24
Active contributors today

## Can anyone help to compute 3 integrals?

Ratnaker Mehta
Featured 2 months ago

$1. \ln | \left(x + 3\right) + \sqrt{{x}^{2} + 6 x + 10} | + {C}_{1}$.

$2. {\sin}^{-} 1 \left(\frac{x - 3}{3}\right) + {C}_{2}$.

$3. {\tan}^{-} 1 \left(2 x + 3\right) + {C}_{3}$.

#### Explanation:

1. $\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} + 6 x + 10}} = {I}_{1} , \text{ say.}$

$\therefore {I}_{1} = \int \frac{\mathrm{dx}}{\sqrt{{\left(x + 3\right)}^{2} + 1}}$.

Letting, $\left(x + 3\right) = u , \mathrm{dx} = \mathrm{du}$.

$\therefore {I}_{1} = \int \frac{\mathrm{du}}{\sqrt{{u}^{2} + {1}^{2}}}$,

$= \ln | \left(u + \sqrt{{u}^{2} + {1}^{2}}\right) |$,

$\Rightarrow {I}_{1} = \ln | \left(x + 3\right) + \sqrt{{x}^{2} + 6 x + 10} | + {C}_{1}$.

2. $\int \frac{\mathrm{dx}}{\sqrt{6 x - {x}^{2}}} = {I}_{2} , \text{ say.}$

$\therefore {I}_{2} = \int \frac{\mathrm{dx}}{\sqrt{9 - 9 + 6 x - {x}^{2}}} = \int \frac{\mathrm{dx}}{\sqrt{{3}^{2} - {\left(x - 3\right)}^{2}}}$.

We subst. $x - 3 = y , \text{ so that, } \mathrm{dx} = \mathrm{dy}$.

$\therefore {I}_{2} = \int \frac{\mathrm{dy}}{\sqrt{{3}^{2} - {y}^{2}}}$,

$= {\sin}^{-} 1 \left(\frac{u}{3}\right)$.

$\Rightarrow {I}_{2} = {\sin}^{-} 1 \left(\frac{x - 3}{3}\right) + {C}_{2}$.

3. $\int \frac{\mathrm{dx}}{2 {x}^{2} + 6 x + 5} = {I}_{3} , \text{ say}$.

:. I_3=intdx/{2(x^2+3x+5/2),

$= \frac{1}{2} \int \frac{\mathrm{dx}}{{x}^{2} + 3 x + \frac{9}{4} + \frac{1}{4}}$,

=1/2intdx/{(x+3/2)^2+(1/2)^2.

Taking $\left(x + \frac{3}{2}\right) = t , \text{ we have, } \mathrm{dx} = \mathrm{dt}$.

:. I_3=1/2intdt/(t^2+(1/2)^2,

$= \frac{1}{2} \left\{\frac{1}{\frac{1}{2}} {\tan}^{-} 1 \left(\frac{t}{\frac{1}{2}}\right)\right\}$,

$= {\tan}^{-} 1 \left(2 t\right)$.

rArr I_3=tan^-1(2(x+3/2)=tan^-1(2x+3)+C_3.

Enjoy Maths., and Spread the Joy!

## What is the limit? lim_(xrarr2) (cos(pi/x))/(x-2)

Jim H
Featured 2 months ago

Without L'Hospital's rule see below.

#### Explanation:

$\cos A = \sin \left(\frac{\pi}{2} - A\right)$ so we have

$\cos \frac{\frac{\pi}{x}}{x - 2} = \sin \frac{\frac{\pi}{2} - \frac{\pi}{x}}{x - 2}$

$= \sin \frac{\frac{\pi \left(x - 2\right)}{2 x}}{x - 2}$

 = pi/(2x) (sin((pi(x-2))/(2x)))/((pi(x-2))/(2x)

As $x \rightarrow 2$, we have $\left(\frac{\pi \left(x - 2\right)}{2 x}\right) \rightarrow 0$ so we get

${\lim}_{x \rightarrow 2} \frac{\pi}{2 x} \frac{\sin \left(\frac{\pi \left(x - 2\right)}{2 x}\right)}{\frac{\pi \left(x - 2\right)}{2 x}} = \frac{\pi}{2 \left(2\right)} \left(1\right) = \frac{\pi}{4}$

## Differentiate cos(x^2+1) using first principle of derivative?

Jim H
Featured 1 month ago

#### Explanation:

$f \left(x\right) = \cos \left({x}^{2} - 1\right)$

We need to find

${\lim}_{h \rightarrow 0} \frac{f \left(x + h\right) - f \left(x\right)}{h} = {\lim}_{h \rightarrow 0} \frac{\cos \left({\left(x + h\right)}^{2} - 1\right) - \cos \left({x}^{2} - 1\right)}{h}$

Let's focus on the expression whose limit we need.

