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The definite integral, $$\int_0^{\pi}3\sin^2t\cos^4t\:dt$$

My question: for the trigonometric integral above the answer is $\frac{3\pi}{16}$. What I want to know is how can I compute these integrals easily. Is there more than one way to solve it? If so, is the key to solving these integrals, just recognizing some trig identities and using u-sub until it looks like a simpler integral?

Here's what I tried (Why doesn't it work!):

I rewrote the integrand as: $3(1-\cos^2t)\cos^4t\:dt$ then foiled it in,

$3\cos^4t-3\cos^6t dt$ , then I used the power rule and multipied through by chain rule and then did

$F(\pi)-F(0)$ and got the answer: $\frac{6}{35}$

Why does this not work?!?

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  • $\begingroup$ I used the double angle identity for sin^2t and it looked even more difficult $\endgroup$ – J the constant Feb 9 at 11:09
  • $\begingroup$ Welcome to MSE. Please edit and use MathJax to properly format math expressions. $\endgroup$ – Lee David Chung Lin Feb 9 at 11:11
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    $\begingroup$ @Jtheconstant, you need to add more detail if you want us to pinpoint where you went wrong. For example what does the antiderivative $F$ you got look like? What are your steps to get that $F$? $\endgroup$ – LHF Feb 9 at 11:19
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    $\begingroup$ for example: 3(cost)^4 became 3/5 * (cost)^5 * -sint $\endgroup$ – J the constant Feb 9 at 11:25
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    $\begingroup$ and - 3 (cost)^6 became -3/7 * (cost)^7 * - sint. Then the integral is ( 3/5 * (cost)^5 * -sint ) + ( 3/5 * (cost)^5 * sint ) and I then plugged pi and 0 in $\endgroup$ – J the constant Feb 9 at 11:29
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Here's a method that can tackle more general problems of this nature.

First, let's consider the integral $I = \int \sin^2 x \cos^4 x dx$ (I just omitted the constant multiplier).

We can write $I = \int \sin x \cdot \sin x \cos^4 x dx$ and use integration by parts with $u = \sin x, dv = \int \sin^2 x \cos^4 x dx$ to get $I = \sin x \cdot (-\frac 15)\cos^5 x - \int\cos x \cdot (-\frac 15)\cos^5 xdx = \sin x \cdot (-\frac 15)\cos^5 x +\frac 15 \int\cos^6 xdx$

Applying the bounds, we get $I_{0}^{\pi} = \frac 15 \int_{0}^{\pi}\cos^6 xdx$

Now we're left with the problem of working out $J = \int \cos^6 xdx$

From Euler's identity and De Moivre's theorem (complex number theory), we can write $\cos x = \frac 12(e^{ix} + e^{-ix})$. We also know that $\cos nx = \frac 12(e^{nix} + e^{-nix})$

Applying binomial theorem and the above, we can write $\cos^6x = (\frac 12(e^{ix} + e^{-ix}))^6= \frac 1{64}(e^{6ix} + 6e^{4ix} + 15e^{2ix} + 20 + 15e^{-2ix} + 6e^{-4ix} + e^{-6ix}) = \frac 1{64}(2\cos 6x + 12\cos 4x + 30\cos 2x + 20)$, and you can then compute $J = \frac 1{64}(\frac 13 \sin 6x + 3\sin 4x + 15\sin 2x + 20x) + c$, and after applying the bounds, you get $J_{0}^{\pi} = \frac {20\pi}{64} = \frac{5\pi}{16}$.

You can now compute $I_{0}^{\pi} = \frac 15 J_{0}^{\pi} = \frac{\pi}{16}$ and the original integral is $3I_{0}^{\pi} =\frac{3\pi}{16}$.

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Use walli's formula given here to do it faster Reduction formula for integral $\sin^m x \cos^n x$ with limits $0$ to $\pi/2$

$I=3\displaystyle\int_{0}^{\pi} \sin^2t\cos^4t\ dt=6 \displaystyle\int_{0}^{\pi/2} \sin^2t\cos^4t\ dt=6\times \dfrac{(2-1)(4-1)(4-3)}{6(6-2)(6-4)}\times \dfrac{\pi}{2}=\dfrac{3\pi}{16}$

your mistake : Use of Power rule in wrong way

we know,

$\displaystyle\int x^m \ dx=\dfrac{x^{m+1}}{m+1}+C$ but this is not analogous to

$3 \displaystyle\int \cos^6t\ dt\neq\dfrac{3}{7}cos^7t + C $ as you've written in comments.

but ,

$\displaystyle\int \cos^6t\ d(cost)\ =-\displaystyle\int \cos^6t \ sint\ dt \ =\dfrac{3}{7}cos^7t + C $

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Approach $(1)$
Let \begin{align} y&=\cos t+i\sin t\implies y^n&=\cos t+i\sin nt\\ \frac{1}{y}&=\cos t-i\sin t\implies \frac{1}{y^n}&=\cos t-i\sin nt\\ \end{align} Then \begin{align} y+\frac{1}{y}=2\cos t \qquad y^n+\frac{1}{y^n}=2\cos nt\\ y-\frac{1}{y}=2i\sin t \qquad y^n+\frac{1}{y^n}=2i\sin nt\\ \end{align} Now you can form your integrand using,
$(2i\sin t)^2=-4\sin^2t$
$(2\cos t)^4=16\cos^4t$
$$3\sin^2t\cos^4t=3\times\frac{1}{-4}\times\frac{1}{16}\left(y-\frac{1}{y}\right)^2\left(y+\frac{1}{y}\right)^4$$ Now use binomial expansion and integrate term by term. Why This approach is better?

