Tried working ahead. Super stuck lol (1 Viewer)

Jojofelyx · Jul 16, 2021

Thanks!

B1andB2 · Jul 16, 2021

differentiate x=5tantheta and rearrange to get dx=5sec^2theta dtheta

replace x's with given sub and dx with above

factorise 25 out of denominator to use trig identity to simplify

then just simplify indices and stuff should cross out, you'll probably be able to follow from there

oh yeah don't forget to replace any lone thetas (if you end up with one) by rearranging given sub if needed/ draw triangle to get stuff back in x form

2 marks is a bit stingy though

Drongoski · Jul 17, 2021

If we can ignore substitution:

$\int \frac {xdx}{(25+x^2)^{\frac{3}{2}}} = \frac{1}{2}\int (25+x^2)^{-\frac {3}{2}}d(25+x^2) \\ \\ = \frac{\frac{1}{2}(25+x^2)^{-\frac{1}{2}}}{-\frac{1}{2}} + C = \frac{-1}{\sqrt{25+x^2}} + C$

Jojofelyx · Jul 17, 2021

Drongoski said:
If we can ignore substitution:

$\int \frac {xdx}{25+x^2} = \frac{1}{2}\int (25+x^2)^{-\frac {3}{2}}d(25+x^2) \\ \\ = \frac{-\frac{1}{2}(25+x^2)^{-\frac{1}{2}}}{-\frac{1}{2}} + C = \frac{1}{\sqrt{25+x^2}} + C$

I did it using a simpler sub of u= 25 + x^2, it was just the specific one which tripped me up

thanks though!

CM_Tutor · Jul 17, 2021

@Jojofelyx , make sure you can do trig substitutions too, questions can require them and there is not always an alternative substitution option for you to use.

CM_Tutor · Jul 17, 2021

In this case:

$\begin{align*} I &= \int \cfrac{x\ dx}{\left(25 + x^2\right)^{\frac{3}{2}}} \\ &= \int \cfrac{5\tan\theta}{\left[25 + \left(5\tan\theta\right)^2\right]^{\frac{3}{2}}} \times 5\sec^2\theta\ d\theta \qquad \text{using the substitution $x = 5\tan\theta \implies dx = 5\sec^2\theta\ d\theta$} \\ &= \int \cfrac{25\tan\theta\sec^2\theta}{\left[25\left(1 + \tan^2\theta\right)\right]^{\frac{3}{2}}}\ d\theta \\ &= \int \cfrac{25\tan\theta\sec^2\theta}{125\sec^3\theta}\ d\theta \quad \text{as $1+\tan^2\theta=\sec^2\theta$} \\ &= \cfrac{1}{5}\int \cfrac{\tan\theta}{\sec\theta}\ d\theta \\ &= \cfrac{1}{5}\int \sin\theta\ d\theta \quad \text{as $\cfrac{\tan\theta}{\sec\theta} = \tan\theta\cos\theta = \sin\theta$} \\ &= \cfrac{1}{5}\bigg[-\cos\theta\bigg] + C \qquad \text{for some constant $C$} \end{align*}$

Now, noting that

$\cos\theta = \cfrac{1}{\sec\theta} = \cfrac{1}{\sqrt{\sec^2\theta}} = \cfrac{1}{\sqrt{1+\tan^2\theta}}$

we can see that

$\begin{align*} I &= -\cfrac{1}{5}\cos\theta + C \\ &= -\cfrac{1}{5} \times \cfrac{1}{\sqrt{1+\tan^2\theta}} + C \\ &= -\cfrac{1}{5} \times \cfrac{1}{\sqrt{1 + \left(\cfrac{x}{5}\right)^2}} + C \quad \text{as $\tan\theta = \cfrac{x}{5}$} \\ &= \cfrac{-1}{\sqrt{25\left(1 + \cfrac{x^2}{25}\right)}} + C \\ &= \cfrac{-1}{\sqrt{25 + x^2}} + C \end{align*}$

Tried working ahead. Super stuck lol (1 Viewer)

Jojofelyx

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oui oui baguette

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