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Evaluate the integral integral (2x)(2x)\,dx

Mathematics

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Answer

sin⁡2(2x)4+C\frac{\sin^2(2x)}{4} + C

Step-by-step solution

6 steps

To evaluate the integral ∫sin⁡(2x)cos⁡(2x) dx\int \sin(2x)\cos(2x)\,dx, we can use a substitution method.

  1. Set up the substitution.

    Let u=sin⁡(2x)u = \sin(2x).

  2. Find the differential dudu.

    Differentiate uu with respect to xx:

    dudx=ddx(sin⁡(2x))=cos⁡(2x)⋅2\frac{du}{dx} = \frac{d}{dx}(\sin(2x)) = \cos(2x) \cdot 2

    So, du=2cos⁡(2x) dxdu = 2\cos(2x)\,dx.

  3. Express cos⁡(2x) dx\cos(2x)\,dx in terms of dudu.

    From the previous step, we have du=2cos⁡(2x) dxdu = 2\cos(2x)\,dx.
    Dividing by 2 gives:

    cos⁡(2x) dx=12du\cos(2x)\,dx = \frac{1}{2}du

  4. Substitute uu and dudu into the integral.

    The integral becomes:

    ∫u⋅12du\int u \cdot \frac{1}{2}du

  5. Integrate with respect to uu.

    12∫u du=12⋅u22+C=u24+C\frac{1}{2} \int u\,du = \frac{1}{2} \cdot \frac{u^2}{2} + C = \frac{u^2}{4} + C

  6. Substitute back u=sin⁡(2x)u = \sin(2x).

    (sin⁡(2x))24+C=sin⁡2(2x)4+C\frac{(\sin(2x))^2}{4} + C = \frac{\sin^2(2x)}{4} + C

    The final answer is sin⁡2(2x)4+C\frac{\sin^2(2x)}{4} + C.

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