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Integrate $$\int \sin^4x \cos^2x dx$$

Now, there's few solutions to this problem already on the internet. For example on yahoo: https://answers.yahoo.com/question/index?qid=20090204203206AAbjUfM and MIT http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-4-techniques-of-integration/part-a-trigonometric-powers-trigonometric-substitution-and-completing-the-square/session-69-integral-of-sin-n-x-cos-m-x-even-exponents/MIT18_01SCF10_ex69sol.pdf

I've tried a slightly different approach and simply wonder if my answer is correct as I don't know how to check it.

\begin{eqnarray} \int \sin^4x \cos^2x dx &=& \int \left( \frac{1-\cos^2x}{2} \right)^2 \cos^2x dx \\ &=& \frac{1}{4} \int (1-2\cos^2x+\cos^4x)\cos^2x dx \\ &=& \frac{1}{4} \int \cos^6x-2\cos^4x+\cos^2xdx \\ &=& \frac{1}{4} \left( \frac{1}{7}\cos^7x-\frac{2}{5}\cos^5x+\frac{1}{3}\cos^3x \right) + C \end{eqnarray}

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  • $\begingroup$ It seems that you did integrate just as if it were a polynomial. $\endgroup$ – Claude Leibovici Jul 19 '14 at 14:26
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    $\begingroup$ You can check your answer by differentiating. It is not correct. It is not true that $\int \cos^k x dx = {1 \over k+1} \cos^{k+1} x$. $\endgroup$ – copper.hat Jul 19 '14 at 14:26
  • $\begingroup$ $\sin^4{x}=\left(\frac{1-\cos{2x}}{2}\right)^2\neq \frac{1-\cos^2{x}}{2}$ $\endgroup$ – SuperAbound Jul 19 '14 at 14:30
  • $\begingroup$ You wrote among other things that $\int \cos^2 x\,dx=\frac{1}{3}\cos^3 x$. This is not true, you can differentiate the "answer" and get $(-\sin x)\cos^2 x$. $\endgroup$ – André Nicolas Jul 19 '14 at 14:37
  • $\begingroup$ Differentiating your last result leads, after simplifications, to $-\frac{1}{4} \sin ^5(x) \cos ^2(x)$ $\endgroup$ – Claude Leibovici Jul 19 '14 at 14:39
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$$\sin^4x\cos^2x=\left(\frac{1-\cos2x}2\right)^2\frac{1+\cos2x}2$$

$$=\frac{(1+\cos2x)(1-2\cos2x+\cos^22x)}8$$

$$=\frac{1 -\cos2x-\cos^22x+\cos^32x}8$$

Again, $\cos^22x=\dfrac{1+\cos4x}2$

and $\cos3y=4\cos^3y-3\cos y\iff 4\cos^3y=\cos3y+3\cos y$ set $y=2x$


Alternatively use Euler Identities

$$2\cos x= e^{ix}+e^{-ix},2i\sin x=e^{ix}-e^{-ix}$$

$$(2i\sin x)^4(2\cos^2x)^2=(e^{ix}-e^{-ix})^4(e^{ix}+e^{-ix})^2$$

$$\implies64\sin^6x\cos^2x=(e^{ix}+e^{-ix})^2\cdot(e^{ix}+e^{-ix})^2(e^{ix}-e^{-ix})^2$$

$$=(e^{i2x}+e^{-i2x}+2)(e^{i2x}-e^{-i2x})^2$$

$$=(e^{i2x}+e^{-i2x}+2)(e^{i4x}+e^{-i4x}-2)$$

$$=e^{i6x}+e^{-i6x}-(e^{i2x}+e^{-i2x})+2(e^{i4x}+e^{-i4x})-4$$

$$=2\cos6x-(2\cos2x)+2(2\cos4x)-4$$

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First, you can rewrite $\sin^4x$ as either $(1-\cos^2x)^2$ or $(\dfrac{1-\cos2x}2)^2$. You seem to have crossed up both.

The second problem is if you try the substitution $u=\cos x,du=-\sin xdx$, you will find that things get messy. This isn't an issue if one of the powers of cosine or sine is odd. For example,

$$\int\sin^5x\cos^2xdx=\int(1-\cos^2x)^2\cos^2x\sin xdx$$

Now if you substitute $u$ as above, you get $\int -u^2(1-u^2)^2du$. But something like this is not an option when both powers are even, which is why we resort to double angle formulas. Your first link is probably the least messy way of doing it. Although if you really insist on getting rid of the trig functions, you could rewrite it as

$$\int\tan^4x\cos^8x\sec^2xdx=\int\dfrac{\tan^4x}{(1+\tan^2x)^4}\sec^2xdx$$

then substitute $u=\tan x$. Have fun with partial fractions. :)

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