc y}{sqrt{8}}dy\=frac{-2cos y+ln|csc y+cot y|}{sqrt{8}}+C=frac{-2cos (x+tfrac{pi}{4})+ln|cot (tfrac{x}{2}+tfrac{pi}{8})|}{sqrt{8}}+C.$$
answered Nov 24 at 16:21
J.G.
21.5k21933
answered Nov 24 at 16:21
J.G.
21.5k21933
answered Nov 24 at 16:21
J.G.
21.5k21933
answered Nov 24 at 16:21
J.G.
21.5k21933
21.5k21933
add a comment |
add a comment |
up vote
2
down vote
Hint:
$$dfrac{sin xcos x}{sin x+cos x}=dfrac{sin2x}{2sqrt2sin(x+pi/4)}$$
Set $x+pi/4=y$
$cos2y=1-2sin^2y$
answered Nov 24 at 16:05
lab bhattacharjee
222k15155273
add a comment |
up vote
2
down vote
Hint:
$$dfrac{sin xcos x}{sin x+cos x}=dfrac{sin2x}{2sqrt2sin(x+pi/4)}$$
Set $x+pi/4=y$
$cos2y=1-2sin^2y$
answered Nov 24 at 16:05
lab bhattacharjee
222k15155273
add a comment |
up vote
2
down vote
up vote
2
down vote
Hint:
$$dfrac{sin xcos x}{sin x+cos x}=dfrac{sin2x}{2sqrt2sin(x+pi/4)}$$
Set $x+pi/4=y$
$cos2y=1-2sin^2y$
answered Nov 24 at 16:05
lab bhattacharjee
222k15155273
Hint:
$$dfrac{sin xcos x}{sin x+cos x}=dfrac{sin2x}{2sqrt2sin(x+pi/4)}$$
Set $x+pi/4=y$
$cos2y=1-2sin^2y$
answered Nov 24 at 16:05
lab bhattacharjee
222k15155273
answered Nov 24 at 16:05
lab bhattacharjee
222k15155273
answered Nov 24 at 16:05
lab bhattacharjee
222k15155273
answered Nov 24 at 16:05
lab bhattacharjee
222k15155273
222k15155273
add a comment |
add a comment |
up vote
2
down vote
For the first one, write numerator as $$dfrac{2sin x}{2sin xcos x+5}=dfrac{sin x-cos x}{2(2sin xcos x+5)}+dfrac{sin x+cos x}{2(2sin xcos x+5)}$$
For the first part, write $$int(sin x-cos x)dx=u, u^2=1+2sin xcos xiffsin xcos x=?$$
For the second, $$int(sin x+cos x)dx=v, v^2=1-2sin xcos xiffsin xcos x=?$$$
answered Nov 24 at 17:49
lab bhattacharjee
222k15155273
Could you expand a Little bit?
– Andres Oropeza
Nov 24 at 19:21
@AndresOropeza, Please find the updated answer
– lab bhattacharjee
Nov 25 at 16:34
add a comment |
up vote
2
down vote
For the first one, write numerator as $$dfrac{2sin x}{2sin xcos x+5}=dfrac{sin x-cos x}{2(2sin xcos x+5)}+dfrac{sin x+cos x}{2(2sin xcos x+5)}$$
For the first part, write $$int(sin x-cos x)dx=u, u^2=1+2sin xcos xiffsin xcos x=?$$
For the second, $$int(sin x+cos x)dx=v, v^2=1-2sin xcos xiffsin xcos x=?$$$
answered Nov 24 at 17:49
lab bhattacharjee
222k15155273
Could you expand a Little bit?
– Andres Oropeza
Nov 24 at 19:21
@AndresOropeza, Please find the updated answer
– lab bhattacharjee
Nov 25 at 16:34
add a comment |
up vote
2
down vote
up vote
2
down vote
For the first one, write numerator as $$dfrac{2sin x}{2sin xcos x+5}=dfrac{sin x-cos x}{2(2sin xcos x+5)}+dfrac{sin x+cos x}{2(2sin xcos x+5)}$$
For the first part, write $$int(sin x-cos x)dx=u, u^2=1+2sin xcos xiffsin xcos x=?$$
For the second, $$int(sin x+cos x)dx=v, v^2=1-2sin xcos xiffsin xcos x=?$$$
answered Nov 24 at 17:49
lab bhattacharjee
222k15155273
For the first one, write numerator as $$dfrac{2sin x}{2sin xcos x+5}=dfrac{sin x-cos x}{2(2sin xcos x+5)}+dfrac{sin x+cos x}{2(2sin xcos x+5)}$$
For the first part, write $$int(sin x-cos x)dx=u, u^2=1+2sin xcos xiffsin xcos x=?$$
For the second, $$int(sin x+cos x)dx=v, v^2=1-2sin xcos xiffsin xcos x=?$$$
answered Nov 24 at 17:49
lab bhattacharjee
222k15155273
edited Nov 25 at 16:34
answered Nov 24 at 17:49
lab bhattacharjee
222k15155273
answered Nov 24 at 17:49
lab bhattacharjee
222k15155273
answered Nov 24 at 17:49
lab bhattacharjee
222k15155273
222k15155273
Could you expand a Little bit?
– Andres Oropeza
Nov 24 at 19:21
@AndresOropeza, Please find the updated answer
– lab bhattacharjee
Nov 25 at 16:34
add a comment |
Could you expand a Little bit?
– Andres Oropeza
Nov 24 at 19:21
@AndresOropeza, Please find the updated answer
– lab bhattacharjee
Nov 25 at 16:34
Could you expand a Little bit?
– Andres Oropeza
Nov 24 at 19:21
Could you expand a Little bit?
– Andres Oropeza
Nov 24 at 19:21
@AndresOropeza, Please find the updated answer
– lab bhattacharjee
Nov 25 at 16:34
@AndresOropeza, Please find the updated answer
– lab bhattacharjee
Nov 25 at 16:34
add a comment |
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Single question please
– lab bhattacharjee
Nov 24 at 16:01
Integrals of these types are quite often easily solved using the Weierstrass Substitution (en.wikipedia.org/wiki/Tangent_half-angle_substitution).
– DavidG
Dec 9 at 10:46