Find Int [cos 2x / (sin 2x)^3 dx]  

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

Let F(x) = intg (cos2x / sin2x)^3 dx.

Let us assume that y = sin2x.

==> dy = 2cos2x. dx ==> dx = dy/2cos2x

Let us substitute.

F(x) = int   cos2x/ y^3 * dy/2cos2x

       = intg dy / 2y^3 .

       = intg (1/2) y^-3 dy

Let us integrate.

==> F(x) = (1/2) y^-2 / -2  + C

==> F(x) = (1/-4) y^-2 + c

              = 1/(-4y^2)  + C

Now we will substitute with y= sin2x

==> F(x) =- 1/[4(sin2x)^2]  + C

Top Answer

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

To find Int [cos 2x / (sin 2x) ^3 dx], substitute sin 2x with y

If y = sin 2x, dy/dx = 2 cos 2x

=> dy / 2 = cos 2x dx

Now Int [cos 2x / (sin 2x) ^3 dx] can be written as

Int [(1/2)/ (y^3) dy]

=> (1/2)* (-1/2) y^-2 +C

=> (-1/4)*y^-2 + C

Replace y = sin 2x

=> (-1/4)*(1/ (sin 2x) ^2) +C

Therefore Int [cos 2x / (sin 2x) ^3 dx] = (-1/4)*(1/ (sin 2x) ^2) +C

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