If $\int e^{x}(\frac{x~\sin^{-1}x}{\sqrt{1-x^{2}}}+\frac{\sin^{-1}x}{(1-x^{2})^{3/2}}+\frac{x}{1-x^{2}})dx=g(x)+C,$ where C is the constant of integration, then $g(\frac{1}{2})$ equals: Options: (A) $\frac{\pi}{4}\sqrt{\frac{e}{3}}$ (B) $\frac{\pi}{6}\sqrt{\frac{e}{3}}$ (C) $\frac{\pi}{4}\sqrt{\frac{e}{2}}$ (D) $\frac{\pi}{6}\sqrt{\frac{e}{2}}$

### Analyzing the Setup When you first glance at the integral $$ \int e^{x} \left( \frac{x \sin^{-1}x}{\sqrt{1-x^{2}}} + \frac{\sin^{-1}x}{(1-x^{2})^{3/2}} + \frac{x}{1-x^{2}} \right) dx $$ your heart might skip a beat. It is messy, algebraic, and involves inverse trigonometric functions. However, in the world of JEE Advanced, whenever you see an exponential function $e^x$ multiplying a sum of terms, you are looking at a pattern recognition test. The identity we are hunting for is the beautiful $$ \int e^x (f(x) + f'(x)) dx = e^x f(x) + C $$ This is the **'Golden Key'** of exponential integration. Our mission is to find the function $f(x)$ hidden inside that bracket. ### Identifying the Function Let's test the most complex-looking term: $$ f(x) = \frac{x \sin^{-1}x}{\sqrt{1-x^2}} $$ Now, we must differentiate this. Using the quotient rule, where $u = x \sin^{-1}x$ and $v = \sqrt{1-x^2}$, we calculate $u'$ and $v'$. The derivative $u'$ involves the product rule, giving us $\sin^{-1}x + \frac{x}{\sqrt{1-x^2}}$. The derivative $v'$ is simply $\frac{-x}{\sqrt{1-x^2}}$. ### The Master Equation When we assemble the derivative $f'(x) = \frac{vu' - uv'}{v^2}$, the magic happens. The terms align, the square roots cancel, and we are left with exactly the remaining terms in our integral. The numerator becomes $$ \sin^{-1}x \left( \sqrt{1-x^2} + \frac{x^2}{\sqrt{1-x^2}} \right) + x $$ which simplifies to $$ \frac{\sin^{-1}x}{\sqrt{1-x^2}} + x $$ Dividing by $v^2 = 1-x^2$, we get $$ f'(x) = \frac{\sin^{-1}x}{(1-x^2)^{3/2}} + \frac{x}{1-x^2} $$ This matches the remaining terms perfectly! We have confirmed our $f(x)$. ### Final Calculation Now, the final step is simply to evaluate $g(x) = e^x f(x)$ at $x = \frac{1}{2}$. Substituting the values, we get $$ g\left(\frac{1}{2}\right) = e^{\frac{1}{2}} \cdot \frac{\frac{1}{2} \sin^{-1}(\frac{1}{2})}{\sqrt{1 - (\frac{1}{2})^2}} $$ With $\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$ and $\sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$, the arithmetic simplifies beautifully. The final result is $$ \frac{\pi}{6} \sqrt{\frac{e}{3}} $$ You see? The monster was just a puzzle waiting to be solved. Keep this pattern in your toolkit, and no integral will ever intimidate you again.

JEE Main 2025 (January)

LEVELJEE Main

Animated Solution for Mathematics - Indefinite Integration: If $\int e^{x} (\frac{x s i n ^{- 1} x}{1 - x ^{2}} + \frac{s i n ^{- 1} x}{( 1 - x ^{2} ) ^{3/2}} + \frac{x}{1 - x ^{2}}) d x = g (x) + C,$ where C is the constant of integration, then $g (\frac{1}{2})$ equals:

Select Answer:

Progressive Hints

Hint 1

Look at the structure of the integral. Whenever you see an exponential function

e^{x}

sitting outside a complex algebraic expression, your brain should immediately scream 'Integration by Parts' or, more specifically, the standard form

\int e^{x} (f (x) + f^{'} (x)) d x

. Do not try to integrate term by term; look for the hidden relationship between the terms inside the bracket.

