How do I know if I should use $\sin(x + \pi/4)$ or $\cos(x - \pi/4)$?

Both are valid! The identity $\sin x + \cos x = \sqrt{2} \sin(x + \pi/4)$ is just a specific form of the harmonic addition theorem. You could also write it as $\sqrt{2} \cos(x - \pi/4)$. As long as you are consistent and careful with your range, the math will lead you to the same result. Choose the one that feels most intuitive to you.

JEE Main 2022 (26 June Shift 1)

LEVELJEE Advanced

Animated Solution for Mathematics - Differential Equations: Let $S = (0, 2 π) - {\frac{π}{2}, \frac{3 π}{4}, \frac{3 π}{2}, \frac{7 π}{4}}$ . Let $y = y (x), x \in S$ , be the solution curve of the differential equation $\frac{d y}{d x} = \frac{1}{1 + s i n 2 x}, y (\frac{π}{4}) = \frac{1}{2}$ . If the sum of abscissas of all the points of intersection of the curve $y = y (x)$ with the curve $y = 2 sin x$ is $\frac{k π}{12}$ , then $k$ is equal to _______.

Q: Why can't I just divide by $\sin x$ when equating the two curves?

This is the most common trap in JEE! If you divide by $\sin x$, you are implicitly assuming $\sin x \neq 0$. If $\sin x = 0$, your division is undefined. By factoring, you explicitly account for the case where $\sin x = 0$, which yields the solution $x = \pi$. Always factor, never divide by a variable expression!

Enter Numerical Value:

Progressive Hints

Hint 1

Look closely at the denominator

1 + sin 2 x

. In the world of JEE Advanced, whenever you see

1

paired with a double-angle sine, your first instinct should be to rewrite

1

sin^{2} x + cos^{2} x

. This transforms the expression into a perfect square,

(sin x + cos x)^{2}

, which is the key to unlocking the entire problem.

Hint 2

Once you have the perfect square, the integration looks daunting. But remember your toolkit: dividing the numerator and denominator by

cos^{2} x

is a classic maneuver. This will introduce

sec^{2} x

in the numerator, which is the derivative of

tan x

. This is your golden ticket to a simple

u

-substitution.

Hint 3

When you reach the intersection step, avoid the common trap of dividing both sides by

sin x

. If you do that, you will lose a valid solution! Instead, bring everything to one side and factor out

sin x

. This ensures you capture all possible intersection points, including the one where

sin x = 0

Visualized Solution

$U n d er s t an d in g t h eD o main S$

Domain $S = (0, 2 π) ∖ {\frac{π}{2}, \frac{3 π}{4}, \frac{3 π}{2}, \frac{7 π}{4}}$
Differential Equation: $\frac{d y}{d x} = \frac{1}{1 + s i n 2 x}$
Initial Condition: $y (\frac{π}{4}) = \frac{1}{2}$

$S im pl i f y in g t h eD e n o mina t or$

Using identity: $1 + sin 2 x = sin^{2} x + cos^{2} x + 2 sin x cos x$
Therefore, $1 + sin 2 x = (sin x + cos x)^{2}$
The DE becomes: $\frac{d y}{d x} = \frac{1}{( s i n x + c o s x ) ^{2}}$

$P r e p a r in g f or I n t e g r a t i o n$

Divide numerator and denominator by $cos^{2} x$ :
$\frac{d y}{d x} = \frac{s e c ^{2} x}{( \frac{s i n x + c o s x}{c o s x} ) ^{2}}$
$\frac{d y}{d x} = \frac{s e c ^{2} x}{( 1 + t a n x ) ^{2}}$

$I n t e g r a t in g t h e E q u a t i o n$

Integrate: $y = \int \frac{s e c ^{2} x}{( 1 + t a n x ) ^{2}} d x$
Let $u = 1 + tan x \Rightarrow d u = sec^{2} x d x$
$y = \int \frac{1}{u ^{2}} d u = - \frac{1}{u} + C$
$y (x) = - \frac{1}{1 + t a n x} + C$

