Question

Solve this integral as soon as possible and please don't give fake answer otherwise I will report you...​

Answer 1

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

Here the concept of Integration has been used. We see that we are given a equation. Firstly we shall different it to make it in a simpler form. Then we will use the formula of integration by part and then integrate it to get our answer.

Let's do it !!

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★ Solution :-

Given,

$\\\;\displaystyle{\bf{\rightarrow\;\;\pink{\int\;\dfrac{x^{2}}{(x\sin\:x\;-\;\cos\:x)^{2}}\:dx}}}$

We already know that,

$\\\;\displaystyle{\tt{:\mapsto\;\;\dfrac{d}{dx}\:(x\sin\:x\;-\;\cos\:x)\;=\;x\cos\:x\;+\;\sin\:x\;-\;\sin\:x}}$

This will give us,

$\\\;\displaystyle{\bf{:\mapsto\;\;\orange{\dfrac{d}{dx}\:(x\sin\:x\;-\;\cos\:x)\;=\;x\cos\:x}}}$

From given equation, we get

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

Since, canceling cos x will give us same term that is x².

Now let,

$\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\int\;\dfrac{x}{\cos\:x}.\dfrac{x\cos\:x}{(x\sin\:x\;-\;\cos\:x)^{2}}\:dx}}}$

This will give us,

$\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\int\;\dfrac{x}{\cos\:x}.\dfrac{x\cos\:x}{(x\sin\:x\;-\;\cos\:x)^{2}}\:dx}}}$

Since both will give us same value.

Now using applying Integration into different parts (since we have two parts here), we get

$\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\dfrac{x}{\cos\:x}.\bigg(\dfrac{-\:1}{x\sin\:x\;+\;\cos\:x}\bigg)\;-\;\int\;\bigg(\dfrac{x\cos\:x\;+\;x\sin\:x}{\cos^{2}\:x}.\dfrac{-\:1}{x\sin\:x\;+\;\cos\:x}\:dx\bigg)}}}$

Now taking the negative sign out, we get

$\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\dfrac{x}{\cos\:x}.\bigg(\dfrac{-\:1}{x\sin\:x\;+\;\cos\:x}\bigg)\;-\;(-)\bigg(\int\;\dfrac{x\cos\:x\;+\;x\sin\:x}{\cos^{2}\:x}.\dfrac{-\:1}{x\sin\:x\;+\;\cos\:x}\:dx\bigg)}}}$

$\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\dfrac{x}{\cos\:x}\:\times\:\bigg(\dfrac{-\:1}{x\sin\:x\;+\;\cos\:x}\bigg)\;+\;\int\;\bigg(\sec^{3}\:x\;\times\;dx\bigg)}}}$

$\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\dfrac{x}{\cos\:x}\:\times\:\bigg(\dfrac{-\:1}{x\sin\:x\;+\;\cos\:x}\bigg)\;+\;\int\;\sec^{3}\:x\:dx}}}$

$\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\dfrac{-\:x}{\cos\:x(x\sin\:x\;+\;\cos\:x)}\;+\;\int\;\sec^{3}\:x\:dx}}}$

On integrating sec³ x dx, we get

$\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\dfrac{-\:x}{\cos\:x(x\sin\:x\;+\;\cos\:x)}\;+\;(\tan\:x\;+\;C)}}}$

We know that,

$\\\;\tt{\mapsto\;\;\tan\theta\;=\;\dfrac{\sin\theta}{\cos\theta}}$

On applying this, we get

$\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\dfrac{-\:x}{\cos\:x(x\sin\:x\;+\;\cos\:x)}\;+\;\bigg(\dfrac{\sin\:x}{\cos\:x}\;+\;C\bigg)}}}$

Now taking the LCM of first two terms, we get

$\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\dfrac{-\:x}{\cos\:x(x\sin\:x\;+\;\cos\:x)}\;+\;\dfrac{\sin\:x(x\sin\:x\;+\;\cos\:x)}{\cos\:x(x\sin\:x\;+\;\cos\:x)}\;+\;C}}}$

Now since denominators are same, we can add them to get,

$\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\dfrac{-\:x\;+\;\sin\:x(x\sin\:x\;+\;\cos\:x)}{\cos\:x(x\sin\:x\;+\;\cos\:x)}\;+\;C}}}$

$\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\dfrac{-\:x\cos^{2}\:x\;+\;\sin\:x\:.\:\cos\:x}{\cos\:x(x\sin\:x\;+\;\cos\:x)}\;+\;C}}}$

Now taking cos x in common we get,

$\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\dfrac{\cos\:x(-\:x\cos\:x\;+\;\sin\:x)}{\cos\:x(x\sin\:x\;+\;\cos\:x)}\;+\;C}}}$