Hint 2

Your toolkit for this problem is the identity

\int e^{x} (f (x) + f^{'} (x)) d x = e^{x} f (x) + C

. The challenge here is not integration, but differentiation. You need to identify which part of the expression inside the bracket acts as

f (x)

and verify if the remaining terms constitute its derivative

f^{'} (x)

Hint 3

Start by testing the most complex-looking term as your candidate for

f (x)

. Let

f (x) = \frac{x s i n ^{- 1} x}{1 - x ^{2}}

. Now, apply the quotient rule carefully. Differentiate this function and see if the resulting expression matches the other two terms in your integral. If it does, you have cracked the code!

Visualized Solution

Analyzing the Integral Structure

The given integral is: $\int e^{x} (\frac{x s i n ^{- 1} x}{1 - x ^{2}} + \frac{s i n ^{- 1} x}{( 1 - x ^{2} ) ^{\frac{3}{2}}} + \frac{x}{1 - x ^{2}}) d x$
Notice the presence of $e^{x}$ multiplied by a large algebraic expression.

The Classic $e^{x}$ Integration Tool

Recall the standard result: $\int e^{x} (f (x) + f^{'} (x)) d x = e^{x} f (x) + C$
Our goal is to split the bracket into a function $f (x)$ and its exact derivative $f^{'} (x)$ .

Identifying the Function $f (x)$

Let's make an educated guess for $f (x)$ .
Let $f (x) = \frac{x s i n ^{- 1} x}{1 - x ^{2}}$
We will differentiate this to see if it generates the remaining terms.

Applying the Quotient Rule

We use the Quotient Rule: $\frac{d}{d x} (\frac{u}{v}) = \frac{v u ^{'} - u v ^{'}}{v ^{2}}$
Here, $u = x sin^{- 1} x$ and $v = 1 - x^{2}$
$u^{'} = sin^{- 1} x + \frac{x}{1 - x ^{2}}$ (using Product Rule)
$v^{'} = \frac{- x}{1 - x ^{2}}$ (using Chain Rule)

Constructing the Numerator of $f^{'} (x)$

The numerator of $f^{'} (x)$ is $v u^{'} - u v^{'}$
$v u^{'} = 1 - x^{2} (sin^{- 1} x + \frac{x}{1 - x ^{2}}) = 1 - x^{2} sin^{- 1} x + x$
$- u v^{'} = - (x sin^{- 1} x) (\frac{- x}{1 - x ^{2}}) = \frac{x ^{2} s i n ^{- 1} x}{1 - x ^{2}}$

Simplifying $f^{'} (x)$

Combine the $sin^{- 1} x$ terms in the numerator:
$sin^{- 1} x (1 - x^{2} + \frac{x ^{2}}{1 - x ^{2}}) = sin^{- 1} x (\frac{1 - x ^{2} + x ^{2}}{1 - x ^{2}}) = \frac{s i n ^{- 1} x}{1 - x ^{2}}$
So, the full numerator is $\frac{s i n ^{- 1} x}{1 - x ^{2}} + x$
Divide by $v^{2} = 1 - x^{2}$ to get $f^{'} (x)$

Matching $f^{'} (x)$ with the Integral

$f^{'} (x) = \frac{\frac{s i n ^{- 1} x}{1 - x ^{2}} + x}{1 - x ^{2}}$
Splitting the fraction: $f^{'} (x) = \frac{s i n ^{- 1} x}{( 1 - x ^{2} ) ^{\frac{3}{2}}} + \frac{x}{1 - x ^{2}}$
This perfectly matches the remaining terms in our original integral!

