integral of $cosxcos(sinx)dx$
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I set $u=cosx$, $du=-sinxdx$, $dx=frac1-sinx$
$ucos(sinx)dx$
$u [-sin(sinx)cosx(1)]+C$
$frac-(sin(sinx)).(cosx)-sinx + C$
tentative answer: $sin(cosx) + C$
Another source says that the answer is $sin(sinx) + C$, and I wanted to know if and where I went wrong.
calculus real-analysis
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up vote
2
down vote
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I set $u=cosx$, $du=-sinxdx$, $dx=frac1-sinx$
$ucos(sinx)dx$
$u [-sin(sinx)cosx(1)]+C$
$frac-(sin(sinx)).(cosx)-sinx + C$
tentative answer: $sin(cosx) + C$
Another source says that the answer is $sin(sinx) + C$, and I wanted to know if and where I went wrong.
calculus real-analysis
1
Why not u=sinx?
â imranfat
May 5 '14 at 21:24
try u = sin(x). Then du = cos(x)dx. Your integral then becomes $int cos(u) du= sin(u) = sin(sin(x))$.
â user137481
May 5 '14 at 21:25
Big hint: $displaystyleint cos(sin x),underbraceBig(cos x,dxBig)_textHINT$. The point is that you want $du=cos x,dx$ so $u=sin x$. $qquad$
â Michael Hardy
May 5 '14 at 22:00
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up vote
2
down vote
favorite
up vote
2
down vote
favorite
I set $u=cosx$, $du=-sinxdx$, $dx=frac1-sinx$
$ucos(sinx)dx$
$u [-sin(sinx)cosx(1)]+C$
$frac-(sin(sinx)).(cosx)-sinx + C$
tentative answer: $sin(cosx) + C$
Another source says that the answer is $sin(sinx) + C$, and I wanted to know if and where I went wrong.
calculus real-analysis
I set $u=cosx$, $du=-sinxdx$, $dx=frac1-sinx$
$ucos(sinx)dx$
$u [-sin(sinx)cosx(1)]+C$
$frac-(sin(sinx)).(cosx)-sinx + C$
tentative answer: $sin(cosx) + C$
Another source says that the answer is $sin(sinx) + C$, and I wanted to know if and where I went wrong.
calculus real-analysis
edited Jun 3 '16 at 5:44
user 170039
10.3k42361
10.3k42361
asked May 5 '14 at 21:20
Gabe Carr
162
162
1
Why not u=sinx?
â imranfat
May 5 '14 at 21:24
try u = sin(x). Then du = cos(x)dx. Your integral then becomes $int cos(u) du= sin(u) = sin(sin(x))$.
â user137481
May 5 '14 at 21:25
Big hint: $displaystyleint cos(sin x),underbraceBig(cos x,dxBig)_textHINT$. The point is that you want $du=cos x,dx$ so $u=sin x$. $qquad$
â Michael Hardy
May 5 '14 at 22:00
add a comment |Â
1
Why not u=sinx?
â imranfat
May 5 '14 at 21:24
try u = sin(x). Then du = cos(x)dx. Your integral then becomes $int cos(u) du= sin(u) = sin(sin(x))$.
â user137481
May 5 '14 at 21:25
Big hint: $displaystyleint cos(sin x),underbraceBig(cos x,dxBig)_textHINT$. The point is that you want $du=cos x,dx$ so $u=sin x$. $qquad$
â Michael Hardy
May 5 '14 at 22:00
1
1
Why not u=sinx?
â imranfat
May 5 '14 at 21:24
Why not u=sinx?
â imranfat
May 5 '14 at 21:24
try u = sin(x). Then du = cos(x)dx. Your integral then becomes $int cos(u) du= sin(u) = sin(sin(x))$.
â user137481
May 5 '14 at 21:25
try u = sin(x). Then du = cos(x)dx. Your integral then becomes $int cos(u) du= sin(u) = sin(sin(x))$.
