Final integrals

**Selim** · February 25th 2010, 03:20 PM

$\int sin(x)^4 * cos(x)^3 * dx$

and

$\int sin(x)^2 * cos(x) ^2 * dx$

**Prove It** · February 25th 2010, 04:05 PM

Originally Posted by Selim

$\int sin(x)^4 * cos(x)^3 * dx$

and

$\int sin(x)^2 * cos(x) ^2 * dx$

1. $\int{\sin^4{x}\cos^3{x}\,dx} = \int{\sin^4{x}\cos^2{x}\cos{x}\,dx}$

$= \int{\sin^4{x}(1 - \sin^2{x})\cos{x}\,dx}$

$= \int{\sin^4{x}\cos{x}\,dx} - \int{\sin^6{x}\cos{x}\,dx}$ .

Now in each integral, make the substitution $u = \sin{x}$ .

**skeeter** · February 25th 2010, 04:10 PM

Originally Posted by Selim

$\int sin(x)^4 * cos(x)^3 * dx$

and

$\int sin(x)^2 * cos(x) ^2 * dx$

$\sin^4{x} \cdot \cos^3{x} =$

$\sin^4{x} \cdot \cos^2{x} \cdot \cos{x} =<br />$

$\sin^4{x}(1-\sin^2{x}) \cdot \cos{x} =$

$(\sin^4{x} - \sin^6{x}) \cdot \cos{x}$

substitution ... let $u = \sin{x}$ , $du = \cos{x} \, dx$

$\sin^2{x} \cdot \cos^2{x} =$

$\left[\frac{1-\cos(2x)}{2}\right] \cdot \left[\frac{1+\cos(2x)}{2}\right] =<br />$

$\frac{1}{4} \left[1 - \cos^2(2x)\right] =$

$\frac{1}{4} \left[\sin^2(2x)\right] =$

$\frac{1}{4} \left[\frac{1 - \cos(4x)}{2}\right] =$

$\frac{1}{8} \left[1 - \cos(4x)\right]$

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