$\frac{\cos \left(\left({x}^{2} - 1\right) + \left(2 x h + {h}^{2}\right)\right) - \cos \left({x}^{2} - 1\right)}{h}$

$= \frac{\cos \left({x}^{2} - 1\right) \cos \left(2 x h + {h}^{2}\right) - \sin \left({x}^{2} - 1\right) \sin \left(2 x h + {h}^{2}\right) - \cos \left({x}^{2} - 1\right)}{h}$

$= \cos \left({x}^{2} - 1\right) \frac{\cos \left(2 x h + {h}^{2}\right) - 1}{h} - \sin \left({x}^{2} - 1\right) \sin \frac{2 x h + {h}^{2}}{h}$

$= \cos \left({x}^{2} - 1\right) \frac{\cos \left(2 x h + {h}^{2}\right) - 1}{h \left(2 x + h\right)} \left(2 x + h\right) - \sin \left({x}^{2} - 1\right) \sin \frac{2 x h + {h}^{2}}{h \left(2 x + h\right)} \left(2 x + h\right)$

We will use the following limits:

${\lim}_{h \rightarrow 0} \frac{\cos \left(2 x h + {h}^{2}\right) - 1}{h \left(2 x + h\right)} = {\lim}_{t \rightarrow 0} \frac{\cos t - 1}{t} = 0$

${\lim}_{h \rightarrow 0} \sin \frac{2 x h + {h}^{2}}{h \left(2 x + h\right)} = {\lim}_{t \rightarrow 0} \sin \frac{t}{t} = 1$

And ${\lim}_{h \rightarrow 0} \left(2 x + h\right) = 2 x$

To evaluate the limit:

$\cos \left({x}^{2} - 1\right) \left(0\right) \left(2 x\right) - \sin \left({x}^{2} - 1\right) \cdot \left(1\right) \cdot \left(2 x\right) = - 2 x \sin \left({x}^{2} - 1\right)$

## Lim x−→2 (x^2 − 4) tan(2x − 4) / (x − 2)^2 Find the following limits. Show all steps. If the limit does not exist, explain why not? Not allow to use L’Hospitals rule.

Andrea S.
Featured 1 month ago

${\lim}_{x \to 2} \frac{\left({x}^{2} - 4\right) \tan \left(2 x - 4\right)}{x - 2} ^ 2 = 8$

#### Explanation:

Evaluate the limit:

${\lim}_{x \to 2} \frac{\left({x}^{2} - 4\right) \tan \left(2 x - 4\right)}{x - 2} ^ 2$

Note that:

$\left({x}^{2} - 4\right) = \left(x + 2\right) \left(x - 2\right)$

so we can simplify:

${\lim}_{x \to 2} \frac{\left({x}^{2} - 4\right) \tan \left(2 x - 4\right)}{x - 2} ^ 2 = {\lim}_{x \to 2} \frac{\left(x + 2\right) \left(x - 2\right) \tan \left(2 x - 4\right)}{x - 2} ^ 2 = {\lim}_{x \to 2} \frac{\left(x + 2\right) \tan \left(2 x - 4\right)}{x - 2}$

Multiply numerator and denominator by $2$:

${\lim}_{x \to 2} \frac{\left({x}^{2} - 4\right) \tan \left(2 x - 4\right)}{x - 2} ^ 2 = {\lim}_{x \to 2} \frac{2 \left(x + 2\right) \tan \left(2 x - 4\right)}{2 x - 4}$

Now consider the limit:

${\lim}_{x \to 2} \tan \frac{2 x - 4}{2 x - 4}$

Substituting $y = 2 x - 4$, as ${\lim}_{x \to 2} 2 x - 4 = 0$ we have:

${\lim}_{x \to 2} \tan \frac{2 x - 4}{2 x - 4} = {\lim}_{y \to 0} \tan \frac{y}{y}$

that is a well known trigonometric limit:

${\lim}_{y \to 0} \tan \frac{y}{y} = {\lim}_{y \to 0} \frac{1}{\cos} y \sin \frac{y}{y} = 1$

So:

${\lim}_{x \to 2} \frac{\left({x}^{2} - 4\right) \tan \left(2 x - 4\right)}{x - 2} ^ 2 = {\lim}_{x \to 2} 2 \left(x + 2\right) \cdot {\lim}_{x \to 2} \tan \frac{2 x - 4}{2 x - 4} = 8 \cdot 1 = 8$

graph{((x^2-4)tan(2x-4))/(x-2)^2 [-4.873, 5.127, 6.26, 11.26]}

## How do you integrate int 1/sqrt(4x+8sqrtx-24)  using trigonometric substitution?

Eric S.
Featured 1 month ago

Use the substitution $\sqrt{x} + 1 = \sqrt{7} \sec \theta$.

#### Explanation:

Let

$I = \int \frac{1}{\sqrt{4 x + 8 \sqrt{x} - 24}} \mathrm{dx}$

Complete the square in the square root:

$I = \frac{1}{2} \int \frac{1}{\sqrt{{\left(\sqrt{x} + 1\right)}^{2} - 7}} \mathrm{dx}$

Apply the substitution $\sqrt{x} + 1 = \sqrt{7} u$:

$I = \int \frac{\sqrt{7} u - 1}{\sqrt{{u}^{2} - 1}} \mathrm{du}$

Rearrange:

$I = \sqrt{7} \int \frac{u}{\sqrt{{u}^{2} - 1}} \mathrm{du} - \int \frac{1}{\sqrt{{u}^{2} - 1}} \mathrm{du}$

Apply the substitution $u = \sec \theta$:

$I = \sqrt{7} \sqrt{{u}^{2} - 1} - \int \sec \theta d \theta$

Hence

$I = \sqrt{7 {u}^{2} - 7} - \ln | \sec \theta + \tan \theta | + C$

Reverse the last substitution:

$I = \sqrt{7 {u}^{2} - 7} - \ln | u + \sqrt{{u}^{2} - 1} | + C$

Reverse the first substitution:

$I = \sqrt{{\left(\sqrt{x} + 1\right)}^{2} - 7} - \ln | \left(\sqrt{x} + 1\right) + \sqrt{{\left(\sqrt{x} + 1\right)}^{2} - 7} | + C$

Simplify:

$I = \sqrt{{x}^{2} + 2 \sqrt{x} - 6} - \ln | \left(\sqrt{x} + 1\right) + \sqrt{{x}^{2} + 2 \sqrt{x} - 6} | + C$

## Evaluate the Limit?

Noah G
Featured 3 weeks ago

The limit is $2$

#### Explanation:

We can rewrite as

$L = {\lim}_{x \to \infty} \tan \frac{\frac{2}{x}}{\frac{1}{x}}$

We see that $\tan \left(\frac{2}{x}\right)$ converges to $0$ as $x \to \infty$ because the larger the value of $x$ the smaller the expression within the tangent becomes and the closer $\tan \left(\frac{2}{x}\right)$ becomes to $0$. Also, the limit ${\lim}_{x \to \infty} \frac{1}{x} = 0$ is commonly used. Therefore, we may use l;Hosptial's rule.

$L = {\lim}_{x \to \infty} \frac{- \frac{2}{x} ^ 2 \cdot {\sec}^{2} \left(\frac{2}{x}\right)}{- \frac{1}{x} ^ 2}$

$L = {\lim}_{x \to \infty} 2 {\sec}^{2} \left(\frac{2}{x}\right)$

The same principal applies with $\frac{2}{x}$ as $\frac{1}{x}$: the limit as $x$ approaches infinity always remains $0$.

$L = 2 {\sec}^{2} \left(0\right)$

$L = 2 \left(1\right)$

$L = 2$

A graphical verification confirms.

In the above graph, the red curve is $x \tan \left(\frac{2}{x}\right)$ and the blue line is $y = 2$. As you can see, the curve converges onto the line.

Hopefully this helps!

##### Questions
• · A minute ago
• · 16 minutes ago
• · 46 minutes ago
• · 2 hours ago
• · 2 hours ago
• · 2 hours ago
• · 2 hours ago · in Power Rule
• · 3 hours ago
• 3 hours ago
• · 4 hours ago
• · 5 hours ago
• · 5 hours ago · in Tangent Line to a Curve
• 5 hours ago · in Chain Rule
• · 5 hours ago
• · 8 hours ago
• · 9 hours ago
• · 9 hours ago
• · 9 hours ago · in Integration by Parts
• · 9 hours ago