$F in d in g t h e C o n s t an tC$

Substitute $x = \frac{π}{4}, y = \frac{1}{2}$ :
$\frac{1}{2} = - \frac{1}{1 + t a n ( \frac{π}{4} )} + C$
$\frac{1}{2} = - \frac{1}{1 + 1} + C \Rightarrow \frac{1}{2} = - \frac{1}{2} + C$
$C = 1$

$F ina l C u r v e E q u a t i o n$

$y (x) = 1 - \frac{1}{1 + t a n x} = \frac{1 + t a n x - 1}{1 + t a n x}$
$y (x) = \frac{t a n x}{1 + t a n x} = \frac{s i n x}{s i n x + c o s x}$

$S e tt in g u pt h e I n t er sec t i o n$

Equate $y (x)$ with $2 sin x$ :
$\frac{s i n x}{s i n x + c o s x} = 2 sin x$
$sin x [\frac{1}{s i n x + c o s x} - 2] = 0$

$C a se 1 : sin x = 0$

Case 1: $sin x = 0$
In $(0, 2 π)$ , $x = π$
Check domain: $π \in S$ (Valid solution)

$C a se 2 : S o l v in g t h e T r i g I d e n t i t y$

Case 2: $\frac{1}{s i n x + c o s x} = 2$
$sin x + cos x = \frac{1}{2}$
Divide by $2$ : $\frac{1}{2} sin x + \frac{1}{2} cos x = \frac{1}{2}$

$F in d in g t h e xv a l u es$

$sin (x + \frac{π}{4}) = \frac{1}{2}$
For $x \in (0, 2 π)$ , $x + \frac{π}{4} \in (\frac{π}{4}, \frac{9 π}{4})$
$x + \frac{π}{4} = \frac{5 π}{6} \Rightarrow x = \frac{7 π}{12}$
$x + \frac{π}{4} = \frac{13 π}{6} \Rightarrow x = \frac{23 π}{12}$

$S u mmin g t h e A b sc i ss a s$

Sum of abscissas $= π + \frac{7 π}{12} + \frac{23 π}{12}$
Sum $= \frac{12 π + 7 π + 23 π}{12}$
Sum $= \frac{42 π}{12}$

$F ina l A n s w er f or k$

Given sum $= \frac{k π}{12}$
Comparing $\frac{42 π}{12} = \frac{k π}{12}$
Final Answer: $k = 42$

00:00 / 00:00

The Sigma Insight: Variable Separable Method

Solution Diagram

Analyzing the Setup

The given differential equation is:

\frac{d y}{d x} = \frac{1}{1 + sin 2 x}

To simplify the denominator, we utilize the trigonometric identities

1 = sin^{2} x + cos^{2} x

and

sin 2 x = 2 sin x cos x

. This transforms the denominator into a perfect square:

\frac{d y}{d x} = \frac{1}{( sin x + cos x ) ^{2}}

The Master Equation

To integrate this expression, we divide both the numerator and the denominator by

cos^{2} x

. This yields:

\frac{d y}{d x} = \frac{sec ^{2} x}{( 1 + tan x ) ^{2}}

We now apply the substitution

u = 1 + tan x

, which implies

d u = sec^{2} x d x

. The integral becomes:

\int \frac{1}{u ^{2}} d u = - \frac{1}{u} + C = - \frac{1}{1 + tan x} + C

Applying Initial Conditions

We are given the initial condition

y (\frac{π}{4}) = \frac{1}{2}

. Substituting

x = \frac{π}{4}

and

tan (\frac{π}{4}) = 1

into our general solution:

\frac{1}{2} = - \frac{1}{1 + 1} + C \Rightarrow \frac{1}{2} = - \frac{1}{2} + C \Rightarrow C = 1

Thus, the specific solution is:

y = 1 - \frac{1}{1 + tan x} = \frac{tan x}{1 + tan x} = \frac{sin x}{sin x + cos x}

Finding the Intersection

We seek the intersection of

y = \frac{s i n x}{s i n x + c o s x}

and

y = 2 sin x

. Setting them equal:

