Mixing
Prove $intcos^n x dx = frac{1}n cos^{n-1}x sin x + frac{n-1}{n}intcos^{n-2} x dx$
$begingroup$
I am trying to prove $$intcos^n x dx = frac{1}n cos^{n-1}x sin x + frac{n-1}{n}intcos^{n-2} x dx$$
This problem is a classic, but I seem to be missing one step or the understanding of two steps which I will outline below.
$$I_n := intcos^n x dx = intcos^{n-1} x cos x dx tag{1}$$
First question: why rewrite the original instead of immediately integrating by parts of $int cos^n x dx$?
Integrate by parts with
$$u = cos^{n-1} x, dv = cos x dx implies du = (n-1)cos^{n-2} x cdot -sin x, v = sin x$$
which leads to
$$I_n = sin x cos^{n-1} x +intsin^2 x (n-1) cos^{n-2} x dx tag{2}$$
Since $(n-1)$ is a constant, we can throw it out front of the integral:
$$I_n = sin x cos^{n-1} x +(n-1)intsin^2 x cos^{n-2} x dxtag{3}$$
I can transform the integral a bit because $sin^2 x + cos^2 x = 1 implies sin^2 x = 1-cos^2 x$
$$I_n = sin x cos^{n-1} x + (n-1)int(1-cos^2 x) cos^{n-2} x dx tag{4}$$
According to Wikipedia as noted here, this simplifies to:
$$I_n = sin x cos^{n-1} x + (n-1) int cos^{n-2} x dx - (n-1)int(cos^n x) dx tag{5}$$
Question 2: How did they simplify the integral of $int(1-cos^2 x) dx$ to $int(cos^n x) dx$?
Assuming knowledge of equation 5, I see how to rewrite it as
$$I_n = sin x cos^{n-1} x + (n-1) I_{n-2} x - (n-1) I_{n} tag{6}$$
and solve for $I_n$. I had tried exploiting the fact that
$$cos^2 x = frac{1}{2} cos(2x) + frac{1}{2} $$
and trying to deal with $int 1 dx - int frac{1}{2} cos (2x) + frac{1}{2} dx$
which left me with $frac{x}{2} - frac{1}{4} sin(2x)$ after integrating those pieces. Putting it all together I have:
$$I_n = sin x cos^{n-1} x + (n-1) I_{n-2} x left(-(n-1) (frac{x}{2} - frac{1}{4} sin 2x) right) tag{7}$$
but I'm unsure how to write the last few terms as an expression of $I_{something}$ to get it to match the usual reduction formula of
$$intcos^n x dx = frac{1}n cos^{n-1}x sin x + frac{n-1}{n}intcos^{n-2} x dx$$
calculus integration indefinite-integrals reduction-formula
edited Jan 28 at 1:26
clathratus
5,2691438
asked Sep 10 '12 at 3:32
JoeJoe
3,85442553
$endgroup$
add a comment |
$begingroup$
I am trying to prove $$intcos^n x dx = frac{1}n cos^{n-1}x sin x + frac{n-1}{n}intcos^{n-2} x dx$$
This problem is a classic, but I seem to be missing one step or the understanding of two steps which I will outline below.
$$I_n := intcos^n x dx = intcos^{n-1} x cos x dx tag{1}$$
First question: why rewrite the original instead of immediately integrating by parts of $int cos^n x dx$?
Integrate by parts with
$$u = cos^{n-1} x, dv = cos x dx implies du = (n-1)cos^{n-2} x cdot -sin x, v = sin x$$
which leads to
$$I_n = sin x cos^{n-1} x +intsin^2 x (n-1) cos^{n-2} x dx tag{2}$$
Since $(n-1)$ is a constant, we can throw it out front of the integral:
$$I_n = sin x cos^{n-1} x +(n-1)intsin^2 x cos^{n-2} x dxtag{3}$$
I can transform the integral a bit because $sin^2 x + cos^2 x = 1 implies sin^2 x = 1-cos^2 x$
$$I_n = sin x cos^{n-1} x + (n-1)int(1-cos^2 x) cos^{n-2} x dx tag{4}$$
According to Wikipedia as noted here, this simplifies to:
$$I_n = sin x cos^{n-1} x + (n-1) int cos^{n-2} x dx - (n-1)int(cos^n x) dx tag{5}$$
Question 2: How did they simplify the integral of $int(1-cos^2 x) dx$ to $int(cos^n x) dx$?
