# Integral Calculus HELP PLZ (1 Viewer)

Thanks!!!

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#### Drdusk

##### π
Moderator

Thanks!!!
There's different ways you can do a)

You can change $\bg_white x^2$ to $\bg_white \bigg((1+x) - 1\bigg)^2$ which after expanding will give you an integral with varying powers of (1+x) which you can integrate straight away.

You can also do a u sub by letting $\bg_white x= \sin^2 (u)$ and then expanding to get varying powers of $\bg_white \sin(x)$ which you can also integrate straight away.

#### Trebla

For the first one, you can use the substitution u=x+1.

For the second one, you will need to differentiate both sides of the substitution with respect to x and use the chain rule
$\bg_white u^2=4-x^2$
$\bg_white \dfrac{du^2}{dx}=-2x$
$\bg_white \dfrac{du^2}{du}\dfrac{du}{dx}=-2x$
$\bg_white x\,dx=-u\,du$

You can then substitute x dx and all the remaining components with x's from there.

#### idkkdi

##### Well-Known Member
There's different ways you can do a)

You can change $\bg_white x^2$ to $\bg_white \bigg((1+x) - 1\bigg)^2$ which after expanding will give you an integral with varying powers of (1+x) which you can integrate straight away.

You can also do a u sub by letting $\bg_white u = \sin^2 (x)$ and then expanding to get varying powers of $\bg_white \sin(x)$ which you can also integrate straight away.
by the second statement he probably means x = sin^2 u

#### Drdusk

##### π
Moderator
Part two would be done as so

$\bg_white I = \int_{0}^{2}x^3\sqrt{4-x^2}dx$

$\bg_white = \int_{0}^{2}x^2 \times x \sqrt{4-x^2}dx$

$\bg_white u^2 = 4 - x^2 \Rightarrow 2udu = -2xdx \Rightarrow udu = -xdx$

$\bg_white (x = 0, u = 2), (x = 2, u = 0)$

$\bg_white \text{Now rearranging the original Integral we can see how this u sub fits in}$

$\bg_white I = \underbrace{-1}_{\text{Cancels -1 in -xdx}}\times \int_{0}^{2}\underbrace{x^2}_{4-u^2}\sqrt{\underbrace{4-x^2}_{u^2}}\times \underbrace{-xdx}_{udu}$

$\bg_white \therefore I = \int_{0}^{2}u^2(4-u^2)du$

$\bg_white \text{-1 becomes +1 because the integral signs are flipped from 2 to 0, into 0 to 2.}$

$\bg_white \text{Trivial from here}$

I suggest going to the textbook and learning how to do u sub integrals because they can get harder from here. I've done it step by step here so you can understand how to do them but you can cut down a few lines.

#### Drongoski

##### Well-Known Member
1)
$\bg_white x^2\sqrt{x+1} = ((x^2+2x+1) -2(x+1) + 1)\sqrt{x+1}\\ \\ dx = d(x+1)\\ \\ \therefore \int x^2\sqrt{x+1}dx = \int[ (x+1)^\frac{5}{2} - 2(x+1)^\frac{3}{2} + (x+1)^\frac{3}{2}]d(x+1) \\ \\ = \frac{2}{7}(x+1)^\frac {7}{2} - \frac {4}{5}(x+1)^\frac{5}{2} + \frac{2}{3}(x+1)^\frac{3}{2} + C$

Evaluating the definite integral will yield: 1696/105

Last edited:
csi

#### Drongoski

##### Well-Known Member
Q2
$\bg_white x^3\sqrt{4-x^2} = [-x(4-x^2) + 4x]\sqrt{4-x^2}\\ -2xdx = d(4-x^2)\\ \\ \int x^3\sqrt{4-x^2}dx = \int [\frac{1}{2}(4-x^2)^\frac {3}{2} - 2(4-x^2)^\frac {1}{2}]d(4-x^2) \\ \\ = [\frac{1}{5}(4-x^2)^\frac {5}{2} - \frac {4}{3} (4-x^2)^\frac{3}{2}] + C$

Evaluating the definite integral will yield: 64/15

Last edited:
csi