# Thread: Integration of sin x cos^3x

1. ## Integration of sin x cos^3x

I need to evaluate $\int sin x cos^3 x dx$

I tried 2 methods: u-substitution and integration using trigonometric identities but ended up with two different answers. Both workings look fine to me. Could anyone pls help out?

Method 1: U-substitution

$u = cos x$
$\frac{du}{dx} = - sin x$

$\int sin x cos^3 x dx$ = $\int sin x. u^3 \frac{du}{-sin x}$
$\int sin x cos^3 x dx$ = $\int - u^3 du$
$\int sin x cos^3 x dx$ = $\frac{-u^4}{4} + C$
$\int sin x cos^3 x dx$ = $\frac{-cos^4x}{4} + C$

Method 2 in new post.

2. Method 2: Integration using trigo identitites

$\int sin x \cos^3 x dx$
$= \int sin x \cos x \cos^2x dx$
$= \int sin x \cos x (1 - sin^2x) dx$

Substitute u = sin x, $\frac{du}{dx} = cos x$
$= \int u \cos x (1-u^2) \frac{du}{cos x}$
$= \int u (1-u^2) du$
$= \frac{u^2}{2} + \frac{u^4}{4} + c$
$= \frac{sin^2 x}{2} + \frac{sin^4 x}{4} + c$
$= 1 - \frac{cos^4 x}{4} + c$

How come both answers are different?!

3. How would you describe the difference between them?

4. Method 2 has an extra term --> 1.

5. What kind of term is that?

6. What I'm trying to say is that why is there an extra "1" from Method 2?

7. Originally Posted by darkvelvet
What I'm trying to say is that why is there an extra "1" from Method 2?
Your mistake is in assuming that the arbitary constant in case 1 is the same as the arbitrary constant in case 2. They are not. Let C be the constant in case 1 and K be the constant in case 2. Then K + 1 is also arbitrary and can be re-labelled as C.