Assuming knowledge of equation 5, I see how to rewrite it as
$$I_n = sin x cos^{n-1} x + (n-1) I_{n-2} x - (n-1) I_{n} tag{6}$$
and solve for $I_n$. I had tried exploiting the fact that
$$cos^2 x = frac{1}{2} cos(2x) + frac{1}{2} $$
and trying to deal with $int 1 dx - int frac{1}{2} cos (2x) + frac{1}{2} dx$
which left me with $frac{x}{2} - frac{1}{4} sin(2x)$ after integrating those pieces. Putting it all together I have:
$$I_n = sin x cos^{n-1} x + (n-1) I_{n-2} x left(-(n-1) (frac{x}{2} - frac{1}{4} sin 2x) right) tag{7}$$
but I'm unsure how to write the last few terms as an expression of $I_{something}$ to get it to match the usual reduction formula of
$$intcos^n x dx = frac{1}n cos^{n-1}x sin x + frac{n-1}{n}intcos^{n-2} x dx$$
calculus integration indefinite-integrals reduction-formula
edited Jan 28 at 1:26
clathratus
5,2691438
asked Sep 10 '12 at 3:32
JoeJoe
3,85442553
$endgroup$
2
$begingroup$
Q1: re-writing it as such is simply a way to make the IBP clear. Q2: $(1-cos^2x)cos^{n-2}=cos^{n-2}- cos^n x$ (the product is simply expanded, since it is easier to deal with the sum of two simple integrals than the integral of a product)
$endgroup$
– user39572
Sep 10 '12 at 3:39
add a comment |
$begingroup$
I am trying to prove $$intcos^n x dx = frac{1}n cos^{n-1}x sin x + frac{n-1}{n}intcos^{n-2} x dx$$
This problem is a classic, but I seem to be missing one step or the understanding of two steps which I will outline below.
$$I_n := intcos^n x dx = intcos^{n-1} x cos x dx tag{1}$$
First question: why rewrite the original instead of immediately integrating by parts of $int cos^n x dx$?
Integrate by parts with
$$u = cos^{n-1} x, dv = cos x dx implies du = (n-1)cos^{n-2} x cdot -sin x, v = sin x$$
which leads to
$$I_n = sin x cos^{n-1} x +intsin^2 x (n-1) cos^{n-2} x dx tag{2}$$
Since $(n-1)$ is a constant, we can throw it out front of the integral:
$$I_n = sin x cos^{n-1} x +(n-1)intsin^2 x cos^{n-2} x dxtag{3}$$
I can transform the integral a bit because $sin^2 x + cos^2 x = 1 implies sin^2 x = 1-cos^2 x$
$$I_n = sin x cos^{n-1} x + (n-1)int(1-cos^2 x) cos^{n-2} x dx tag{4}$$
According to Wikipedia as noted here, this simplifies to:
$$I_n = sin x cos^{n-1} x + (n-1) int cos^{n-2} x dx - (n-1)int(cos^n x) dx tag{5}$$
Question 2: How did they simplify the integral of $int(1-cos^2 x) dx$ to $int(cos^n x) dx$?
Assuming knowledge of equation 5, I see how to rewrite it as
$$I_n = sin x cos^{n-1} x + (n-1) I_{n-2} x - (n-1) I_{n} tag{6}$$
and solve for $I_n$. I had tried exploiting the fact that
$$cos^2 x = frac{1}{2} cos(2x) + frac{1}{2} $$
and trying to deal with $int 1 dx - int frac{1}{2} cos (2x) + frac{1}{2} dx$
which left me with $frac{x}{2} - frac{1}{4} sin(2x)$ after integrating those pieces. Putting it all together I have:
$$I_n = sin x cos^{n-1} x + (n-1) I_{n-2} x left(-(n-1) (frac{x}{2} - frac{1}{4} sin 2x) right) tag{7}$$
but I'm unsure how to write the last few terms as an expression of $I_{something}$ to get it to match the usual reduction formula of
$$intcos^n x dx = frac{1}n cos^{n-1}x sin x + frac{n-1}{n}intcos^{n-2} x dx$$
calculus integration indefinite-integrals reduction-formula
edited Jan 28 at 1:26
clathratus
5,2691438
asked Sep 10 '12 at 3:32
JoeJoe
3,85442553
$endgroup$
I am trying to prove $$intcos^n x dx = frac{1}n cos^{n-1}x sin x + frac{n-1}{n}intcos^{n-2} x dx$$
This problem is a classic, but I seem to be missing one step or the understanding of two steps which I will outline below.
$$I_n := intcos^n x dx = intcos^{n-1} x cos x dx tag{1}$$
First question: why rewrite the original instead of immediately integrating by parts of $int cos^n x dx$?
