I have the folllwoing integral ∫ 0 π / 4 ∫ 0 2 π sin 2 ⁡ ( ρ ) sin ⁡ ( θ ) − cos ⁡ ( ρ ) sin 2 ⁡ ( ρ ) sin ⁡ ( θ ) d θ d ρand I would like to know why exaclty one can rewrite it as follows ( ∫ 0 π / 4 sin 2 ⁡ ( ρ ) − cos ⁡ ( ρ ) sin 2 ⁡ ( ρ ) d ρ ) ∫ 0 2 π sin ⁡ θ d θ .

Question

I have the folllwoing integral      ∫    0          π              /            4            ∫    0          2      π            sin    2    ⁡  (  ρ  )  sin  ⁡  (  θ  )  −  cos  ⁡  (  ρ  )      sin    2    ⁡  (  ρ  )  sin  ⁡  (  θ  )     d  θ  d  ρand I would like to know why exaclty one can rewrite it as follows      (          ∫      0              π                  /                4                    sin      2        ⁡    (    ρ    )    −    cos    ⁡    (    ρ    )          sin      2        ⁡    (    ρ    )        d    ρ    )        ∫    0          2      π        sin  ⁡  θ    d  θ  .

Julien Carrillo · Accepted Answer

First, factor out the sinθ:      ∫    0          π              /            4            ∫    0          2      π        (      sin    2    ⁡  (  ρ  )  −  cos  ⁡  (  ρ  )      sin    2    ⁡  (  ρ  )  )  sin  ⁡  θ    d  θ    d  ρ  .Since   ρ does not depend on   θ, it can be factored out of the inner   θ integral (just as a constant can be factored out of an integral):      ∫    0          π              /            4        (      sin    2    ⁡  (  ρ  )  −  cos  ⁡  (  ρ  )      sin    2    ⁡  (  ρ  )  )      ∫    0          2      π        sin  ⁡  θ    d  θ    d  ρ  .The quantity       ∫    0          2      π        sin  ⁡  θ    d  θ is just a number (in this case, it is zero), so it can be factored out of the outer   ρ integral just as any number can:      (          ∫      0              π                  /                4              (          sin      2        ⁡    (    ρ    )    −    cos    ⁡    (    ρ    )          sin      2        ⁡    (    ρ    )    )        d    ρ    )        ∫    0          2      π        sin  ⁡  θ    d  θ  .In general, the following holds via the same logic:      ∫    a    b        ∫    c    d    f  (  x  )  g  (  y  )    d  y    d  x  =      (          ∫      a      b        f    (    x    )        d    x    )        (          ∫      c      d        g    (    y    )        d    y    )    .

I have the folllwoing integral <msubsup> ∫ 0 <mrow class="MJX-TeXAt

Answered question

Answer & Explanation

New Questions in Integral Calculus