4

I am not sure where I am going wrong but this problem gets more complicated

$$\int \cos^4 x\;dx $$

$u =\cos^3 x$

$du = -3\cos^2 x \sin x \;dx$

$dv = \cos x\;dx$

$v = \sin x$

$$\sin x \cos^3 x + 3\int \sin^2x \cos^2 x \;dx$$

Now I have two squared trig functions, I can take either in the form of 1 + squared trig function but either way I get a power of 4 and nothing has gotten more simple. What is wrong?

Thomas Andrews
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Dan the man
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    If you write $\sin^2 x = 1-\cos^2 x$ in the last, you get $\int \cos^4 x,dx$ back, but with a factor $\neq 1$. That helps. – Daniel Fischer Jul 01 '13 at 19:10
  • I dont understand. – Dan the man Jul 01 '13 at 19:13
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    You get $A = B + c\cdot A$. Then $(1-c)\cdot A = B$ and $A = \frac{1}{1-c}B$. – Daniel Fischer Jul 01 '13 at 19:15
  • Are you trying to confuse me? I don't follow at all. – Dan the man Jul 01 '13 at 19:19
  • $A$ stands for $\int \cos^4 x,dx$, the thing you want to find. $B$ is here $\sin x\cos^3 x + 3\int \cos^2 x,dx$, and $c = -3$. If you substitute $\sin^2 x = 1 - \cos^2 x$ in the equation $\int \cos^4 x,dx = \sin x\cos^3 x +3\int\sin^2 x\cos^2 x,dx$ you got from partial integration, you get $\int\cos^4 x,dx = \sin x\cos^3 x + 3\int \cos^2 x, dx - 3\int \cos^4 x, dx$. Move the last term to the left hand side to obtain $4\int\cos^4 x,dx = ...$. Divide by $4$. – Daniel Fischer Jul 01 '13 at 19:23
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    @Dantheman Best not to insult people trying to help you, and "Are you trying to confuse me?" might be considered insulting. – Thomas Andrews Jul 01 '13 at 19:24

3 Answers3

4

$$\int \cos^4 x=\int{\cos^2x}\cos^2x\;dx=\int\left(\frac{1+\cos2x}{2}\right)\left(\frac{1+\cos2x}{2}\right)\;dx=\int\frac14(1+2\cos2x+\cos^22x)\;dx=\int\frac14\left(1+2\cos2x+\left(\frac{1+\cos4x}{2}\right)\right)\;dx$$

$$\frac38x+\frac14sin(2x)+\frac{1}{32}sin(4x)+c$$

M.H
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  • Heh, I got this same answer by writing the $z^4 = \frac{T_4(z)}{8} + \frac{T_2(z)}{2} + \frac{3}{8}$, there $T_n$ are the Chebyshev polynomials of the first kind. But that was not basic calculus, so I wasn't going to post it as an answer. This computes that directly... – Thomas Andrews Jul 01 '13 at 19:23
  • @Thomas Andrews:i like see your answer – M.H Jul 01 '13 at 19:40
  • It's not much more than my comment and the fact that $T_n(\cos x)=\cos nx$. – Thomas Andrews Jul 01 '13 at 19:41
  • @Thomas Andrews: Thanks i got it – M.H Jul 01 '13 at 19:44
4

Write $$I_n=\int \cos^n xdx$$ Integrate by parts with $u=\cos ^{n-1}x,dv=\cos x dx$. We get that $$I_n={\cos^{n-1}x\sin x}+(n-1)\int \cos^{n-2}x\sin^2 xdx$$

Using $\sin^2 x=1-\cos^2x$, we obtain $$I_n={\cos^{n-1}x\sin x}+(n-1)\int \cos^{n-2}x dx-(n-1)\int \cos^{n}x dx$$ $$I_n={\cos^{n-1}x\sin x}+(n-1)\int \cos^{n-2}x dx-(n-1)I_n $$ which gives the recursion formula $$I_n=\frac{\cos^{n-1}x\sin x}n+\frac{n-1}n I_{n-2}$$

Use this for $n=4$.

Pedro
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Try using this identity:

$$2\sin(x)\cos(x) = \sin(2x)$$

rurouniwallace
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