Now let's cancel cos x from both numerator and denominator, which will give us

$\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\dfrac{(-\:x\cos\:x\;+\;\sin\:x)}{(x\sin\:x\;+\;\cos\:x)}\;+\;C}}}$

On rearranging, we get

$\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\dfrac{\sin\:x\;-\;x\cos\:x}{(\cos\:x\;+\;x\sin\:x)}\;+\;C}}}$

$\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\red{\bigg(\dfrac{\sin\:x\;-\;x\cos\:x}{\cos\:x\;+\;x\sin\:x}\bigg)\;+\;C}}}}$

Now equating the value of A, we get

$\\\;\displaystyle{\sf{\rightarrow\;\;\int\;\dfrac{x^{2}}{(x\sin\:x\;-\;\cos\:x)^{2}}\:dx\;=\;\bf{\blue{\bigg(\dfrac{\sin\:x\;-\;x\cos\:x}{\cos\:x\;+\;x\sin\:x}\bigg)\;+\;C}}}}$

$\\\;\underline{\boxed{\tt{Required\;\:Integral\;=\;\bf{\purple{\bigg(\dfrac{\sin\:x\;-\;x\cos\:x}{\cos\:x\;+\;x\sin\:x}\bigg)\;+\;C}}}}}$

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★ More to know :-

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

• This is the Product Rule of Integration.

$\\\;\displaystyle{\sf{\leadsto\;\;\int\:(f\:-\:g)\:dx\;=\;\int\:f\:dx\;-\;\int\:g\:dx}}$

• This is the Difference Rule of Integration.

$\\\;\displaystyle{\sf{\leadsto\;\;\int\:c\:f(x)\:dx\;=\;c\int\:f(x)\:dx}}$

• This is the Multiplication by Constant Rule.

Answer 2

Given,

\begin{gathered}\\\;\displaystyle{\bf{\rightarrow\;\;\pink{\int\;\dfrac{x^{2}}{(x\sin\:x\;-\;\cos\:x)^{2}}\:dx}}}\end{gathered}

→∫

(xsinx−cosx)

2

x

2

dx

We already know that,

\begin{gathered}\\\;\displaystyle{\tt{:\mapsto\;\;\dfrac{d}{dx}\:(x\sin\:x\;-\;\cos\:x)\;=\;x\cos\:x\;+\;\sin\:x\;-\;\sin\:x}}\end{gathered}

:↦

dx

d

(xsinx−cosx)=xcosx+sinx−sinx

This will give us,

\begin{gathered}\\\;\displaystyle{\bf{:\mapsto\;\;\orange{\dfrac{d}{dx}\:(x\sin\:x\;-\;\cos\:x)\;=\;x\cos\:x}}}\end{gathered}

:↦

dx

d

(xsinx−cosx)=xcosx

From given equation, we get

\begin{gathered}\\\;\displaystyle{\bf{\rightarrow\;\;\int\;\dfrac{x}{\cos\:x}.\dfrac{x\cos\:x}{(x\sin\:x\;-\;\cos\:x)^{2}}\:dx}}\end{gathered}

→∫

cosx

x

.

(xsinx−cosx)

2

xcosx

dx

Since, canceling cos x will give us same term that is x².

Now let,

\begin{gathered}\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\int\;\dfrac{x}{\cos\:x}.\dfrac{x\cos\:x}{(x\sin\:x\;-\;\cos\:x)^{2}}\:dx}}}\end{gathered}

→A=∫

cosx

x

.

(xsinx−cosx)

2

xcosx

dx

This will give us,

\begin{gathered}\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\int\;\dfrac{x}{\cos\:x}.\dfrac{x\cos\:x}{(x\sin\:x\;-\;\cos\:x)^{2}}\:dx}}}\end{gathered}

→A=∫

cosx

x

.

(xsinx−cosx)

2

xcosx

dx

Since both will give us same value.

Now using applying Integration into different parts (since we have two parts here), we get

\begin{gathered}\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\dfrac{x}{\cos\:x}.\bigg(\dfrac{-\:1}{x\sin\:x\;+\;\cos\:x}\bigg)\;-\;\int\;\bigg(\dfrac{x\cos\:x\;+\;x\sin\:x}{\cos^{2}\:x}.\dfrac{-\:1}{x\sin\:x\;+\;\cos\:x}\:dx\bigg)}}}\end{gathered}

→A=

cosx

x

.(

xsinx+cosx

−1

)−∫(

cos

2

x

xcosx+xsinx

.

xsinx+cosx

−1

dx)