Polynomials are much easier to integrate

Approach $(2)$

Instead of convert the integrand into polynomial, you can convert it to exponential using Euler formula.


It seems you struggling with $\int\cos^nt\:dt$. Don't use power rule here direction. You can use reduction formula which came using integration by parts.

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  • $\begingroup$ Why can't I use power rule? Oh, I know is it because chain rule is for 'derivatives' and not 'integrals'? I think this may be the error I made. If so, why can't I just multiply by the reciprocal of the expression that gets multiplied when using chain rule $\endgroup$ – J the constant Feb 9 at 11:53
  • $\begingroup$ Take a look at here @Jtheconstant $\endgroup$ – emonHR Feb 9 at 12:04
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Note :-The solution is specifically for the problem in question

The given function is even

So let $$I = 3\int_0^{\pi} \sin^{2}t \cos^{4}t dt$$ And it is equal to $$I = 6\int_0^{\pi/2} \sin^{2}t \cos^{4}t dt\tag{1}$$

Now, we know that $$\int_a^b f(x) dx = \int_a^b f(b+a-x) dx$$

Therefore We write the integral as $$I = 6\int_0^{\pi/2} \sin^{2}(\frac{\pi}{2}-t) \cos^{4}(\frac{\pi}{2}-t) $$

That is equal to $$I = 6\int_0^{\pi/2} \cos^{2}t \sin^{4}t dt\tag{2}$$

adding equation (1) and (2) we get $$2I =6 \int_0^{\pi/2} \sin^{2}t \cos^{4}t + \cos^{2}t \sin^{4}t dt$$ Further $$2I = 6\int_0^{\pi/2} \cos^{2}t \sin^{2}t[\sin^{2}t+\cos^{2}t] $$ $$I = 3\int_0^{\pi/2} \sin^{2}t \cos^{2}t dt$$ $$I = \frac{3}{4} \int_0^{\pi/2} (2\cos t \sin t)^2 dt $$

Integrating, $$I = \frac{3}{4} [( \frac{\pi}{4})-(\frac{1}{2} -\frac{1}{2})]$$

So,

$$I= \frac {3\pi}{16}$$

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Use $$\frac{1+\cos\left(ax\right)}{2}=\cos^{2}\left(\frac{ax}{2}\right)$$

Then we have:

$$\int_{0}^{\pi}3\cos^{4}\left(t\right)dt-\int_{0}^{\pi}3\cos^{6}\left(t\right)dt=\frac{3}{4}\int_{0}^{\pi}\left(1+\cos\left(2t\right)\right)^{2}dt-\frac{3}{8}\int_{0}^{\pi}\left(1+\cos\left(2t\right)\right)^{3}dt$$

Can you take it from here?

Or using reduction formula for $\int_{0}^{\pi}3\left(\cos\left(x\right)\right)^{4}dx$ follows:

$$=3\left(\frac{1}{4}\cos^{3}\left(x\right)\sin\left(x\right)\Big|_0^\pi+\frac{3}{4}\int_{0}^{\pi}\cos^{2}\left(x\right)dx)\right)$$$$=3\left(\frac{3}{8}\int_{0}^{\pi}1+\cos\left(2x\right)dx\right)$$$$=\frac{9\pi}{8}$$

Use this for the other one.

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  • $\begingroup$ The answer is 3π/16 $\endgroup$ – J the constant Feb 9 at 11:42
  • $\begingroup$ Can any substitution I make or identity I use work as long as the rewritten integrand matches the original one? What is the trick to solve these integrals? $\endgroup$ – J the constant Feb 9 at 11:49
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I see $2$ quick methods. Firstly, one could use $\sin t\cos t=\frac12\sin2t$ and $\cos^2t=\frac12(1+\cos2t)$ to get $$\int_0^{\pi}3\sin^2t\cos^4t\,dt=\frac38\int_0^{\pi}\left(\sin^22t+\sin^22t\cos2t\right)dt$$ The period of $\sin2t$ and $\cos2t$ is $\pi$ so the integral is the average value of the integrand times the length of the interval. Now $\langle\sin^22t\rangle=\langle\cos^22t\rangle=\frac12\langle\sin^22t+\cos^22t\rangle=\frac12\langle1\rangle=\frac12$ and $\langle\sin^22t\cos2t\rangle=0$ because it has on odd power of $\cos2t$ so $$\int_0^{\pi}3\sin^2t\cos^4t\,dt=\frac38\pi\left(\frac12+0\right)=\frac3{16}$$ The other easy approach is via the Beta and Gamma functions, but it's a little higher level math: $$\begin{align}\int_0^{\pi}3\sin^2t\cos^4t\,dt&=3\times2\int_0^{\pi/2}\sin^2t\cos^4t\,dt=3\operatorname{B}\left(\frac32,\frac52\right)\\ &=3\frac{\operatorname{\Gamma}\left(\frac32\right)\operatorname{\Gamma}\left(\frac52\right)}{\operatorname{\Gamma}\left(4\right)}=3\frac{\sqrt{\pi}\left(\frac12\right)\sqrt{\pi}\left(\frac12\right)\left(\frac32\right)}{3!}=\frac{3\pi}{16}\end{align}$$

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