Integrate by parts with
$$u = cos^{n-1} x, dv = cos x dx implies du = (n-1)cos^{n-2} x cdot -sin x, v = sin x$$
which leads to
$$I_n = sin x cos^{n-1} x +intsin^2 x (n-1) cos^{n-2} x dx tag{2}$$
Since $(n-1)$ is a constant, we can throw it out front of the integral:
$$I_n = sin x cos^{n-1} x +(n-1)intsin^2 x cos^{n-2} x dxtag{3}$$
I can transform the integral a bit because $sin^2 x + cos^2 x = 1 implies sin^2 x = 1-cos^2 x$
$$I_n = sin x cos^{n-1} x + (n-1)int(1-cos^2 x) cos^{n-2} x dx tag{4}$$
According to Wikipedia as noted here, this simplifies to:
$$I_n = sin x cos^{n-1} x + (n-1) int cos^{n-2} x dx - (n-1)int(cos^n x) dx tag{5}$$
Question 2: How did they simplify the integral of $int(1-cos^2 x) dx$ to $int(cos^n x) dx$?
Assuming knowledge of equation 5, I see how to rewrite it as
$$I_n = sin x cos^{n-1} x + (n-1) I_{n-2} x - (n-1) I_{n} tag{6}$$
and solve for $I_n$. I had tried exploiting the fact that
$$cos^2 x = frac{1}{2} cos(2x) + frac{1}{2} $$
and trying to deal with $int 1 dx - int frac{1}{2} cos (2x) + frac{1}{2} dx$
which left me with $frac{x}{2} - frac{1}{4} sin(2x)$ after integrating those pieces. Putting it all together I have:
$$I_n = sin x cos^{n-1} x + (n-1) I_{n-2} x left(-(n-1) (frac{x}{2} - frac{1}{4} sin 2x) right) tag{7}$$
but I'm unsure how to write the last few terms as an expression of $I_{something}$ to get it to match the usual reduction formula of
$$intcos^n x dx = frac{1}n cos^{n-1}x sin x + frac{n-1}{n}intcos^{n-2} x dx$$
calculus integration indefinite-integrals reduction-formula
calculus integration indefinite-integrals reduction-formula
edited Jan 28 at 1:26
clathratus
5,2691438
asked Sep 10 '12 at 3:32
JoeJoe
3,85442553
edited Jan 28 at 1:26
clathratus
5,2691438
asked Sep 10 '12 at 3:32
JoeJoe
3,85442553
edited Jan 28 at 1:26
clathratus
5,2691438
edited Jan 28 at 1:26
clathratus
5,2691438
edited Jan 28 at 1:26
clathratus
5,2691438
5,2691438
asked Sep 10 '12 at 3:32
JoeJoe
3,85442553
asked Sep 10 '12 at 3:32
JoeJoe
3,85442553
asked Sep 10 '12 at 3:32
JoeJoe
3,85442553
3,85442553
2
$begingroup$
Q1: re-writing it as such is simply a way to make the IBP clear. Q2: $(1-cos^2x)cos^{n-2}=cos^{n-2}- cos^n x$ (the product is simply expanded, since it is easier to deal with the sum of two simple integrals than the integral of a product)
$endgroup$
– user39572
Sep 10 '12 at 3:39
add a comment |
2
$begingroup$
Q1: re-writing it as such is simply a way to make the IBP clear. Q2: $(1-cos^2x)cos^{n-2}=cos^{n-2}- cos^n x$ (the product is simply expanded, since it is easier to deal with the sum of two simple integrals than the integral of a product)
$endgroup$
– user39572
Sep 10 '12 at 3:39
2
2
$begingroup$
Q1: re-writing it as such is simply a way to make the IBP clear. Q2: $(1-cos^2x)cos^{n-2}=cos^{n-2}- cos^n x$ (the product is simply expanded, since it is easier to deal with the sum of two simple integrals than the integral of a product)
$endgroup$
– user39572
Sep 10 '12 at 3:39
$begingroup$
Q1: re-writing it as such is simply a way to make the IBP clear. Q2: $(1-cos^2x)cos^{n-2}=cos^{n-2}- cos^n x$ (the product is simply expanded, since it is easier to deal with the sum of two simple integrals than the integral of a product)
$endgroup$
– user39572
Sep 10 '12 at 3:39
add a comment |
2 Answers
2
active
oldest
votes
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I guess your problem is step four:
$$I_n = sin x cos^{n-1} x + (n-1)int(1-cos^2 x) cos^{n-2} x dx tag{4}$$
Note that
$$begin{align}int(1-cos^2 x) cos^{n-2} x dx & = int cos^{n-2} x dx-int cos^2 x; cos^{n-2} x dx \ & =int cos^{n-2} x dx-int cos^{n} x dx end{align}$$
so we get
$$begin{align}I_n & = sin x cos^{n-1} x + (n-1)int cos^{n-2} x dx-(n-1)int cos^{n} x dx \ I_n & = sin x cos^{n-1} x + (n-1)I_{n-2}-(n-1)I_n end{align}$$
or
$$nI_n = sin x cos^{n-1} x + (n-1)I_{n-2}$$
$$I_n =frac{ sin x cos^{n-1} x}n + frac{n-1}nI_{n-2}$$
edited Jul 9 '18 at 17:23
JB-Franco
3201320
answered Sep 10 '12 at 3:41
Pedro Tamaroff♦Pedro Tamaroff
97.5k10153297
$endgroup$
$begingroup$
Ah, just what I was looking for. Thanks Peter.
