Scary Integration!

Calculus Level 4

∫ x 2 ( x sin ⁡ x + cos ⁡ x ) 2 d x \large \displaystyle \int \frac { { x }^{ 2 } }{ ({ x\sin { x } +\cos { x } ) }^{ 2 } } dx

Which of the choices is equivalent to the indefinite integral above (ignoring arbitrary constant)?

A . I = sin ⁡ x + x cos ⁡ x x sin ⁡ x − cos ⁡ x \displaystyle A.\quad I= \frac { \sin { x } +x\cos { x } }{ x\sin { x } -\cos { x } }

B . I = sin ⁡ x − x cos ⁡ x x sin ⁡ x + cos ⁡ x \displaystyle B.\quad I= \frac { \sin { x } -x\cos { x } }{ x\sin { x } +\cos { x } }

C . I = x sec ⁡ x x sin ⁡ x + cos ⁡ x − ∫ sec ⁡ x ( 1 + x tan ⁡ x ) x sin ⁡ x + cos ⁡ x d x \displaystyle C.\quad I= \frac { x\sec { x } }{ x\sin { x } +\cos { x } } -\int \frac { \sec { x } (1+x\tan { x) } }{ x\sin { x } +\cos { x } } dx

D . I = − x sec ⁡ x x sin ⁡ x + cos ⁡ x + ∫ sec ⁡ x ( 1 + x tan ⁡ x ) x sin ⁡ x + cos ⁡ x d x \displaystyle D.\quad I= -\frac { x\sec { x } }{ x\sin { x } +\cos { x } } +\int \frac { \sec { x } (1+x\tan { x) } }{ x\sin { x } +\cos { x } } dx

This was a question asked in IITJEE when the paper was subjective type. Please share if you like the problem.
B,D D,C A A,C B,C

Samarth Agarwal by Samarth Agarwal , 21, India

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1 solution

Ravi Dwivedi
Jul 10, 2015

Let I = ∫ x 2 ( x s i n x + c o s x ) 2 d x I=\int \frac{x^2}{(xsinx+cosx)^2} \mathrm{d}x\\

Writing x 2 = x s e c x ⋅ x c o s x x^2=xsecx \cdot xcosx

I = ∫ x s e c x ⋅ x c o s x ( x s i n x + c o s x ) 2 d x I=\int \frac{xsecx \cdot xcosx}{(xsinx+cosx)^2}\mathrm{d}x

Integrating by parts taking x s e c x xsecx as first function and x c o s x ( x s i n x + c o s x ) 2 \frac{xcosx}{(xsinx+cosx)^2} as second function we get

I = x s e c x ∫ x c o s x ( x s i n x + c o s x ) 2 + ∫ ( ( x s e c x t a n x + s e c x ) ( ∫ x c o s x ( x s i n x + c o s x ) 2 d x ) d x ) . . . ( 1 ) I=x secx \int \frac{xcosx}{(xsinx+cosx)^2} + \int( {(xsecxtanx + secx)}( \int \frac{xcosx}{(xsinx+cosx)^2}\mathrm{d}x)\mathrm{d}x) ...(1)

Put I 1 = ∫ x c o s x ( x s i n x + c o s x ) 2 d x I_{1}=\int \frac{xcosx}{(xsinx+cosx)^2}\mathrm{d}x\\

Put x s i n x + c o s x = t xsinx+cosx=t ⟹ x c o s x d x = d t \implies xcosx \mathrm{d}x= \mathrm{d}t

⟹ I 1 = ∫ 1 t 2 d t \implies I_{1}=\int \frac{1}{t^2}\mathrm{d}t ⟹ I 1 = − 1 t \implies I_{1}= -\frac{1}{t} (Ignoring arbitrary constant as per the question)

⟹ I 1 = − 1 x c o s x + s i n x \implies I_{1}=-\frac{1}{xcosx+sinx}

Putting value of I 1 I_{1} in ( 1 ) (1) we get

I = − x s e c x x s i n x + c o s x + ∫ s e c x ( 1 + t a n x ) x s i n x + c o s x d x which is the option D I=-\frac{xsecx}{xsinx+cosx} + \int \frac{secx(1+tanx)}{xsinx+cosx}\mathrm{d}x \\ \text{which is the option D}

Simplifying this further by taking c o s x cosx common in denominator in the second term we get

I = − x s e c x x s i n x + c o s x + ∫ s e c x ( 1 + x t a n x ) c o s x ( 1 + x t a n x ) d x I=-\frac{xsecx}{xsinx+cosx} + \int \frac{secx(1+xtanx)}{cosx(1+xtanx)}\mathrm{d}x

I = − x s e c x x s i n x + c o s x + ∫ s e c x c o s x d x I=-\frac{xsecx}{xsinx+cosx} + \int \frac{secx}{cosx}\mathrm{d}x

I = − x s e c x x s i n x + c o s x + ∫ sec ⁡ 2 x d x I=-\frac{xsecx}{xsinx+cosx} + \int \sec^2 x\mathrm{d}x

I = − x s e c x x s i n x + c o s x + t a n x I=-\frac{xsecx}{xsinx+cosx} + tanx

I = − x x c o s x s i n x + cos ⁡ 2 x + t a n x I=-\frac{x}{xcosxsinx+\cos^2 x} +tanx (Dividing by secx both num and denominator in first term)

I = − x + x sin ⁡ 2 x + s i n x c o s x c o s x ( x s i n x + c o s x ) I=\frac{-x+x\sin^2 x+sinxcosx}{cosx(xsinx+cosx)}

I = − x cos ⁡ 2 x + s i n x c o s x c o s x ( x s i n x + c o s x ) I=\frac{-x\cos^2 x+sinxcosx}{cosx(xsinx+cosx)}

I = c o s x ( − x c o s x + s i n x ) c o s x ( x s i n x + c o s x ) I=\frac{cosx(-xcosx+sinx)}{cosx(xsinx+cosx)}

I = s i n x − x c o s x x s i n x + c o s x I=\frac{sinx-xcosx}{xsinx+cosx} which is the option B.

So B and D are correct.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Since this is a JEE problem, a quicker approach would be to differentiate all of the options, to see if they match up with what we want.

Otherwise, all that you know is B and D are true, but have not verified if A, C, E could be true.

First of all I want to say that in the options of this question the only combination in which B and D appears together is B,D.So no need to check for A and C (note there is no E in this).

Also it is clear from my method that A and C are not the answers as that we can say by comparing the answers B and D with A and C.

I would like to say here that differentiation is not the shorter approach as JEE exam don't have that much time to differentiate all these options,then simplify and then check.

IIT has given this question not to differentiate the options but by viewing the options you can have idea what to do. I mean differentiating won't work

Here I come up with the thinking x 2 = x s e c x ⋅ x c o s x x^2= xsecx \cdot xcosx (which I can tell you how) after which all follows immediately and can be done in a JEE exam and _this method is advisable to post on brilliant_

By the way can you tell me which year paper was it?

Ravi Dwivedi - 5 years, 11 months ago

This question was asked when the paper was subjective with no options

Samarth Agarwal - 5 years, 11 months ago

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Which year was it?

Ravi Dwivedi - 5 years, 11 months ago

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I dont know that...sorry:(

Samarth Agarwal - 5 years, 11 months ago

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@Samarth Agarwal – OK...By the way.....enjoyed solving it!!

Ravi Dwivedi - 5 years, 11 months ago

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