# Help to solve: $\int 1/(x\sqrt{25-x^2})\ dx$

I'm a brand new student. Need some help to integrate this.

Perform this integration: $$\int \frac{1}{x\sqrt{25-x^2}}\ dx$$

I'm able to obtain in theta terms like this:

$$\frac 15 \ln⁡|\cscθ-\cotθ|+C$$

But I have problems to convert in terms of "$x$"

Thanks a lot.

This is what is suggested by this image:

Using the substitution $x=5\sin{\theta}$ and $dx=5\cos{\theta}$, we obtain: $$\int \frac{1}{5\sin{\theta}(5\cos{\theta})}\cdot 5\cos{\theta}~d\theta$$ Simplifying, we obtain: $$\frac{1}{5}\int \frac{d\theta}{\sin{\theta}}$$ Using trigonometric identities: $$\frac{1}{5}\int \csc{\theta}~d\theta$$ Integrating: $$\frac{1}{5}\ln|\csc{\theta}-\cot{\theta}|+C$$

• Please use: math.meta.stackexchange.com/questions/5020/… – Arnaldo Mar 17 '17 at 15:25
• It helps if you explain the steps you took to get at the solution in $\theta$, e.g. what was the substitution you used? – StackTD Mar 17 '17 at 15:27
• Yes Exactly. See the problem image..I believe I have the half of the solution....thanks.. – El_Master Mar 17 '17 at 15:27
• @El_Master Did you use the substitution $x=5\sin(\theta)$ and $dx=5\cos(\theta)~d\theta$ ? – projectilemotion Mar 17 '17 at 15:33
• you can use relationships between trigonometric functions to change it back, so if $x = sen \theta$ then $cos \theta = \sqrt{1 - sen^2 \theta} = \sqrt{1 - x^2}$ and other identities that you can use – Cato Mar 17 '17 at 15:33

HINT:

If trigonometric substitution is not mandatory, set $\sqrt{a^2-x^2}=y$

$$\int\dfrac{2x\ dx}{2x^2\sqrt{a^2-x^2}}=\int\dfrac{dy}{y^2-a^2}=?$$

Else setting $x=5\sin y$ where $-\dfrac\pi2\le y\le\dfrac\pi2\implies\cos y\ge0$

$\cos y=\sqrt{1-\left(\dfrac x5\right)^2}=\dfrac{\sqrt{25-x^2}}5$

$$\implies\int\dfrac{dx}{x\sqrt{25-x^2}}=\int\dfrac{5\cos y\ dy}{5\sin y(5\cos y)}=\dfrac{\ln|\csc y-\cot y|}5+K$$

• I appreciete the help... I understand is not mandatory... but in this case I need it (rules are not mine). – El_Master Mar 17 '17 at 15:59

$$x=5\sin\theta$$ $$\frac x5=\sin\theta$$ Therefore $$\frac5x=\csc\theta$$ And since $$\sin\theta=\frac x5$$, we know that $$\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\frac{x^2}{25}}$$ And since $$\cot\theta=\frac{\cos\theta}{\sin\theta}$$, $$\cot\theta=\frac{5\sqrt{1-\frac{x^2}{25}}}{x}=\frac{\sqrt{25-x^2}}x$$ Which gives your result as $$\frac15\ln\bigg|\frac{5-\sqrt{25-x^2}}x\bigg|+C$$