$endgroup$
– Joe
Sep 10 '12 at 3:58
add a comment |
$begingroup$
How about verifying using differentiation? We are asked to show
$$int_0^x cos^n t dt = frac{1}{n} cos^{n-1} x sin x + frac{n-1}{n} int^x_0 cos^{n-2} t dt + C$$
for some constant $C$. This is true iff the derivatives of each side are equal.
Differentiating the right side, we get using the product rule
$$RHS = frac{n-1}{n} cos^{n-2} x (-sin x) sin x + frac{1}{n} cos^{n-1} x cos x + frac{n-1}{n} cos^{n-2} x,$$
then combining,
$$RHS = frac{n-1}{n} (1-sin^2 x) cos^{n-2} x + frac{1}{n} cos^n x$$
and finally using $cos^2 x + sin^2 x = 1$,
$$RHS = frac{n-1}{n}cos^n x+frac{1}{n} cos^nx = cos^n x,$$
which is the what we want.
answered Sep 10 '12 at 3:52
abnryabnry
11.7k22366
$endgroup$
$begingroup$
Nice idea, thanks.
$endgroup$
– Joe
Sep 10 '12 at 3:58
add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
I guess your problem is step four:
$$I_n = sin x cos^{n-1} x + (n-1)int(1-cos^2 x) cos^{n-2} x dx tag{4}$$
Note that
$$begin{align}int(1-cos^2 x) cos^{n-2} x dx & = int cos^{n-2} x dx-int cos^2 x; cos^{n-2} x dx \ & =int cos^{n-2} x dx-int cos^{n} x dx end{align}$$
so we get
$$begin{align}I_n & = sin x cos^{n-1} x + (n-1)int cos^{n-2} x dx-(n-1)int cos^{n} x dx \ I_n & = sin x cos^{n-1} x + (n-1)I_{n-2}-(n-1)I_n end{align}$$
or
$$nI_n = sin x cos^{n-1} x + (n-1)I_{n-2}$$
$$I_n =frac{ sin x cos^{n-1} x}n + frac{n-1}nI_{n-2}$$
edited Jul 9 '18 at 17:23
JB-Franco
3201320
answered Sep 10 '12 at 3:41
Pedro Tamaroff♦Pedro Tamaroff
97.5k10153297
$endgroup$
$begingroup$
Ah, just what I was looking for. Thanks Peter.
$endgroup$
– Joe
Sep 10 '12 at 3:58
add a comment |
$begingroup$
I guess your problem is step four:
$$I_n = sin x cos^{n-1} x + (n-1)int(1-cos^2 x) cos^{n-2} x dx tag{4}$$
Note that
$$begin{align}int(1-cos^2 x) cos^{n-2} x dx & = int cos^{n-2} x dx-int cos^2 x; cos^{n-2} x dx \ & =int cos^{n-2} x dx-int cos^{n} x dx end{align}$$
so we get
$$begin{align}I_n & = sin x cos^{n-1} x + (n-1)int cos^{n-2} x dx-(n-1)int cos^{n} x dx \ I_n & = sin x cos^{n-1} x + (n-1)I_{n-2}-(n-1)I_n end{align}$$
or
$$nI_n = sin x cos^{n-1} x + (n-1)I_{n-2}$$
$$I_n =frac{ sin x cos^{n-1} x}n + frac{n-1}nI_{n-2}$$
edited Jul 9 '18 at 17:23
JB-Franco
3201320
answered Sep 10 '12 at 3:41
Pedro Tamaroff♦Pedro Tamaroff
97.5k10153297
$endgroup$
$begingroup$
Ah, just what I was looking for. Thanks Peter.
