Question

$∫ {cos}^{ - 1}x ( \frac{1}{x})dx$
Please solve this ASAP and those who don't know answer, don't answer ...
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Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the concept of integration has been used. Firstly we can find the value of cos inverse to change expression from fractional form to normal form. Then firstly we will integrate the expression and then differentiate it to find our answer.

Let's do it !!

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★ Correct Question :-

Find the integral product of :-

$\;\displaystyle{\bf{\mapsto\;\;\blue{\int\:\cos^{-1}\:\dfrac{1}{x}\:dx}}}$

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★ Solution :-

Given,

$\\\;\displaystyle{\bf{\mapsto\;\;\red{\int\:\cos^{-1}\:\dfrac{1}{x}\:dx}}}$

We already know that,

$\\\;\rm{\leadsto\;\;\orange{\cos^{-1}\:\dfrac{1}{x}\;=\;\sec^{-1}\:x}}$

Now applying this value in the given expression, we get

$\\\;\displaystyle{\bf{\rightarrow\;\;\int\:\sec^{-1}\:x\:dx}}$

This can be written as,

$\\\;\displaystyle{\sf{\Longrightarrow\;\;\int\:\cos^{-1}\:\dfrac{1}{x}\:dx\;=\;\int\:\sec^{-1}\:x\:dx}}$

This also can be written as,

$\\\;\displaystyle{\sf{\Longrightarrow\;\;\int\:\cos^{-1}\:\dfrac{1}{x}\:dx\;=\;\int\:\sec^{-1}\:x.1\:dx}}$

Now on differentiating this we get,

$\\\;\displaystyle{\sf{\Longrightarrow\;\;\int\:\cos^{-1}\:\dfrac{1}{x}\:dx\;=\;\bf{\sec^{-1}\:x\:\int\:dx\:-\:\int\:\bigg(\dfrac{d}{dx}\:(\sec^{-1}\:x)\:.\:\int\:dx\bigg)dx}}}$

$\\\;\displaystyle{\sf{\Longrightarrow\;\;\int\:\cos^{-1}\:\dfrac{1}{x}\:dx\;=\;\bf{x\:\sec^{-1}\:x\:\int\:dx\:-\:\int\:\dfrac{1}{x\sqrt{x^{2}\;-\;1}}\:x\:dx}}}$

Cancelling x, we get

$\\\;\displaystyle{\sf{\Longrightarrow\;\;\int\:\cos^{-1}\:\dfrac{1}{x}\:dx\;=\;\bf{x\:\sec^{-1}\:x\:\int\:dx\:-\:\int\:\dfrac{1}{\sqrt{x^{2}\;-\;1}}\:dx}}}$

Using integral property, now this can be written as,

$\\\;\displaystyle{\bf{\Longrightarrow\;\;\int\:\cos^{-1}\:\dfrac{1}{x}\:dx\;=\;\bf{\green{x\:\sec^{-1}\:x\:-\:log\:\big|x\;+\;\sqrt{x^{2}\;-\;1}\big|\;+\;C}}}}$

This is the required answer.

$\\\;\;\underline{\boxed{\tt{Required\;\:Integral\;=\;\bf{\purple{x\:\sec^{-1}\:x\:-\:log\:\big|x\;+\;\sqrt{x^{2}\;-\;1}\big|\;+\;C}}}}}$

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★ More to know :-

• Constant Function of Integration ::

$\\\;\displaystyle{\tt{\leadsto\;\;\int\:a\:dx\;=\;ax\;+\;C}}$

• Variable Function of Integration ::

$\\\;\displaystyle{\tt{\leadsto\;\;\int\:x\:dx\;=\;\dfrac{x^{2}}{2}\;+\;C}}$

• Square Function of Integration ::

$\\\;\displaystyle{\tt{\leadsto\;\;\int\:x^{2}\:dx\;=\;\dfrac{x^{3}}{3}\;+\;C}}$

• Reciprocal Function of Integration ::

$\\\;\displaystyle{\tt{\leadsto\;\;\int\:\dfrac{1}{x}\:dx\;=\;In|x|\;+\;C}}$

• Trigonometric Function of Integration ::

$\\\;\displaystyle{\tt{\leadsto\;\;\int\:\cos(x)\:dx\;=\;\sin(x)\;+\;C}}$

$\\\;\displaystyle{\tt{\leadsto\;\;\int\:\sin(x)\:dx\;=\;-\:\cos(x)\;+\;C}}$

Answer 2

Step-by-step explanation:

Width of rectangle is 25 cm.

Step-by-step explanation:

Given :-

A wire bend in form of square of side 30 cm.

Then wire is again bend in form of rectangle of length 35 cm.

To find :-

Width of the rectangle.

Solution :-

Here, Concept is : If we are bending wire in form of square than again bending it in rectangle. Than, perimeter of square will equal to perimeter of rectangle because we are not increasing length of wire by one measure we are bending it in square and rectangular shape.

So,

Perimeter of square = 4 × side

⟶ Perimeter = 4 × 30

⟶ Perimeter = 120

Thus,

Perimeter of square is 120 cm.

According to concept, Perimeter of square and perimeter of rectangle are equal.

So, Perimeter of rectangle is 120 cm.

Let, Breadth or width or rectangle be x cm.

We know,

Perimeter of rectangle = 2(Length + Breadth)

⟶ 120 = 2×(35 + x)

⟶ 120 = 70 + 2x

\⟶ 120 - 70 = 2x

⟶ 50 = 2x

⟶ 50/2 = x

⟶ x = 25

We take, Width of rectangle be x.

Therefore,

Width of rectangle is 25 cm.