â user137481
May 5 '14 at 21:25
Big hint: $displaystyleint cos(sin x),underbraceBig(cos x,dxBig)_textHINT$. The point is that you want $du=cos x,dx$ so $u=sin x$. $qquad$
â Michael Hardy
May 5 '14 at 22:00
Big hint: $displaystyleint cos(sin x),underbraceBig(cos x,dxBig)_textHINT$. The point is that you want $du=cos x,dx$ so $u=sin x$. $qquad$
â Michael Hardy
May 5 '14 at 22:00
add a comment |Â
3 Answers
3
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up vote
1
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Hint: Let $u=sin(x) implies fracdudx=cos(x)iff du=cos(x)dx$.
Then we have the integral $intcos(u)du$.
add a comment |Â
up vote
0
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Hint:
$$(sin(cos(x)))'=-sin(x)cos(cos(x)),$$
$$(sin(sin(x)))'=cos(x)cos(sin(x)).$$
add a comment |Â
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0
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$$intcos xcos (sin x)dx=intcos (sin x)dleft( sin x right)=sin (sin x)+C$$
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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
Hint: Let $u=sin(x) implies fracdudx=cos(x)iff du=cos(x)dx$.
Then we have the integral $intcos(u)du$.
add a comment |Â
up vote
1
down vote
Hint: Let $u=sin(x) implies fracdudx=cos(x)iff du=cos(x)dx$.
Then we have the integral $intcos(u)du$.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint: Let $u=sin(x) implies fracdudx=cos(x)iff du=cos(x)dx$.
Then we have the integral $intcos(u)du$.
Hint: Let $u=sin(x) implies fracdudx=cos(x)iff du=cos(x)dx$.
Then we have the integral $intcos(u)du$.
edited Jun 25 '14 at 13:52
answered May 5 '14 at 21:24
beep-boop
7,87452856
7,87452856
add a comment |Â
add a comment |Â
up vote
0
down vote
Hint:
$$(sin(cos(x)))'=-sin(x)cos(cos(x)),$$
$$(sin(sin(x)))'=cos(x)cos(sin(x)).$$
add a comment |Â
up vote
0
down vote
Hint:
$$(sin(cos(x)))'=-sin(x)cos(cos(x)),$$
$$(sin(sin(x)))'=cos(x)cos(sin(x)).$$
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint:
$$(sin(cos(x)))'=-sin(x)cos(cos(x)),$$
$$(sin(sin(x)))'=cos(x)cos(sin(x)).$$
Hint:
$$(sin(cos(x)))'=-sin(x)cos(cos(x)),$$
$$(sin(sin(x)))'=cos(x)cos(sin(x)).$$
answered Jul 4 '16 at 11:55
Yves Daoust
112k665206
112k665206
add a comment |Â
add a comment |Â
up vote
0
down vote
$$intcos xcos (sin x)dx=intcos (sin x)dleft( sin x right)=sin (sin x)+C$$
add a comment |Â
up vote
0
down vote
$$intcos xcos (sin x)dx=intcos (sin x)dleft( sin x right)=sin (sin x)+C$$
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$intcos xcos (sin x)dx=intcos (sin x)dleft( sin x right)=sin (sin x)+C$$
$$intcos xcos (sin x)dx=intcos (sin x)dleft( sin x right)=sin (sin x)+C$$
answered Aug 10 at 19:11
Vincent Law
1
1
add a comment |Â
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1
Why not u=sinx?
â imranfat
May 5 '14 at 21:24
try u = sin(x). Then du = cos(x)dx. Your integral then becomes $int cos(u) du= sin(u) = sin(sin(x))$.
â user137481
May 5 '14 at 21:25
Big hint: $displaystyleint cos(sin x),underbraceBig(cos x,dxBig)_textHINT$. The point is that you want $du=cos x,dx$ so $u=sin x$. $qquad$
â Michael Hardy
May 5 '14 at 22:00