$endgroup$
– Joe
Sep 10 '12 at 3:58
add a comment |
$begingroup$
I guess your problem is step four:
$$I_n = sin x cos^{n-1} x + (n-1)int(1-cos^2 x) cos^{n-2} x dx tag{4}$$
Note that
$$begin{align}int(1-cos^2 x) cos^{n-2} x dx & = int cos^{n-2} x dx-int cos^2 x; cos^{n-2} x dx \ & =int cos^{n-2} x dx-int cos^{n} x dx end{align}$$
so we get
$$begin{align}I_n & = sin x cos^{n-1} x + (n-1)int cos^{n-2} x dx-(n-1)int cos^{n} x dx \ I_n & = sin x cos^{n-1} x + (n-1)I_{n-2}-(n-1)I_n end{align}$$
or
$$nI_n = sin x cos^{n-1} x + (n-1)I_{n-2}$$
$$I_n =frac{ sin x cos^{n-1} x}n + frac{n-1}nI_{n-2}$$
edited Jul 9 '18 at 17:23
JB-Franco
3201320
answered Sep 10 '12 at 3:41
Pedro Tamaroff♦Pedro Tamaroff
97.5k10153297
$endgroup$
I guess your problem is step four:
$$I_n = sin x cos^{n-1} x + (n-1)int(1-cos^2 x) cos^{n-2} x dx tag{4}$$
Note that
$$begin{align}int(1-cos^2 x) cos^{n-2} x dx & = int cos^{n-2} x dx-int cos^2 x; cos^{n-2} x dx \ & =int cos^{n-2} x dx-int cos^{n} x dx end{align}$$
so we get
$$begin{align}I_n & = sin x cos^{n-1} x + (n-1)int cos^{n-2} x dx-(n-1)int cos^{n} x dx \ I_n & = sin x cos^{n-1} x + (n-1)I_{n-2}-(n-1)I_n end{align}$$
or
$$nI_n = sin x cos^{n-1} x + (n-1)I_{n-2}$$
$$I_n =frac{ sin x cos^{n-1} x}n + frac{n-1}nI_{n-2}$$
edited Jul 9 '18 at 17:23
JB-Franco
3201320
answered Sep 10 '12 at 3:41
Pedro Tamaroff♦Pedro Tamaroff
97.5k10153297
edited Jul 9 '18 at 17:23
JB-Franco
3201320
edited Jul 9 '18 at 17:23
JB-Franco
3201320
edited Jul 9 '18 at 17:23
JB-Franco
3201320
3201320
answered Sep 10 '12 at 3:41
Pedro Tamaroff♦Pedro Tamaroff
97.5k10153297
answered Sep 10 '12 at 3:41
Pedro Tamaroff♦Pedro Tamaroff
97.5k10153297
answered Sep 10 '12 at 3:41
Pedro Tamaroff♦Pedro Tamaroff
97.5k10153297
97.5k10153297
$begingroup$
Ah, just what I was looking for. Thanks Peter.
$endgroup$
– Joe
Sep 10 '12 at 3:58
add a comment |
$begingroup$
Ah, just what I was looking for. Thanks Peter.
$endgroup$
– Joe
Sep 10 '12 at 3:58
$begingroup$
Ah, just what I was looking for. Thanks Peter.
$endgroup$
– Joe
Sep 10 '12 at 3:58
$begingroup$
Ah, just what I was looking for. Thanks Peter.
$endgroup$
– Joe
Sep 10 '12 at 3:58
add a comment |
$begingroup$
How about verifying using differentiation? We are asked to show
$$int_0^x cos^n t dt = frac{1}{n} cos^{n-1} x sin x + frac{n-1}{n} int^x_0 cos^{n-2} t dt + C$$
for some constant $C$. This is true iff the derivatives of each side are equal.
Differentiating the right side, we get using the product rule
$$RHS = frac{n-1}{n} cos^{n-2} x (-sin x) sin x + frac{1}{n} cos^{n-1} x cos x + frac{n-1}{n} cos^{n-2} x,$$
then combining,
$$RHS = frac{n-1}{n} (1-sin^2 x) cos^{n-2} x + frac{1}{n} cos^n x$$
and finally using $cos^2 x + sin^2 x = 1$,
$$RHS = frac{n-1}{n}cos^n x+frac{1}{n} cos^nx = cos^n x,$$
which is the what we want.
answered Sep 10 '12 at 3:52
abnryabnry
11.7k22366
$endgroup$
$begingroup$
Nice idea, thanks.
