Integrate Sinx^4: Solutions with Integration by Parts

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Homework Statement


Integration of sinx^4


Homework Equations


integration by parts


The Attempt at a Solution


I tried to break the function into 2 parts like (sin^3x)(sinx) and solve it using integration by parts but kept getting a function (sin^2x)(cos^2x) over and over again.
 
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Does it specifically ask for you to use integration by parts? Maybe using trig identities for ## \sin^2 (x) ## will be easier.

If you get a recurring integral, try something like this:

\displaystyle \int \sin^2 (x) \cos ^2 (x) = [\text{Integral stuff...}] - \int \sin^2 (x) \cos ^2 (x)

\displaystyle \int \sin^2 (x) \cos ^2 (x) + \int \sin^2 (x) \cos ^2 (x) = [\text{Integral stuff...}]

\displaystyle 2 \int \sin^2 (x) \cos ^2 (x) = [\text{Integral stuff...}]

\displaystyle \int \sin^2 (x) \cos ^2 (x) = \dfrac{[\text{Integral stuff...}]}{2}
 
It doesn't specifically say anything. I actually tried to use the trig identity sin^2x but got a very messy function to evaluate something like (1/2- 1/2cos2x)^2. I am posting what I actually got by applying integration by parts.

∫sin^3xsinx= sin^3xcosx- 3∫sin^2xcos^2x ...1
=sin^3xcosx - 3[2sin^3xcosx -4∫cos^2xsin^2x]

Let y= ∫sin^3xsinx
from 1 we get, sin^3xcosx -3∫sin^2xcos^2x= sin^3xcosx-6sin^3xcosx+12∫cos^2xsin^2x
→ 6sin^3xcosx= 15∫sin^2xcos^2x
→ 5∫sin^2xcos^2x= 2sin^3xcosx
→ ∫sin^2xcos^2x=2sin^3xcosx/5+C
Please check if there are any reckless mistakes and any step is wrong.
 
Last edited:
First, please note that "sin x^4, which is what you wrote, is NOT the same as "sin^4(x) what is what I am sure was intended.

The simplest way to integrate it is to use the fact that sin^4(x)= (1/2- (1/2)cos(2x))^2 as you say you started to use but apparently stopped at that point. Continuing, that is equal to (1/4)- cos(2x)+ (1/4)cos^2(2x)= 1/4- cos(2x)+ (1/4)(1/2+ (1/2)cos(4x)= 3/8- cos(2x)+ 1/8 cos(4x).

Now as for your integration by parts, I have no idea what such things as
→ 6sin^3xcosx= 15∫sin^2xcos^2x
mean.
 
Dumbledore211 said:

Homework Statement


Integration of sinx^4


Homework Equations


integration by parts


The Attempt at a Solution


I tried to break the function into 2 parts like (sin^3x)(sinx) and solve it using integration by parts but kept getting a function (sin^2x)(cos^2x) over and over again.

If you know how to do ##F(t) = \int \sin^2 (t)\, dt##, then you can write
\sin^4 (x) = \sin^2 (x)[1-\cos^2 (x)] = \sin^2 (x) - (\sin(x) \cos(x))^2 = <br /> \sin^2 (x) - \frac{1}{4} \sin^2 (2x) and get the solution in terms of ##F(x)## and ##F(2x)##.
 
Last edited:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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