four basic/easy integrals

**golfman44** · Dec 4th 2009, 01:59 AM

1 $\int (x^4+3)^4x^3dx$

$u = x^4+3$

$du = 4x^3dx$

$\frac{1}{4}du = x^3dx$

$\frac{1}{4}\int (u)^4 du$

$\frac{1}{4}\int \frac{1}{5}(u)^5 + c$

final answer: $\frac{1}{20}(x^4+3)^5 + C$

THE NEXT THREE I AM STRUGGLING WITH... HELP IS APPRECIATED!

2. $\int (x^2-2)^2dx$

3. $\int sin^5(2x)cos(2x)dx$

4. $\int \frac{9x^6+5x^4+7}{x^3}dx$

**Prove It** · Dec 4th 2009, 02:57 AM

Originally Posted by golfman44

1 $\int (x^4+3)^4x^3dx$

$u = x^4+3$

$du = 4x^3dx$

$\frac{1}{4}du = x^3dx$

$\frac{1}{4}\int (u)^4 du$

$\frac{1}{4}\int \frac{1}{5}(u)^5 + c$

final answer: $\frac{1}{20}(x^4+3)^5 + C$

THE NEXT THREE I AM STRUGGLING WITH... HELP IS APPRECIATED!

2. $\int (x^2-2)^2dx$

3. $\int sin^5(2x)cos(2x)dx$

4. $\int \frac{9x^6+5x^4+7}{x^3}dx$

2. Expand.

$\int{(x^2 - 2)^2\,dx} = \int{x^4 - 4x^2 + 4\,dx}$

$= \frac{1}{5}x^5 - \frac{4}{3}x^3 + 4x + C$ .

3. Use a $u$ substitution.

$\int{\sin^5{(2x)}\cos{(2x)}\,dx} = \frac{1}{2}\int{\sin^5{(2x)}\cdot 2\cos{(2x)}\,dx}$ .

Let $u = \sin{(2x)}$ so that $\frac{du}{dx} = 2\cos{(2x)}$ .

So the integral becomes

$\int{u^5\,\frac{du}{dx}\,dx} = \int{u^5\,du}$

$= \frac{1}{6}u^6 + C$

$= \frac{1}{6}\sin^6{(2x)} + C$ .

4. Divide everything by the denominator.

$\int{\frac{9x^6 + 5x^4 + 7}{x^3}\,dx} = \int{9x^3 + 5x + 7x^{-3}\,dx}$

$= \frac{9}{4}x^4 + \frac{5}{2}x^2 - \frac{7}{2}x^{-2} + C$

$= \frac{9x^4}{4} + \frac{5x^2}{2} - \frac{7}{2x^2} + C$

$= \frac{9x^6 + 10x^4 - 14}{4x^2} + C$ .

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