Question

please solve this and please don't post fake answer.​

Answer 1

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

Here the Concept of Integration has been used. We see we are given a equation. Firstly we shall reduce the given equation in form of sin and cos function. Then we shall integrate it and find the answer.

Let's do it !!

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★ Solution :-

$\\\;\displaystyle{\bf{\mapsto\;\;\green{\int\:\dfrac{1\;+\;\cos\:4x}{\cot\:x\;-\;\tan\:x}\:dx}}}$

Now firstly let's simplify this equation in the form of sin and cos function.

We know that,

$\\\;\tt{\leadsto\;\;\orange{\cot\:\theta\;=\;\dfrac{\cos\theta}{\sin\theta}}}$

$\\\;\tt{\leadsto\;\;\orange{\tan\:\theta\;=\;\dfrac{\sin\theta}{\cos\theta}}}$

Now applying this with the given expression, we get

$\\\;\displaystyle{\bf{\rightarrow\;\;\int\:\dfrac{1\;+\;\cos\:4x}{\bigg(\dfrac{\cos\:x}{\sin\:x}\bigg)\;-\;\bigg(\dfrac{\sin\:x}{\cos\:x}\bigg)}\:dx}}$

$\\\;\displaystyle{\bf{\rightarrow\;\;\int\:\dfrac{1\;+\;\cos\:4x}{\dfrac{\cos\:x}{\sin\:x}\;-\;\dfrac{\sin\:x}{\cos\:x}}\:dx}}$

$\\\;\displaystyle{\bf{\rightarrow\;\;\int\:\dfrac{1\;+\;\cos\:4x}{\dfrac{\cos^{2}\:x\;-\;\sin^{2}\:x}{\sin\:x\;\cos\:x}}\:dx}}$

On integrating, this can be written as,

$\\\;\displaystyle{\bf{\rightarrow\;\;\int\:\dfrac{2\:\cos^{2}\:2x}{\cos\:2x}\:\sin\:x\:.\:\cos\:x\:dx}}$

$\\\;\displaystyle{\bf{\rightarrow\;\;\int\:\dfrac{1\;+\;\cos\:4x}{\cot\:x\;-\;\tan\:x}\:dx\;=\;\bf{\int\:\dfrac{2\:\cos^{2}\:2x}{\cos\:2x}\:\sin\:x\:.\:\cos\:x\:dx}}}$

Finally this will give us,

$\\\;\displaystyle{\bf{\rightarrow\;\;\int\:\dfrac{1\;+\;\cos\:4x}{\cot\:x\;-\;\tan\:x}\:dx\;=\;\bf{\int\:\cos\:2x\:.\:\sin\:2x\:dx}}}$

Now taking out the ½ , we get

$\\\;\displaystyle{\bf{\rightarrow\;\;\int\:\dfrac{1\;+\;\cos\:4x}{\cot\:x\;-\;\tan\:x}\:dx\;=\;\bf{\dfrac{1}{2}\:\int\:\sin\:4x\:dx}}}$

This will eventually give us,

$\\\;\displaystyle{\bf{\rightarrow\;\;\int\:\dfrac{1\;+\;\cos\:4x}{\cot\:x\;-\;\tan\:x}\:dx\;=\;\bf{\red{-\:\dfrac{1}{8}\:\cos\:4x\:+\:C}}}}$

This is the required answer.

$\\\;\underline{\boxed{\tt{Required\;Integral\;=\;\bf{\purple{-\:\dfrac{1}{8}\:\cos\:4x\:+\:C}}}}}$

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★ Laws of integration :-

$\\\;\displaystyle{\sf{\leadsto\;\;\int\:x^{n}\:dx\;=\;\dfrac{x^{n\:+\:1}}{n\;+\;1}\:+\:C}}$

$\\\;\displaystyle{\sf{\leadsto\;\;\int\:a^{x}\:dx\;=\;\dfrac{a^{x}}{log_{e}\:a}\:+\:C}}$

$\\\;\displaystyle{\sf{\leadsto\;\;\int\:e^{x}\:dx\;=\;e^{x}\:+\:C}}$