Extracting $g (x)$

Since the integral is exactly $\int e^{x} (f (x) + f^{'} (x)) d x$ , the result is $e^{x} f (x) + C$ .
The problem states the result is $g (x) + C$ .
Therefore, $g (x) = e^{x} f (x) = e^{x} (\frac{x s i n ^{- 1} x}{1 - x ^{2}})$

Substituting $x = \frac{1}{2}$

We need to find the value of $g (\frac{1}{2})$ .
Substitute $x = \frac{1}{2}$ into $g (x)$ :
$g (\frac{1}{2}) = e^{\frac{1}{2}} \cdot \frac{\frac{1}{2} s i n ^{- 1} ( \frac{1}{2} )}{1 - ( \frac{1}{2} ) ^{2}}$

Evaluating the Trigonometric and Algebraic Terms

$sin^{- 1} (\frac{1}{2}) = \frac{π}{6}$
$1 - (\frac{1}{2})^{2} = 1 - \frac{1}{4} = \frac{3}{4} = \frac{3}{2}$
Now, plug these values back into the expression.

Final Calculation

$g (\frac{1}{2}) = e \cdot \frac{\frac{1}{2} \cdot \frac{π}{6}}{\frac{3}{2}}$
$g (\frac{1}{2}) = e \cdot \frac{\frac{π}{12}}{\frac{3}{2}} = e \cdot \frac{π}{12} \cdot \frac{2}{3}$
$g (\frac{1}{2}) = \frac{π}{6} \frac{e}{3} = \frac{π}{6} \frac{e}{3}$

00:00 / 00:00

The Sigma Insight: Evaluation of Special Integral Forms

Solution Diagram

Analyzing the Setup

When you first glance at the integral

\int e^{x} (\frac{x sin ^{- 1} x}{1 - x ^{2}} + \frac{sin ^{- 1} x}{( 1 - x ^{2} ) ^{3/2}} + \frac{x}{1 - x ^{2}}) d x

your heart might skip a beat. It is messy, algebraic, and involves inverse trigonometric functions.

However, in the world of JEE Advanced, whenever you see an exponential function

e^{x}

multiplying a sum of terms, you are looking at a pattern recognition test. The identity we are hunting for is the beautiful

\int e^{x} (f (x) + f^{'} (x)) d x = e^{x} f (x) + C

This is the 'Golden Key' of exponential integration. Our mission is to find the function

f (x)

hidden inside that bracket.

Identifying the Function

Let's test the most complex-looking term:

f (x) = \frac{x sin ^{- 1} x}{1 - x ^{2}}

Now, we must differentiate this. Using the quotient rule, where

u = x sin^{- 1} x

and

v = 1 - x^{2}

, we calculate

u^{'}

and

v^{'}

The derivative

u^{'}

involves the product rule, giving us

sin^{- 1} x + \frac{x}{1 - x ^{2}}

. The derivative

v^{'}

is simply

\frac{- x}{1 - x ^{2}}

The Master Equation

When we assemble the derivative

f^{'} (x) = \frac{v u ^{'} - u v ^{'}}{v ^{2}}

, the magic happens. The terms align, the square roots cancel, and we are left with exactly the remaining terms in our integral.

The numerator becomes

sin^{- 1} x (1 - x^{2} + \frac{x ^{2}}{1 - x ^{2}}) + x

which simplifies to

\frac{sin ^{- 1} x}{1 - x ^{2}} + x

Dividing by

v^{2} = 1 - x^{2}

, we get

f^{'} (x) = \frac{sin ^{- 1} x}{( 1 - x ^{2} ) ^{3/2}} + \frac{x}{1 - x ^{2}}

This matches the remaining terms perfectly! We have confirmed our

f (x)

Final Calculation

Now, the final step is simply to evaluate

g (x) = e^{x} f (x)

x = \frac{1}{2}

. Substituting the values, we get

g (\frac{1}{2}) = e^{\frac{1}{2}} \cdot \frac{\frac{1}{2} sin ^{- 1} ( \frac{1}{2} )}{1 - ( \frac{1}{2} ) ^{2}}

With

sin^{- 1} (\frac{1}{2}) = \frac{π}{6}

and

1 - \frac{1}{4} = \frac{3}{2}

, the arithmetic simplifies beautifully.

