Trigonometric Integrals...1

May 2016
368
5
NYC
Integrate cos^3 x * sin x dx

I broke the cosine function into two parts:

cos^2 x * cos x

The integral now looks like this:

cos^2 x * sin x * cos x dx

I then let cos^2 x = 1 - sin^2 x.

Here is the integral now:

(1 - sin^2 x) * sin x * cos x dx

I let u = sin x making du = cos x dx.

The Integral became very easy: (1 - u^2)u du.

After integrating, I back-substituted for u.

My final answer is (1/4)(2 sin^2 x - sin^4 x) + C.

The textbook answer is (-1/4)cos^4 x + C.

Who is right?

What did I do wrong?

Is my answer equivalent to the textbook answer?
 

romsek

MHF Helper
Nov 2013
6,744
3,037
California
there's an easier way and we'll check if the answers are the same.

Have you been taught integration by substitution yet? The way I would do this integral is as follows

$\int \cos^3(x) \sin(x)~dx$

$u=\cos(x),~~~~du = -\sin(x)~dx$

so our original integral becomes

$-\int u^3 ~du = -\dfrac 1 4 u^4 + C = -\dfrac 1 4 \cos^4(x) + C$

The textbook is correct.

I'm trying to figure out where you went wrong and will post when I do.
 
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romsek

MHF Helper
Nov 2013
6,744
3,037
California
ok I see what's going on

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

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

the text book answer is $-\dfrac 1 4 \cos^4 + C$

your answer is equivalent to $-\dfrac 1 4 \cos^4 + \dfrac 1 4 + C^\prime = -\dfrac 1 4 \cos^4 + C$

so the two answers are ultimately the same.
 
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May 2016
368
5
NYC
ok I see what's going on

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

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

the text book answer is $-\dfrac 1 4 \cos^4 + C$

your answer is equivalent to $-\dfrac 1 4 \cos^4 + \dfrac 1 4 + C^\prime = -\dfrac 1 4 \cos^4 + C$

so the two answers are ultimately the same.
It is good to know that both answers are correct. I did every step correctly. Thank you for showing the U-Substitution method which was done several chapters ago.
 
May 2016
368
5
NYC
Most answers in the back of the book are simplified completely. The three authors do not show how to simplify the answers to match the textbook. For this question and the next set of questions, I am interested in solving via the trigonometric integrals method.