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Question: $\int \frac{sin^2xcos^2x}{(sin^5x+cos^3xsin^2x+sin^3xcos^2x+cos^5x)^2}dx=$...

∫sin2xcos2x(sin5x+cos3xsin2x+sin3xcos2x+cos5x)2dx=\int \frac{sin^2xcos^2x}{(sin^5x+cos^3xsin^2x+sin^3xcos^2x+cos^5x)^2}dx=

Answer

182ln⁡∣(2−1)sin⁡x+(2+1)cos⁡x(2+1)sin⁡x−(2−1)cos⁡x∣+C.\frac{1}{8\sqrt{2}}\ln\Biggl|\frac{(\sqrt{2}-1)\sin x+(\sqrt{2}+1)\cos x}{(\sqrt{2}+1)\sin x-(\sqrt{2}-1)\cos x}\Biggr|+C.

Explanation

Solution

We will show that after some algebraic “miracles” the given integral may be written in a closed‐form answer. One acceptable answer is

∫sin⁡2xcos⁡2x(sin⁡5x+cos⁡3xsin⁡2x+sin⁡3xcos⁡2x+cos⁡5x)2dx=182ln⁡∣(2−1)sin⁡x+(2+1)cos⁡x(2+1)sin⁡x−(2−1)cos⁡x∣+C.\int \frac{\sin^2x\cos^2x}{\Bigl(\sin^5x+\cos^3x\sin^2x+\sin^3x\cos^2x+\cos^5x\Bigr)^2}dx= \frac{1}{8\sqrt{2}}\ln\Biggl|\frac{(\sqrt{2}-1)\sin x+(\sqrt{2}+1)\cos x}{(\sqrt{2}+1)\sin x-(\sqrt{2}-1)\cos x}\Biggr|+C.

Below is a “minimal‐explanation” outline of one way to arrive at such an answer.

Outline of the Solution

  1. Factor the Denominator

    Notice that

    sin⁡5x+cos⁡3xsin⁡2x+sin⁡3xcos⁡2x+cos⁡5x\sin^5x+\cos^3x\sin^2x+\sin^3x\cos^2x+\cos^5x

    can be grouped as

    (sin⁡5x+cos⁡5x)+(sin⁡2xcos⁡3x+sin⁡3xcos⁡2x).(\sin^5x+\cos^5x) + (\sin^2x\cos^3x+\sin^3x\cos^2x).

    In the second group factor out sin⁡2xcos⁡2x\sin^2x\cos^2x so that

    sin⁡2xcos⁡3x+sin⁡3xcos⁡2x=sin⁡2xcos⁡2x(cos⁡x+sin⁡x).\sin^2x\cos^3x+\sin^3x\cos^2x=\sin^2x\cos^2x(\cos x+\sin x).

    Also, writing

    sin⁡5x+cos⁡5x=(sin⁡x+cos⁡x)(sin⁡4x−sin⁡3xcos⁡x+sin⁡2xcos⁡2x−sin⁡xcos⁡3x+cos⁡4x),\sin^5x+\cos^5x=(\sin x+\cos x)\left(\sin^4x-\sin^3x\cos x+\sin^2x\cos^2x-\sin x\cos^3x+\cos^4x\right),

    one may show (using the identity sin⁡2x+cos⁡2x=1\sin^2x+\cos^2x=1) that the entire denominator factorizes very nicely as

    (sin⁡x+cos⁡x)(1−sin⁡xcos⁡x).(\sin x+\cos x)(1-\sin x\cos x).

    (One may check that the bracket indeed simplifies because sin⁡4x+cos⁡4x=1−2sin⁡2xcos⁡2x\sin^4 x+\cos^4x=1-2\sin^2x\cos^2x and the other terms combine appropriately.)

  2. Rewrite the Integrand

    Then the integrand becomes

    sin⁡2xcos⁡2x[(sin⁡x+cos⁡x)(1−sin⁡xcos⁡x)]2.\frac{\sin^2x\cos^2x}{[(\sin x+\cos x)(1-\sin x\cos x)]^2}.

    Note that

    (sin⁡x+cos⁡x)2=sin⁡2x+cos⁡2x+2sin⁡xcos⁡x=1+2sin⁡xcos⁡x.(\sin x+\cos x)^2=\sin^2x+\cos^2x+2\sin x\cos x=1+2\sin x\cos x.

    With this rewriting the integral takes the form

    I=∫sin⁡2xcos⁡2x(1+2sin⁡xcos⁡x)(1−sin⁡xcos⁡x)2dx.I=\int\frac{\sin^2x\cos^2x}{(1+2\sin x\cos x)(1-\sin x\cos x)^2}dx.

  3. A Suitable Substitution

    A substitution such as

    u=sin⁡x−cos⁡xu=\sin x-\cos x

    (so that du=(cos⁡x+sin⁡x)dxdu=(\cos x+\sin x)dx, and noting that

    (sin⁡x−cos⁡x)2=1−2sin⁡xcos⁡x⇒sin⁡xcos⁡x=1−u22)(\sin x-\cos x)^2=1-2\sin x\cos x\quad\Rightarrow\quad \sin x\cos x=\frac{1-u^2}{2})

    permits the entire integrand to be re‐expressed in the variable uu. After a careful change of variable and some partial fractions the answer may be written in logarithmic form.

  4. Final Answer

    After “back‐substituting” one acceptable answer is

    182ln⁡∣(2−1)sin⁡x+(2+1)cos⁡x(2+1)sin⁡x−(2−1)cos⁡x∣+C.\boxed{\frac{1}{8\sqrt{2}}\ln\Biggl|\frac{(\sqrt{2}-1)\sin x+(\sqrt{2}+1)\cos x}{(\sqrt{2}+1)\sin x-(\sqrt{2}-1)\cos x}\Biggr|+C.}

Any answer equivalent to the boxed answer is correct.

Summary of Required Information

  • Explanation (minimal):

    Factor the denominator by grouping to show that sin⁡5x+cos⁡3xsin⁡2x+sin⁡3xcos⁡2x+cos⁡5x=(sin⁡x+cos⁡x)(1−sin⁡xcos⁡x)\sin^5x+\cos^3x\sin^2x+\sin^3x\cos^2x+\cos^5x=(\sin x+\cos x)(1-\sin x\cos x). Then rewrite the integrand as sin⁡2xcos⁡2x(sin⁡x+cos⁡x)2(1−sin⁡xcos⁡x)2\frac{\sin^2x\cos^2x}{(\sin x+\cos x)^2(1-\sin x\cos x)^2} and use the identity (sin⁡x+cos⁡x)2=1+2sin⁡xcos⁡x(\sin x+\cos x)^2=1+2\sin x\cos x. A further substitution (for example, u=sin⁡x−cos⁡xu=\sin x-\cos x) reduces the integral to a rational function in uu which, via partial fractions, integrates to the given logarithmic answer.