Evalutating an indefinit integral
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How can I evaluate the following integral?
$$
intfraccosx1+sin2x dx
$$
I tried the following way, but I was not able to proceed further:
$$
begingather
I&=intfraccosxleft(sinx+cosxright)^2 dx\
&= intfracsecxleft(1+tanxright)^2 dx
endgather
$$
calculus trigonometry indefinite-integrals
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up vote
0
down vote
favorite
How can I evaluate the following integral?
$$
intfraccosx1+sin2x dx
$$
I tried the following way, but I was not able to proceed further:
$$
begingather
I&=intfraccosxleft(sinx+cosxright)^2 dx\
&= intfracsecxleft(1+tanxright)^2 dx
endgather
$$
calculus trigonometry indefinite-integrals
Hope you'll satisty with the solution.
â Arpit Yadav
Aug 12 at 3:01
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How can I evaluate the following integral?
$$
intfraccosx1+sin2x dx
$$
I tried the following way, but I was not able to proceed further:
$$
begingather
I&=intfraccosxleft(sinx+cosxright)^2 dx\
&= intfracsecxleft(1+tanxright)^2 dx
endgather
$$
calculus trigonometry indefinite-integrals
How can I evaluate the following integral?
$$
intfraccosx1+sin2x dx
$$
I tried the following way, but I was not able to proceed further:
$$
begingather
I&=intfraccosxleft(sinx+cosxright)^2 dx\
&= intfracsecxleft(1+tanxright)^2 dx
endgather
$$
calculus trigonometry indefinite-integrals
edited Aug 12 at 1:42
Math Lover
12.5k21232
12.5k21232
asked Aug 12 at 0:17
Hussien Mohamed
634112
634112
Hope you'll satisty with the solution.
â Arpit Yadav
Aug 12 at 3:01
add a comment |Â
Hope you'll satisty with the solution.
â Arpit Yadav
Aug 12 at 3:01
Hope you'll satisty with the solution.
â Arpit Yadav
Aug 12 at 3:01
Hope you'll satisty with the solution.
â Arpit Yadav
Aug 12 at 3:01
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
Hint:
Write numerator as $$2cos x=cos x+sin x +(cos x-sin x)$$
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Using lab bhattacharjee result,$$I=int fraccos x+ sin x(cos x+ sin x)^2dxspace +int fraccos x-sin x(cos x+sin x)^2dx$$
$$I=I_1+I_2$$
$$I_1=int frac1cos x+sin xdx$$
$$I_1=intfrac1sqrt2(cos xcdotcosfracpi4+sin xcdot sinfracpi4)dx$$
$$I_1=intfracsec(x-pi/4)sqrt2dx$$
$$I_1=frac1sqrt2cdotln[(sec(x-pi/4)+tan(x-pi/4)]+C_1$$
Now$$I_2=int fraccos x-sin x(cos x+sin x)^2dx$$
$$sin x+cos x=timplies(cos x-sin x)dx=dt$$
$$I_2=intfrac 1t^2space dt$$
$$I_2=-frac1t+C_2=-frac1sin x+cos x+C_2$$
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Hint:
Write numerator as $$2cos x=cos x+sin x +(cos x-sin x)$$
add a comment |Â
up vote
2
down vote
Hint:
Write numerator as $$2cos x=cos x+sin x +(cos x-sin x)$$
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint:
Write numerator as $$2cos x=cos x+sin x +(cos x-sin x)$$
Hint:
Write numerator as $$2cos x=cos x+sin x +(cos x-sin x)$$
answered Aug 12 at 1:34
lab bhattacharjee
215k14152264
215k14152264
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up vote
0
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Using lab bhattacharjee result,$$I=int fraccos x+ sin x(cos x+ sin x)^2dxspace +int fraccos x-sin x(cos x+sin x)^2dx$$
$$I=I_1+I_2$$
$$I_1=int frac1cos x+sin xdx$$
$$I_1=intfrac1sqrt2(cos xcdotcosfracpi4+sin xcdot sinfracpi4)dx$$
$$I_1=intfracsec(x-pi/4)sqrt2dx$$
$$I_1=frac1sqrt2cdotln[(sec(x-pi/4)+tan(x-pi/4)]+C_1$$
Now$$I_2=int fraccos x-sin x(cos x+sin x)^2dx$$
$$sin x+cos x=timplies(cos x-sin x)dx=dt$$
$$I_2=intfrac 1t^2space dt$$
$$I_2=-frac1t+C_2=-frac1sin x+cos x+C_2$$
add a comment |Â
up vote
0
down vote
Using lab bhattacharjee result,$$I=int fraccos x+ sin x(cos x+ sin x)^2dxspace +int fraccos x-sin x(cos x+sin x)^2dx$$
$$I=I_1+I_2$$
$$I_1=int frac1cos x+sin xdx$$
$$I_1=intfrac1sqrt2(cos xcdotcosfracpi4+sin xcdot sinfracpi4)dx$$
$$I_1=intfracsec(x-pi/4)sqrt2dx$$
$$I_1=frac1sqrt2cdotln[(sec(x-pi/4)+tan(x-pi/4)]+C_1$$
Now$$I_2=int fraccos x-sin x(cos x+sin x)^2dx$$
$$sin x+cos x=timplies(cos x-sin x)dx=dt$$
$$I_2=intfrac 1t^2space dt$$
$$I_2=-frac1t+C_2=-frac1sin x+cos x+C_2$$
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Using lab bhattacharjee result,$$I=int fraccos x+ sin x(cos x+ sin x)^2dxspace +int fraccos x-sin x(cos x+sin x)^2dx$$
$$I=I_1+I_2$$
$$I_1=int frac1cos x+sin xdx$$
$$I_1=intfrac1sqrt2(cos xcdotcosfracpi4+sin xcdot sinfracpi4)dx$$
$$I_1=intfracsec(x-pi/4)sqrt2dx$$
$$I_1=frac1sqrt2cdotln[(sec(x-pi/4)+tan(x-pi/4)]+C_1$$
Now$$I_2=int fraccos x-sin x(cos x+sin x)^2dx$$
$$sin x+cos x=timplies(cos x-sin x)dx=dt$$
$$I_2=intfrac 1t^2space dt$$
$$I_2=-frac1t+C_2=-frac1sin x+cos x+C_2$$
Using lab bhattacharjee result,$$I=int fraccos x+ sin x(cos x+ sin x)^2dxspace +int fraccos x-sin x(cos x+sin x)^2dx$$
$$I=I_1+I_2$$
$$I_1=int frac1cos x+sin xdx$$
$$I_1=intfrac1sqrt2(cos xcdotcosfracpi4+sin xcdot sinfracpi4)dx$$
$$I_1=intfracsec(x-pi/4)sqrt2dx$$
$$I_1=frac1sqrt2cdotln[(sec(x-pi/4)+tan(x-pi/4)]+C_1$$
Now$$I_2=int fraccos x-sin x(cos x+sin x)^2dx$$
$$sin x+cos x=timplies(cos x-sin x)dx=dt$$
$$I_2=intfrac 1t^2space dt$$
$$I_2=-frac1t+C_2=-frac1sin x+cos x+C_2$$
edited Aug 12 at 3:00
answered Aug 12 at 2:54
Arpit Yadav
331215
331215
add a comment |Â
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Hope you'll satisty with the solution.
â Arpit Yadav
Aug 12 at 3:01