$endgroup$
– Joe
Sep 10 '12 at 3:58
add a comment |
$begingroup$
How about verifying using differentiation? We are asked to show
$$int_0^x cos^n t dt = frac{1}{n} cos^{n-1} x sin x + frac{n-1}{n} int^x_0 cos^{n-2} t dt + C$$
for some constant $C$. This is true iff the derivatives of each side are equal.
Differentiating the right side, we get using the product rule
$$RHS = frac{n-1}{n} cos^{n-2} x (-sin x) sin x + frac{1}{n} cos^{n-1} x cos x + frac{n-1}{n} cos^{n-2} x,$$
then combining,
$$RHS = frac{n-1}{n} (1-sin^2 x) cos^{n-2} x + frac{1}{n} cos^n x$$
and finally using $cos^2 x + sin^2 x = 1$,
$$RHS = frac{n-1}{n}cos^n x+frac{1}{n} cos^nx = cos^n x,$$
which is the what we want.
answered Sep 10 '12 at 3:52
abnryabnry
11.7k22366
$endgroup$
$begingroup$
Nice idea, thanks.
$endgroup$
– Joe
Sep 10 '12 at 3:58
add a comment |
$begingroup$
How about verifying using differentiation? We are asked to show
$$int_0^x cos^n t dt = frac{1}{n} cos^{n-1} x sin x + frac{n-1}{n} int^x_0 cos^{n-2} t dt + C$$
for some constant $C$. This is true iff the derivatives of each side are equal.
Differentiating the right side, we get using the product rule
$$RHS = frac{n-1}{n} cos^{n-2} x (-sin x) sin x + frac{1}{n} cos^{n-1} x cos x + frac{n-1}{n} cos^{n-2} x,$$
then combining,
$$RHS = frac{n-1}{n} (1-sin^2 x) cos^{n-2} x + frac{1}{n} cos^n x$$
and finally using $cos^2 x + sin^2 x = 1$,
$$RHS = frac{n-1}{n}cos^n x+frac{1}{n} cos^nx = cos^n x,$$
which is the what we want.
answered Sep 10 '12 at 3:52
abnryabnry
11.7k22366
$endgroup$
How about verifying using differentiation? We are asked to show
$$int_0^x cos^n t dt = frac{1}{n} cos^{n-1} x sin x + frac{n-1}{n} int^x_0 cos^{n-2} t dt + C$$
for some constant $C$. This is true iff the derivatives of each side are equal.
Differentiating the right side, we get using the product rule
$$RHS = frac{n-1}{n} cos^{n-2} x (-sin x) sin x + frac{1}{n} cos^{n-1} x cos x + frac{n-1}{n} cos^{n-2} x,$$
then combining,
$$RHS = frac{n-1}{n} (1-sin^2 x) cos^{n-2} x + frac{1}{n} cos^n x$$
and finally using $cos^2 x + sin^2 x = 1$,
$$RHS = frac{n-1}{n}cos^n x+frac{1}{n} cos^nx = cos^n x,$$
which is the what we want.
answered Sep 10 '12 at 3:52
abnryabnry
11.7k22366
edited Aug 21 '13 at 16:07
answered Sep 10 '12 at 3:52
abnryabnry
11.7k22366
answered Sep 10 '12 at 3:52
abnryabnry
11.7k22366
answered Sep 10 '12 at 3:52
abnryabnry
11.7k22366
11.7k22366
$begingroup$
Nice idea, thanks.
$endgroup$
– Joe
Sep 10 '12 at 3:58
add a comment |
$begingroup$
Nice idea, thanks.
$endgroup$
– Joe
Sep 10 '12 at 3:58
$begingroup$
Nice idea, thanks.
$endgroup$
– Joe
Sep 10 '12 at 3:58
$begingroup$
Nice idea, thanks.
$endgroup$
– Joe
Sep 10 '12 at 3:58
add a comment |
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Q1: re-writing it as such is simply a way to make the IBP clear. Q2: $(1-cos^2x)cos^{n-2}=cos^{n-2}- cos^n x$ (the product is simply expanded, since it is easier to deal with the sum of two simple integrals than the integral of a product)
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– user39572
Sep 10 '12 at 3:39