Question

Evaluate
$\displaystyle \sf \int \bigg(\dfrac{x}{x \: sinx + cosx} \bigg)^{2} dx$
​

Answer 1

SOLUTION:-

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•We have $\displaystyle \bf{\int \bigg(\dfrac{x}{x \: sinx + cosx} \bigg)^{2} dx}$

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$\displaystyle \sf = \int \bigg( \dfrac{x}{cos \: x} \bigg) \dfrac{x \: cosx \: dx}{(x \: sinx + cos\:x ) ^{2} }$

Note that

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$\displaystyle \sf \star \: \frac{d}{dx} (x \: sinx + cosx) = x \: cos \: x$

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$\displaystyle \sf = \int {\bigg( \dfrac{x}{cos \: x} \bigg)}^{2} \{{(x \: sin x + cosx) }^{ - 2} (x \: cosx) \}dx$

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$\displaystyle \sf = \dfrac{x}{cosx} . \dfrac{(x \: sinx + cos \: x)^{ - 1} }{ - 1} - \int \dfrac{cos \: x.1 + x \: sinx}{{cos}^{2}x } . \dfrac{ {(x \: sinx + cosx)}^{ - 1} }{ - 1}dx$

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$\displaystyle \sf = \dfrac{ - x}{cos \: x(x \: sinx + cosx)} + \int {sec}^{2} x \: dx$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\displaystyle \sf \boxed{\boxed{ \sf= \dfrac{ - x}{cos \: x(x \: sinx + cosx)} + tanx + c} }\to \mathfrak{answer}$

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Answer 2

Answer:

${\tt{\blue{\bold{\huge{Answer}}}}}$

Step-by-step explanation:

To solve this, the first thing you have to keep in mind about integral of

∫ XcosX/(XsinX+cosX)^2.dx, this intuition will come to you by practicing more and more problems of integration's.

Anyway, this can be calculated simply by assuming (XsinX + cosX) as 't' and solve for 'dt'.

t = (XsinX + cosX)

dt = (XcosX + sinX - sinX).dx

dt = XcosX.dx

Now,

= ∫ XcosX/(XsinX+cosX)^2.dx, will become,

= ∫ 1/t^2.dt, which will finally be equal to,

= -1/t, where t = (XsinX + cosX)

In short we can say;

∫ XcosX/(XsinX+cosX)^2.dx = -1/(XsinX + cosX)

Now lets come to main question:

= ∫ X^2/(XcosX + sinX)^2.dx

Make numerator to XcosX somehow, since you already know the integration of

∫ XcosX/(XsinX+cosX)^2.dx in advance, don't you ?

And definitely you know how to make numerator equal to XcosX.

YES you are right!! multiply and divide by cosX, so our integral becomes:

= ∫ cosX.X^2/cosX.(XcosX + sinX)^2.dx, which simplifies to

= ∫ XsecX.XcosX/(XcosX + sinX)^2.dx,

How do I evaluate ∫x2(xsinx+cosx)2dx∫x2(xsin⁡x+cos⁡x)2dx?

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18 Answers



Abhishek Mishra, Maths is awesome :)

Answered December 30, 2015

Originally Answered: How do I integrate x^2/ (xsinx+cosx) ^2 dx?

To solve this, the first thing you have to keep in mind about integral of

∫ XcosX/(XsinX+cosX)^2.dx, this intuition will come to you by practicing more and more problems of integration's.

Anyway, this can be calculated simply by assuming (XsinX + cosX) as 't' and solve for 'dt'.

t = (XsinX + cosX)

dt = (XcosX + sinX - sinX).dx

dt = XcosX.dx

Now,

= ∫ XcosX/(XsinX+cosX)^2.dx, will become,

= ∫ 1/t^2.dt, which will finally be equal to,

= -1/t, where t = (XsinX + cosX)

In short we can say;

∫ XcosX/(XsinX+cosX)^2.dx = -1/(XsinX + cosX)

Now lets come to main question:

= ∫ X^2/(XcosX + sinX)^2.dx

Make numerator to XcosX somehow, since you already know the integration of

∫ XcosX/(XsinX+cosX)^2.dx in advance, don't you ?

And definitely you know how to make numerator equal to XcosX.

YES you are right!! multiply and divide by cosX, so our integral becomes:

= ∫ cosX.X^2/cosX.(XcosX + sinX)^2.dx, which simplifies to

= ∫ XsecX.XcosX/(XcosX + sinX)^2.dx,

Now it makes sense to multiply and divide by cosX right ?

Now you know what to do, yes apply formula for integration for multiplication of 2 terms, and YES keep in mind that we have to choose 2nd term in such a way whose integration is not indefinite, of course secX will not be a good choice

THE formula:

1st INTO integration of 2nd MINUS integration of differential of 1st INTO integration of 2nd.

Yay!! I still remember it along with the tone :D

Hence,

after making secX as first term, further our integral simplifies to:

One clarification about notation: sin^2X is 'sin squared x' and not 'sin to the power 2X'

= XsecX.∫ XcosX/(XcosX + sinX)^2.dx - ∫ (secX + XsecXtanX). ∫ XcosX/(XcosX + sinX)^2.dx

and from here on, you can figure it out ;)

= -XsecX/(XsinX + cosX) + ∫ (secX + XsecXtanX)/(XsinX + cosX).dx

= -XsecX/(XsinX + cosX) + ∫ (1/cosX + XsinX/cos^2X)/(XsinX + cosX).dx

= -XsecX/(XsinX + cosX) + ∫ 1/cos^2X (cosX + XsinX)/(XsinX + cosX).dx

= -XsecX/(XsinX + cosX) + ∫ sec^2X.dx

= -(XsecX/(XsinX + cosX)) + tanX

= -(X/cosX.(XsinX + cosX)) + sinX/cosX

= -(X + sinX.(XsinX + cosX))/cosX.(XsinX + cosX))

= -(X + Xsin^2X + sinXcosX)/cosX.(XsinX + cosX)

= (sinXcosX - Xcos^2X)/cosX.(XsinX + cosX)

= (sinX - XcosX)/(XsinX + cosX)

So,

that's your final answer, is it ? No you forgot to add constant, C

Therefore,

final answer is => (sinX - XcosX)/(XsinX + cosX) + C

P.S.: Solving such problem after 7 years, my 12th class came in front of my eyes and yes L.H.S. = R.H.S