Question

Please solve this integral problem ASAP
and please no fake answers
​

Answer 1

$\bold{\boxed{\color{red}{\bf{\underline{Hope. it's .helpful .to .you}}}}}$

Answer 2

★ Concept ::

Here the concept of Integration has been used. We see that we are given a equation to be integrated. Firstly we shall integrate with respect to x to simplify this equation. Then we will integrate and then using trigonometric identity we will integrate further to get our answer

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{\pink{\sin(A\:+\:B)\;=\;\bf{\sin\:A\cos\:B\;+\;\cos\:A\sin\:B}}}}$

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★ Solution :-

Given,

$\\\;\displaystyle{\sf{\mapsto\;\;\red{\int\:\dfrac{\sin\:x}{\sin(x\:-\;a)}\:dx}}}$

Let (x - a) be t.

Then,

$\\\;\displaystyle{\sf{\rightarrow\;\;\int\:\dfrac{\sin\:(t\;+\;a)}{\sin\:t}\:dx}}$

(since x - a + a = t + a = x)

Now let's DIFFERENTIATE t with respect to x.

→ dt/dx = [d(x - a)] / dx

→ dt/dx = 1 (since a is constant)

→ dt = dx

Now let's integrate the main equation we have.

$\\\;\displaystyle{\sf{\rightarrow\;\;\int\:\dfrac{\sin\:(t\;+\;a)}{\sin\:t}\:dt}}$

We know that,

$\\\;\tt{\rightarrow\;\;\sin(A\:+\:B)\;=\;\sin\:A\cos\:B\;+\;\cos\:A\sin\:B}$

• Here A = t
• B = a

By applying this in the above equation, we get

$\\\;\displaystyle{\sf{\rightarrow\;\;\int\:\dfrac{\sin\:t\cos\:a\;+\;\cos\:a\sin\:t}{\sin\:t}\:dt}}$

Now separating the equation, we get

$\\\;\displaystyle{\sf{\rightarrow\;\;\int\:\bigg[\bigg(\dfrac{\sin\:t\cos\:a}{\sin\:t}\bigg)\;+\;\bigg(\dfrac{\cos\:a\sin\:t}{\sin\:t}\bigg)\bigg]\:dt}}$

$\\\;\displaystyle{\sf{\rightarrow\;\;\int\:(\cos\:a\:dt)\;+\;\int\;(\cot\:t\sin\:a\:dt)}}$

This will give us,

$\\\;\displaystyle{\sf{\rightarrow\;\;\int\:\cos\:a\:dt\;+\;\int\;\cot\:t\sin\:a\:dt}}$

$\\\;\displaystyle{\sf{\rightarrow\;\;\cos\:a\int\:dt\;+\;\sin\:a\int\;\cot\:t\:dt}}$

(We already know that sin and cos functions are constants. )

$\\\;\displaystyle{\sf{\rightarrow\;\;\cos\:a\;\times\;t\;+\;\sin\:a\:\log\bigg|\sin\:t\bigg|\;+\;C}}$

Now using the value of t = x - a, we get

$\\\;\displaystyle{\sf{\rightarrow\;\;\cos\:a\;\times\;(x\:-\:a)\;+\;\sin\:a\:\log\bigg|\sin\:(x\:-\:a)\bigg|\;+\;C}}$

$\\\;\displaystyle{\sf{\rightarrow\;x\cos\:a\;-\;a\cos\:a\;+\;\sin\:a\:\log\bigg|\sin(x\:-\:a)\bigg|\;+\;C}}$

By rearranging, we get

$\\\;\displaystyle{\sf{\rightarrow\;\;\sin\:a\:\log\bigg|\sin(x\:-\:a)\bigg|\;+\;x\cos\:a\;-\;a\cos\:a\;+\;C}}$

• Let, - a cos a + C = D

Then,

$\\\;\displaystyle{\sf{\rightarrow\;\;\orange{\sin\:a\:\log\bigg|\sin(x\:-\:a)\bigg|\;+\;x\cos\:a\;+\;D}}}$

By applying intial equation, we get

$\\\;\displaystyle{\sf{\rightarrow\;\;\int\:\dfrac{\sin\:x}{\sin(x\:-\;a)}\:dx\;=\;\blue{\sin\:a\:\log|\sin(x\:-\:a)|\;+\;x\cos\:a\;+\;D}}}$

This is the required answer.

$\\\;\underline{\boxed{\tt{Required\;\:answer\;=\;\bf{\purple{\sin\:a\:\log|\sin(x\:-\:a)|\;+\;x\cos\:a\;+\;D}}}}}$