\frac{sin x}{sin x + cos x} = 2 sin x

To avoid losing roots, we rearrange the equation:

sin x (\frac{1}{sin x + cos x} - 2) = 0

This yields two distinct cases for

x \in (0, 2 π)

Case 1:

sin x = 0 \Rightarrow x = π

Case 2:

sin x + cos x = \frac{1}{2}

. Multiplying by

\frac{1}{2}

, we get:

sin (x + \frac{π}{4}) = \frac{1}{2}

Solving for

x

in the given interval:

x + \frac{π}{4} = \frac{5 π}{6} \Rightarrow x = \frac{7 π}{12}

x + \frac{π}{4} = \frac{13 π}{6} \Rightarrow x = \frac{23 π}{12}

Final Calculation

The sum of the abscissas is:

π + \frac{7 π}{12} + \frac{23 π}{12} = \frac{12 π + 7 π + 23 π}{12} = \frac{42 π}{12}

Comparing this to the form

\frac{k π}{12}

, we conclude that $k = 42$ .

Common Traps & FAQs

Why can't I just divide by

sin x

when equating the two curves?

This is the most common trap in JEE! If you divide by

sin x

, you are implicitly assuming

sin x \neq = 0

. If

sin x = 0

, your division is undefined. By factoring, you explicitly account for the case where

sin x = 0

, which yields the solution

x = π

. Always factor, never divide by a variable expression!

How do I know if I should use

sin (x + π /4)

cos (x - π /4)

Both are valid! The identity

sin x + cos x = 2 sin (x + π /4)

is just a specific form of the harmonic addition theorem. You could also write it as

2 cos (x - π /4)

. As long as you are consistent and careful with your range, the math will lead you to the same result. Choose the one that feels most intuitive to you.

The domain

S

excludes several points. Do I need to check my answers against these?

Absolutely. In JEE Advanced, the domain is not just a formality; it is a boundary condition. If your calculated

x

value falls on an excluded point, it is an extraneous solution. Always perform a quick sanity check to ensure your roots lie within the valid set

S

Enter Numerical Value:

Visualized Solution

$U n d er s t an d in g t h eD o main S$

$S im pl i f y in g t h eD e n o mina t or$

$P r e p a r in g f or I n t e g r a t i o n$

$I n t e g r a t in g t h e E q u a t i o n$

$F in d in g t h e C o n s t an tC$

$F ina l C u r v e E q u a t i o n$

$S e tt in g u pt h e I n t er sec t i o n$

$C a se 1 : sin x = 0$

$C a se 2 : S o l v in g t h e T r i g I d e n t i t y$

$F in d in g t h e xv a l u es$

$S u mmin g t h e A b sc i ss a s$

$F ina l A n s w er f or k$

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Master Equation

Applying Initial Conditions

Finding the Intersection

Final Calculation

Similar Questions

Let a curve $y = f (x)$ pass through the points $(0, 5)$ and $(lo g_{e} 2, k)$ . If the curve satisfies the differential equation $2 (3 + y) e^{2 x} d x - (7 + e^{2 x}) d y = 0$ , then k is equal to

If $y (x)$ is the solution of the differential equation $x d y - (y^{2} - 4 y) d x = 0$ for $x > 0$ , $y (1) = 2$ , and the slope of the curve $y = y (x)$ is never zero, then the value of $10 y (2)$ is ______.

Let a curve $y = y (x)$ be given by the solution of the differential equation $cos (\frac{1}{2} cos^{- 1} (e^{- x})) d x = e^{2 x} - 1 d y$ . If it intersects $y$ -axis at $y = - 1$ , and the intersection point of the curve with $x$ -axis is $(α, 0)$ , then $e^{α}$ is equal to

If the solution curve $y = y (x)$ of the differential equation $(1 + y^{2}) (1 + lo g_{e} x) d x + x d y = 0, x > 0$ passes through the point $(1, 1)$ and $y (e) = \frac{α - t a n ( 3/2 )}{β + t a n ( 3/2 )}$ , then $α + 2 β$ is