The final result is

\frac{π}{6} \frac{e}{3}

You see? The monster was just a puzzle waiting to be solved. Keep this pattern in your toolkit, and no integral will ever intimidate you again.

Common Traps & FAQs

Why can't I just use Integration by Parts (IBP) on each term separately?

You could, but it would be a nightmare of circular integrals and endless algebraic loops. The identity

\int e^{x} (f (x) + f^{'} (x)) d x

is a powerful shortcut derived from IBP. Using it saves you 10 minutes of algebra and prevents the sign errors that often plague students during long calculations. Always look for the pattern first!

How do I know which part is

f (x)

and which is

f^{'} (x)

Look for the term that looks like a product of functions. Usually, the more complex term is

f (x)

, and its derivative will naturally produce the simpler terms. If your derivative doesn't match, try a different grouping, but usually, the first term is the best bet. In this problem, the presence of

x sin^{- 1} x

strongly suggests it is the parent function.

What if I get a sign wrong during differentiation? It happens all the time!

Sign errors are the silent killers of JEE ranks. When differentiating

\frac{u}{v}

, remember the formula is

\frac{v u ^{'} - u v ^{'}}{v ^{2}}

. The negative sign in the derivative of the denominator often flips the sign of the second term. Be meticulous! Write down

u

v

u^{'}

, and

v^{'}

separately before combining them. Slow down to speed up.

Animated Solution for Mathematics - Indefinite Integration: If $\int e^{x} (\frac{x s i n ^{- 1} x}{1 - x ^{2}} + \frac{s i n ^{- 1} x}{( 1 - x ^{2} ) ^{3/2}} + \frac{x}{1 - x ^{2}}) d x = g (x) + C,$ where C is the constant of integration, then $g (\frac{1}{2})$ equals:

Select Answer:

Visualized Solution

Analyzing the Integral Structure

The Classic $e^{x}$ Integration Tool

Identifying the Function $f (x)$

Applying the Quotient Rule

Constructing the Numerator of $f^{'} (x)$

Simplifying $f^{'} (x)$

Matching $f^{'} (x)$ with the Integral

Extracting $g (x)$

Substituting $x = \frac{1}{2}$

Evaluating the Trigonometric and Algebraic Terms

Final Calculation

The Sigma Insight: Evaluation of Special Integral Forms

Analyzing the Setup

Identifying the Function

The Master Equation

Final Calculation

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Select Answer:

Visualized Solution

Analyzing the Integral Structure

The Classic .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ex Integration Tool

Identifying the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Applying the Quotient Rule

Constructing the Numerator of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Matching .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x) with the Integral

Extracting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=21​

Evaluating the Trigonometric and Algebraic Terms

Final Calculation

The Sigma Insight: Evaluation of Special Integral Forms

Analyzing the Setup

Identifying the Function

The Master Equation

Final Calculation

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The Classic $e^{x}$ Integration Tool

Identifying the Function $f (x)$

Constructing the Numerator of $f^{'} (x)$

Simplifying $f^{'} (x)$

Matching $f^{'} (x)$ with the Integral

Extracting $g (x)$

Substituting $x = \frac{1}{2}$

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Let $f (t) = \int (\frac{1 - s i n ( l o g _{e} t )}{1 - c o s ( l o g _{e} t )}) d t, t > 1$ . If $f (e^{π /2}) = - e^{π /2}$ and $f (e^{π /4}) = α e^{π /4}$ , then $α$ equals

Let $I = \int \frac{e ^{x}}{e ^{4 x} + e ^{2 x} + 1} d x, J = \int \frac{e ^{- x}}{e ^{- 4 x} + e ^{- 2 x} + 1} d x$ . Then, for an arbitrary constant $C$ , the value of $J - I$ equals

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