Python, 29 bytes

lambda n:int(bin(n)[:1:-1],2)

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Python, 29 bytes

lambda n:int(bin(n)[:1:-1],2)

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JavaScript (ES6), 30 bytes

f=(n,q=0)=>n?f(n>>1,q*2+n%2):q

This basically calculates the reverse one bit at a time: We start with q = 0; while n is positive, we multiply q by 2, sever the last bit off of n with n>>1, and add it to q with +n%2. When n reaches 0, the number has been successfully reversed, and we return q.

Thanks to JS's long built-in names, solving this challenge the easy way takes 44 bytes:

n=>+[...n.toString(2),'0b'].reverse().join``

Using recursion and a string, you can get an alternate 32 byte solution:

f=(n,q='0b')=>n?f(n>>1,q+n%2):+q

J, 6 bytes

|.&.#:

|. reverse

&. under

#: base 2

Java 8, 53 47 46 bytes

-4 bytes thanks to Titus

This is a lambda expression which has the same principle as ETH's answer (although recursion would have been too verbose in Java, so we loop instead):

x->{int t=0;for(;x>0;x/=2)t=t*2+x%2;return t;}

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Expanded, ungolfed and commented, it looks like this:

x -> { 
    int t = 0;           // Initialize result holder   
    while (x > 0) {      // While there are bits left in input:
        t <<= 1;         //   Add a 0 bit at the end of result
        t += x%2;        //   Set it to the last bit of x
        x >>= 1;         //   Hack off the last bit of x
    }              
    return t;            // Return the final result
};

05AB1E, 3 bytes

bRC

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Jelly, 3 bytes

BUḄ

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B   # convert to binary
 U  # reverse
  Ḅ # convert to decimal

Japt, 5 bytes

¢w n2

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Bash/Unix utilities, 24 bytes

dc<<<2i`dc<<<2o$1p|rev`p

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Ruby, 29 bytes

->n{("%b"%n).reverse.to_i(2)}

CJam, 8 bytes

ri2bW%2b

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Explanation

ri          e# Read integer
  2b        e# Convert to binary array
    W%      e# Reverse array
      2b    e# Convert from binary array to number. Implicitly display

Perl 6, 19 bytes

{:2(.base(2).flip)}

Mathematica, 19 bytes

#~IntegerReverse~2&

Java 8, 84 79 73 72 bytes

a->a.valueOf(new StringBuffer(a.toString(a,2)).reverse().toString(),2);

Lambda expression. Converts to binary representation, reverses, parses the string in base 2.

-5 bytes thanks to Poke!

-6 more bytes thanks to Poke!

-1, Poke has done more golfing on this than me by now.

Haskell, 36 bytes

0!b=b
a!b=div a 2!(b+b+mod a 2)
(!0)

Same algorithm (and length!) as ETHproductions’ JavaScript answer.

MATL, 4 bytes

BPXB

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Explanation

B     % Input a number implicitly. Convert to binary array
P     % Reverse array
XB    % Convert from binary array to number. Display implicitly

Mathematica, 38 bytes

#+##&~Fold~Reverse[#~IntegerDigits~2]&

Pyth, 6 bytes

i_.BQ2

Test suite available here.

Explanation

i_.BQ2
    Q     eval(input())
  .B      convert to binary
 _        reverse
i    2    convert from base 2 to base 10

Groovy, 46 bytes

{0.parseInt(0.toBinaryString(it).reverse(),2)}

k, 18 bytes

{2/:|X@&|\X:0b\:x}

Example:

k){2/:|X@&|\X:0b\:x}6
3

Batch, 62 bytes

@set/an=%1/2,r=%2+%1%%2
@if %n% gtr 0 %0 %n% %r%*2
@echo %r%

Explanation: On the first pass, %1 contains the input parameter while %2 is empty. We therefore evaluate n as half of %1 and r as +%1 modulo 2 (the % operator has to be doubled to quote it). If n is not zero, we then call ourselves tail recursively passing in n and an expression that gets evaluated on the next pass effectively doubling r each time.

C#, 98 bytes

using System.Linq;using b=System.Convert;a=>b.ToInt64(string.Concat(b.ToString(a,2).Reverse()),2);

PHP, 44 bytes

for(;$n=&$argv[1];$n>>=1)$r+=$r+$n%2;echo$r;

iterative: while input has set bits, pop a bit from input and push it to output. Run with -nr.

reursive, 52 bytes

function r($n,$r=0){return$n?r($n>>1,$r*2+$n%2):$r;}

PHP, 36 bytes

<?=bindec(strrev(decbin($argv[1])));

using builtins: convert to base2, reverse string, convert to decimal
(same as Robert Lerner´s answer, but using a parameter)

Perl 5 (5.12+), 34 bytes

say oct"0b".reverse sprintf"%b",<>

(Input on stdin, output on stdout, requires -E or -M5.012 at no cost).

Straightforward but not all that short. Too bad "reverse" and "sprintf" are such weighty keywords.

oct is an odd legacy name; oct "123" gives the decimal equivalent of octal 123 as you'd expect, but oct "0x123" interprets its input as hex and oct "0b101" interprets it as binary.

C, 48 44 bytes

r;f(n){r=n&1;while(n/=2)r=2*r+n%2;return r;}

Previous 44 bytes solution:

r;f(n){r=0;while(n)r=2*(r+n%2),n/=2;return r/2;}

Ungolfed and usage:

r;
f(n){
 r=n&1;
  while(n/=2)
   r=2*r+n%2;
 return r;
}


main() {
#define P(x) printf("%d %d\n",x,f(x))
P(1);
P(2);
P(3);
P(4);
P(5);
P(6);
P(7);
P(8);
P(9);
P(10);
}

PHP, 29 bytes

<?=bindec(strrev(decbin(1)));

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Explanation

Above evaluates "1", base converts to binary, performs string reversal,
converts back to base 10.