If the solution $y (x)$ of the given differential equation $(e^{y} + 1) cos x d x + e^{y} sin x d y = 0$ passes through the point $(\frac{π}{2}, 0)$ , then the value of $e^{y (\frac{π}{6})}$ is equal to_________

Let the curve $y = y (x)$ be the solution of the differential equation, $\frac{d y}{d x} = 2 (x + 1)$ . If the numerical value of area bounded by the curve $y = y (x)$ and $x$ -axis is $\frac{4 8}{3}$ , then the value of $y (1)$ is equal to ____

If $y = y (x)$ satisfies the differential equation $16 x + 9 x (4 + 9 + x) cos y d y = (1 + 2 sin y) d x, x > 0$ and $y (256) = \frac{π}{2}, y (49) = α$ , then $2 sin α$ is equal to :

If $y = y (x)$ is the solution curve of the differential equation $(x^{2} - 4) d y - (y^{2} - 3 y) d x = 0, x > 2, y (4) = \frac{3}{2}$ and the slope of the curve is never zero, then the value of $y (10)$ equals :

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } UnderstandingtheDomainS

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SimplifyingtheDenominator

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PreparingforIntegration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IntegratingtheEquation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FindingtheConstantC

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FinalCurveEquation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SettinguptheIntersection

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case1:sinx=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case2:SolvingtheTrigIdentity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Findingthexvalues

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SummingtheAbscissas

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FinalAnswerfork

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Master Equation

Applying Initial Conditions

Finding the Intersection

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a curve y=f(x) pass through the points (0,5) and (loge​2,k). If the curve satisfies the differential equation 2(3+y)e2xdx−(7+e2x)dy=0, then k is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the solution curve y=y(x) of the differential equation (1+y2)(1+loge​x)dx+xdy=0,x>0 passes through the point (1,1) and y(e)=β+tan(3/2)α−tan(3/2)​, then α+2β is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the solution y(x) of the given differential equation (ey+1)cosxdx+eysinxdy=0 passes through the point (2π​,0), then the value of ey(6π​) is equal to_________

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=y(x) satisfies the differential equation 16x+9x​​(4+9+x​​)cosydy=(1+2siny)dx,x>0 and y(256)=2π​,y(49)=α, then 2sinα is equal to :

$U n d er s t an d in g t h eD o main S$

$S im pl i f y in g t h eD e n o mina t or$

$P r e p a r in g f or I n t e g r a t i o n$

$I n t e g r a t in g t h e E q u a t i o n$

$F in d in g t h e C o n s t an tC$

$F ina l C u r v e E q u a t i o n$

$S e tt in g u pt h e I n t er sec t i o n$

$C a se 1 : sin x = 0$

$C a se 2 : S o l v in g t h e T r i g I d e n t i t y$

$F in d in g t h e xv a l u es$

$S u mmin g t h e A b sc i ss a s$

$F ina l A n s w er f or k$

Let a curve $y = f (x)$ pass through the points $(0, 5)$ and $(lo g_{e} 2, k)$ . If the curve satisfies the differential equation $2 (3 + y) e^{2 x} d x - (7 + e^{2 x}) d y = 0$ , then k is equal to

If the solution curve $y = y (x)$ of the differential equation $(1 + y^{2}) (1 + lo g_{e} x) d x + x d y = 0, x > 0$ passes through the point $(1, 1)$ and $y (e) = \frac{α - t a n ( 3/2 )}{β + t a n ( 3/2 )}$ , then $α + 2 β$ is

If the solution $y (x)$ of the given differential equation $(e^{y} + 1) cos x d x + e^{y} sin x d y = 0$ passes through the point $(\frac{π}{2}, 0)$ , then the value of $e^{y (\frac{π}{6})}$ is equal to_________

If $y = y (x)$ satisfies the differential equation $16 x + 9 x (4 + 9 + x) cos y d y = (1 + 2 sin y) d x, x > 0$ and $y (256) = \frac{π}{2}, y (49) = α$ , then $2 sin α$ is equal to :