Now taking the negative sign out, we get

\begin{gathered}\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\dfrac{x}{\cos\:x}.\bigg(\dfrac{-\:1}{x\sin\:x\;+\;\cos\:x}\bigg)\;-\;(-)\bigg(\int\;\dfrac{x\cos\:x\;+\;x\sin\:x}{\cos^{2}\:x}.\dfrac{-\:1}{x\sin\:x\;+\;\cos\:x}\:dx\bigg)}}}\end{gathered}

→A=

cosx

x

.(

xsinx+cosx

−1

)−(−)(∫

cos

2

x

xcosx+xsinx

.

xsinx+cosx

−1

dx)

\begin{gathered}\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\dfrac{x}{\cos\:x}\:\times\:\bigg(\dfrac{-\:1}{x\sin\:x\;+\;\cos\:x}\bigg)\;+\;\int\;\bigg(\sec^{3}\:x\;\times\;dx\bigg)}}}\end{gathered}

→A=

cosx

x

×(

xsinx+cosx

−1

)+∫(sec

3

x×dx)

\begin{gathered}\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\dfrac{x}{\cos\:x}\:\times\:\bigg(\dfrac{-\:1}{x\sin\:x\;+\;\cos\:x}\bigg)\;+\;\int\;\sec^{3}\:x\:dx}}}\end{gathered}

→A=

cosx

x

×(

xsinx+cosx

−1

)+∫sec

3

xdx

\begin{gathered}\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\dfrac{-\:x}{\cos\:x(x\sin\:x\;+\;\cos\:x)}\;+\;\int\;\sec^{3}\:x\:dx}}}\end{gathered}

→A=

cosx(xsinx+cosx)

−x

+∫sec

3

xdx

On integrating sec³ x dx, we get

\begin{gathered}\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\dfrac{-\:x}{\cos\:x(x\sin\:x\;+\;\cos\:x)}\;+\;(\tan\:x\;+\;C)}}}\end{gathered}

→A=

cosx(xsinx+cosx)

−x

+(tanx+C)

We know that,

\begin{gathered}\\\;\tt{\mapsto\;\;\tan\theta\;=\;\dfrac{\sin\theta}{\cos\theta}}\end{gathered}

↦tanθ=

cosθ

sinθ

On applying this, we get

\begin{gathered}\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\dfrac{-\:x}{\cos\:x(x\sin\:x\;+\;\cos\:x)}\;+\;\bigg(\dfrac{\sin\:x}{\cos\:x}\;+\;C\bigg)}}}\end{gathered}

→A=

cosx(xsinx+cosx)

−x

+(

cosx

sinx

+C)

Now taking the LCM of first two terms, we get

\begin{gathered}\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\dfrac{-\:x}{\cos\:x(x\sin\:x\;+\;\cos\:x)}\;+\;\dfrac{\sin\:x(x\sin\:x\;+\;\cos\:x)}{\cos\:x(x\sin\:x\;+\;\cos\:x)}\;+\;C}}}\end{gathered}

→A=

cosx(xsinx+cosx)

−x

+

cosx(xsinx+cosx)

sinx(xsinx+cosx)

+C

Now since denominators are same, we can add them to get,

\begin{gathered}\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\dfrac{-\:x\;+\;\sin\:x(x\sin\:x\;+\;\cos\:x)}{\cos\:x(x\sin\:x\;+\;\cos\:x)}\;+\;C}}}\end{gathered}

→A=

cosx(xsinx+cosx)

−x+sinx(xsinx+cosx)

+C

\begin{gathered}\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\dfrac{-\:x\cos^{2}\:x\;+\;\sin\:x\:.\:\cos\:x}{\cos\:x(x\sin\:x\;+\;\cos\:x)}\;+\;C}}}\end{gathered}

→A=

cosx(xsinx+cosx)

−xcos

2

x+sinx.cosx

+C

Now taking cos x in common we get,

\begin{gathered}\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\dfrac{\cos\:x(-\:x\cos\:x\;+\;\sin\:x)}{\cos\:x(x\sin\:x\;+\;\cos\:x)}\;+\;C}}}\end{gathered}

→A=

cosx(xsinx+cosx)

cosx(−xcosx+sinx)

+C

Now let's cancel cos x from both numerator and denominator, which will give us

\begin{gathered}\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\dfrac{(-\:x\cos\:x\;+\;\sin\:x)}{(x\sin\:x\;+\;\cos\:x)}\;+\;C}}}\end{gathered}

→A=

(xsinx+cosx)

(−xcosx+sinx)

+C

On rearranging, we get

\begin{gathered}\\\;\displaystyle{\sf{\rightarrow\;\;A\;=\;\bf{\dfrac{\sin\:x\;-\;x\cos\:x}{(\cos\:x\;+\;x\sin\:x)}\;+\;C}}}\end{gathered}

→A=

(cosx+xsinx)

sinx−xcosx

+C