$\begingroup$

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

Now, there's few solutions to this problem already on the internet. For example on yahoo: and MIT

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}

$\endgroup$ 6

2 Answers

$\begingroup$

$$\sin^4x\cos^2x=\left(\frac{1-\cos2x}2\right)^2\frac{1+\cos2x}2$$

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

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

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

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

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

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

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

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

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

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

$\endgroup$ $\begingroup$

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. :)

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy