If $\int \frac{\cos x \, dx}{\sin^2x(1+\sin x)^{2/3}} = f(x) (1 + \sin^6x)^{1/\lambda} + c$ where $c$ is a constant of integration, then $\lambda f\left(\frac{\pi}{3}\right)$ is equal to : Options: (A) 4 (B) -2 (C) 8 (D) -4

### Analyzing the Setup The given integral is: $$I = \int \frac{\cos x \, dx}{\sin^3 x (1+\sin^6 x)^{2/3}}$$ We observe that $\cos x \, dx$ is the derivative of $\sin x$. This suggests the substitution $t = \sin x$, which transforms the integral into: $$I = \int \frac{dt}{t^3(1+t^6)^{2/3}}$$ ### Algebraic Manipulation To simplify the term $(1+t^6)^{2/3}$, we factor out $t^6$ from within the parenthesis. Since $(t^6)^{2/3} = t^4$, the expression becomes: $$I = \int \frac{dt}{t^3 \cdot t^4 (1+t^{-6})^{2/3}} = \int \frac{dt}{t^7(1+t^{-6})^{2/3}}$$ ### The Master Substitution We now introduce a second substitution to handle the fractional power. Let $1+t^{-6} = r^3$. Differentiating both sides with respect to $t$ yields: $$-6t^{-7} \, dt = 3r^2 \, dr \implies \frac{dt}{t^7} = -\frac{1}{2} r^2 \, dr$$ Substituting these into our integral, the $r^2$ terms cancel out: $$I = \int \frac{-\frac{1}{2} r^2 \, dr}{r^2} = -\frac{1}{2} \int dr = -\frac{1}{2} r + c$$ ### Final Calculation Back-substituting $r = (1+t^{-6})^{1/3}$ and $t = \sin x$, we obtain: $$I = -\frac{1}{2} (1+\sin^{-6} x)^{1/3} + c = -\frac{1}{2} \left( \frac{\sin^6 x + 1}{\sin^6 x} \right)^{1/3} + c$$ Simplifying the expression, we get: $$I = -\frac{1}{2} \csc^2 x (1+\sin^6 x)^{1/3} + c$$ Comparing this to the form $f(x)(1+\sin^6 x)^{1/\lambda} + c$, we identify $f(x) = -\frac{1}{2} \csc^2 x$ and $\lambda = 3$. Evaluating at $x = \pi/3$: $$f(\pi/3) = -\frac{1}{2} \csc^2(\pi/3) = -\frac{1}{2} \left( \frac{2}{\sqrt{3}} \right)^2 = -\frac{1}{2} \cdot \frac{4}{3} = -\frac{2}{3}$$ The final result is: $$\lambda f(\pi/3) = 3 \times \left( -\frac{2}{3} \right) = **-2**$$

JEE Main 2020 (8 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Indefinite Integration: If $\int \frac{c o s x d x}{s i n ^{2} x ( 1 + s i n x ) ^{2/3}} = f (x) (1 + sin^{6} x)^{1/ λ} + c$ where $c$ is a constant of integration, then $λ f (\frac{π}{3})$ is equal to :

Q: Wait, why did we factor out $t^6$ from the bracket? Wouldn't it be easier to just leave it?

That is a fantastic question! If you leave it as $(1+t^6)^{2/3}$, you are essentially stuck because the power prevents easy integration. By factoring out $t^6$, we create a term $t^{-6}$ inside the bracket. Why is this genius? Because the derivative of $t^{-6}$ is $-6t^{-7}$, which perfectly matches the $t^7$ we get in the denominator. It turns a dead-end into a highway.

Q: How did we know to choose $r^3$ for the second substitution? Why not just $r$ or $r^2$?

We chose $r^3$ specifically to annihilate the fractional power of $2/3$. When you raise $(r^3)^{2/3}$, the exponents multiply: $3 \times (2/3) = 2$. It leaves you with a clean $r^2$. Always look at the denominator of the fractional power—it is a roadmap for your substitution. We want to clear the fraction, so we pick a power that is a multiple of the denominator.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Look closely at the numerator. You see

cos x d x

. This is the derivative of

sin x

. This is your first major clue. Do not be intimidated by the complex denominator; start by simplifying the variable space using the substitution

t = sin x

Hint 2

Hint 2: The Arsenal. Once you have the integral in terms of

t

, you will face a term like

(1 + t^{6})^{2/3}

. The trick here is to 'force' a factor of

t^{6}

out of the bracket. This will transform the expression into

t^{4} (1 + t^{- 6})^{2/3}

, which aligns perfectly with the

t^{3}

already in the denominator.

Hint 3

Hint 3: The Execution. Now you have an integral involving

(1 + t^{- 6})^{2/3}

. To eliminate that fractional power, set

1 + t^{- 6} = r^{3}

. Differentiating this will give you a term that perfectly cancels the remaining

t

terms. It is a beautiful, elegant collapse.

Visualized Solution

Identify the Integral $I$

Given Integral: $I = \int \frac{c o s x d x}{s i n ^{3} x ( 1 + s i n ^{6} x ) ^{\frac{2}{3}}}$
Target: Find the value of $λ f (\frac{π}{3})$
Note: We use the corrected form of the integral based on the RHS structure.

First Substitution: $t = sin x$

Let $t = sin x$
Differentiating both sides: $cos x d x = d t$

Transform the Integral

Substitute $t$ and $d t$ into the integral.
The integral becomes: $I = \int \frac{d t}{t ^{3} ( 1 + t ^{6} ) ^{\frac{2}{3}}}$

Algebraic Manipulation: Factoring $t^{6}$

Focus on the bracket: $(1 + t^{6})^{\frac{2}{3}}$
Factor out $t^{6}$ : $(t^{6} (t^{- 6} + 1))^{\frac{2}{3}}$
Apply the fractional power: $t^{6 \cdot \frac{2}{3}} (1 + t^{- 6})^{\frac{2}{3}} = t^{4} (1 + t^{- 6})^{\frac{2}{3}}$

Rewrite the Integral

Combine the $t$ terms in the denominator.
$t^{3} \cdot t^{4} = t^{7}$
Rewrite the integral: $I = \int \frac{d t}{t ^{7} ( 1 + t ^{- 6} ) ^{\frac{2}{3}}}$

Second Substitution: $1 + t^{- 6} = r^{3}$

Let $1 + t^{- 6} = r^{3}$
We choose $r^{3}$ to cancel the $\frac{2}{3}$ fractional power.

Differentiate the Second Substitution

Differentiating: $- 6 t^{- 7} d t = 3 r^{2} d r$
Rearrange to isolate $\frac{d t}{t ^{7}}$ : $\frac{d t}{t ^{7}} = - \frac{3}{6} r^{2} d r = - \frac{1}{2} r^{2} d r$

Substitute into the Integral

Replace $\frac{d t}{t ^{7}}$ with $- \frac{1}{2} r^{2} d r$
Replace $(1 + t^{- 6})^{\frac{2}{3}}$ with $(r^{3})^{\frac{2}{3}} = r^{2}$
$I = \int \frac{- \frac{1}{2} r ^{2} d r}{r ^{2}}$

Integrate with respect to $r$

Cancel $r^{2}$ in numerator and denominator.
$I = - \frac{1}{2} \int d r$
Integrate: $I = - \frac{1}{2} r + c$

Reverse Substitution for $r$

Recall $r^{3} = 1 + t^{- 6} ⟹ r = (1 + t^{- 6})^{\frac{1}{3}}$
Substitute back: $I = - \frac{1}{2} (1 + t^{- 6})^{\frac{1}{3}} + c$
Simplify: $1 + t^{- 6} = \frac{t ^{6} + 1}{t ^{6}}$
$r = \frac{( 1 + t ^{6} ) ^{\frac{1}{3}}}{t ^{2}}$

Reverse Substitution for $t$

Recall $t = sin x$
Substitute back: $I = - \frac{1}{2} \frac{( 1 + s i n ^{6} x ) ^{\frac{1}{3}}}{s i n ^{2} x} + c$
Rewrite using $csc^{2} x$ : $I = - \frac{1}{2} csc^{2} x (1 + sin^{6} x)^{\frac{1}{3}} + c$

Compare and Identify $f (x)$ and $λ$

Compare with given RHS: $f (x) (1 + sin^{6} x)^{\frac{1}{λ}} + c$
Identify $f (x) = - \frac{1}{2} csc^{2} x$
Identify $\frac{1}{λ} = \frac{1}{3} ⟹ λ = 3$

Evaluate $f (\frac{π}{3})$

Calculate $f (\frac{π}{3}) = - \frac{1}{2} csc^{2} (\frac{π}{3})$
We know $sin (\frac{π}{3}) = \frac{3}{2}$ , so $csc (\frac{π}{3}) = \frac{2}{3}$
Substitute: $f (\frac{π}{3}) = - \frac{1}{2} (\frac{2}{3})^{2} = - \frac{1}{2} \cdot \frac{4}{3} = - \frac{2}{3}$

Final Calculation: $λ f (\frac{π}{3})$

Final Target: $λ f (\frac{π}{3})$
Substitute $λ = 3$ and $f (\frac{π}{3}) = - \frac{2}{3}$
Result: $3 \times (- \frac{2}{3}) = - 2$
The correct option is -2.

00:00 / 00:00

The Sigma Insight: Integration by Substitution

Analyzing the Setup

The given integral is:

I = \int \frac{cos x d x}{sin ^{3} x ( 1 + sin ^{6} x ) ^{2/3}}

We observe that

cos x d x

is the derivative of

sin x

. This suggests the substitution

t = sin x

, which transforms the integral into:

I = \int \frac{d t}{t ^{3} ( 1 + t ^{6} ) ^{2/3}}

Algebraic Manipulation

To simplify the term

(1 + t^{6})^{2/3}

, we factor out

t^{6}

from within the parenthesis. Since

(t^{6})^{2/3} = t^{4}

, the expression becomes:

I = \int \frac{d t}{t ^{3} \cdot t ^{4} ( 1 + t ^{- 6} ) ^{2/3}} = \int \frac{d t}{t ^{7} ( 1 + t ^{- 6} ) ^{2/3}}

The Master Substitution

We now introduce a second substitution to handle the fractional power. Let

1 + t^{- 6} = r^{3}

Differentiating both sides with respect to

t

yields:

- 6 t^{- 7} d t = 3 r^{2} d r ⟹ \frac{d t}{t ^{7}} = - \frac{1}{2} r^{2} d r

Substituting these into our integral, the

r^{2}

terms cancel out:

I = \int \frac{- \frac{1}{2} r ^{2} d r}{r ^{2}} = - \frac{1}{2} \int d r = - \frac{1}{2} r + c

Final Calculation

Back-substituting

r = (1 + t^{- 6})^{1/3}

and

t = sin x

, we obtain:

I = - \frac{1}{2} (1 + sin^{- 6} x)^{1/3} + c = - \frac{1}{2} (\frac{sin ^{6} x + 1}{sin ^{6} x})^{1/3} + c

Simplifying the expression, we get:

I = - \frac{1}{2} csc^{2} x (1 + sin^{6} x)^{1/3} + c

Comparing this to the form

f (x) (1 + sin^{6} x)^{1/ λ} + c

, we identify

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

and

λ = 3

Evaluating at

x = π /3

f (π /3) = - \frac{1}{2} csc^{2} (π /3) = - \frac{1}{2} (\frac{2}{3})^{2} = - \frac{1}{2} \cdot \frac{4}{3} = - \frac{2}{3}

The final result is:

λ f (π /3) = 3 \times (- \frac{2}{3}) = * * - 2 * *

Common Traps & FAQs

Wait, why did we factor out

t^{6}

from the bracket? Wouldn't it be easier to just leave it?

That is a fantastic question! If you leave it as

(1 + t^{6})^{2/3}

, you are essentially stuck because the power prevents easy integration. By factoring out

t^{6}

, we create a term

t^{- 6}

inside the bracket. Why is this genius? Because the derivative of

t^{- 6}

- 6 t^{- 7}

, which perfectly matches the

t^{7}

we get in the denominator. It turns a dead-end into a highway.

How did we know to choose

r^{3}

for the second substitution? Why not just

r

r^{2}

We chose

r^{3}

specifically to annihilate the fractional power of

2/3

. When you raise

(r^{3})^{2/3}

, the exponents multiply:

3 \times (2/3) = 2

. It leaves you with a clean

r^{2}

. Always look at the denominator of the fractional power—it is a roadmap for your substitution. We want to clear the fraction, so we pick a power that is a multiple of the denominator.

The question has a typo in the integral power. How are we supposed to handle that in an exam?

In JEE Advanced, sometimes the structure of the RHS is your best guide. If the RHS is

f (x) (1 + sin^{6} x)^{1/ λ}

, it strongly implies the integral must result in a form involving

(1 + sin^{6} x)

. Use the RHS as a 'North Star' to correct the LHS. It shows you are thinking like a mathematician, not just a calculator. Trust the structure!

Animated Solution for Mathematics - Indefinite Integration: If $\int \frac{c o s x d x}{s i n ^{2} x ( 1 + s i n x ) ^{2/3}} = f (x) (1 + sin^{6} x)^{1/ λ} + c$ where $c$ is a constant of integration, then $λ f (\frac{π}{3})$ is equal to :

Select Answer:

Visualized Solution

Identify the Integral $I$

First Substitution: $t = sin x$

Transform the Integral

Algebraic Manipulation: Factoring $t^{6}$

Rewrite the Integral

Second Substitution: $1 + t^{- 6} = r^{3}$

Differentiate the Second Substitution

Substitute into the Integral

Integrate with respect to $r$

Reverse Substitution for $r$

Reverse Substitution for $t$

Compare and Identify $f (x)$ and $λ$

Evaluate $f (\frac{π}{3})$

Final Calculation: $λ f (\frac{π}{3})$

The Sigma Insight: Integration by Substitution

Analyzing the Setup

Algebraic Manipulation

The Master Substitution

Final Calculation

Similar Questions

If $\int \frac{c o s x}{s i n ^{3} x ( 1 + s i n ^{6} x ) ^{2/3}} d x = f (x) (1 + sin^{6} x)^{1/3} + c$ , where $c$ is a constant of integration, then $λ f (\frac{π}{3})$ is equal to :

If $\int \frac{d x}{x ^{3} ( 1 + x ^{6} ) ^{2/3}} = x f (x) (1 + x^{6})^{1/3} + C$ where C is a constant of integration, then the function f(x) is equal to-

The value of the integral $\int \frac{s i n θ - s i n 2 θ ( s i n ^{3} θ + s i n ^{2} θ + s i n θ )}{2 s i n ^{4} θ + 3 s i n ^{2} θ + 6} \frac{d θ}{1 - c o s 2 θ}$ is (where $c$ is a constant of integration)

If $\int \frac{s i n ^{3/2} x + c o s ^{3/2} x}{s i n ^{3} x c o s ^{3} x s i n ( x - θ )} d x = A cos θ tan x - sin θ + B cos θ - sin θ cot x + C$ where C is the integration constant, then AB is equal to

The integral $\int sec^{2/3} x csc^{4/3} x d x$ is equal to (Hence C is a constant of integration)

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Select Answer:

Visualized Solution

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

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

Transform the Integral

Algebraic Manipulation: Factoring .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t6

Rewrite the Integral

Second Substitution: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1+t−6=r3

Differentiate the Second Substitution

Substitute into the Integral

Integrate with respect to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

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

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

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

Evaluate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(3π​)

Final Calculation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λf(3π​)

The Sigma Insight: Integration by Substitution

Analyzing the Setup

Algebraic Manipulation

The Master Substitution

Final Calculation

Similar Questions

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Identify the Integral $I$

First Substitution: $t = sin x$

Algebraic Manipulation: Factoring $t^{6}$

Second Substitution: $1 + t^{- 6} = r^{3}$

Integrate with respect to $r$

Reverse Substitution for $r$

Reverse Substitution for $t$

Compare and Identify $f (x)$ and $λ$

Evaluate $f (\frac{π}{3})$

Final Calculation: $λ f (\frac{π}{3})$

If $\int \frac{c o s x}{s i n ^{3} x ( 1 + s i n ^{6} x ) ^{2/3}} d x = f (x) (1 + sin^{6} x)^{1/3} + c$ , where $c$ is a constant of integration, then $λ f (\frac{π}{3})$ is equal to :

If $\int \frac{d x}{x ^{3} ( 1 + x ^{6} ) ^{2/3}} = x f (x) (1 + x^{6})^{1/3} + C$ where C is a constant of integration, then the function f(x) is equal to-

The value of the integral $\int \frac{s i n θ - s i n 2 θ ( s i n ^{3} θ + s i n ^{2} θ + s i n θ )}{2 s i n ^{4} θ + 3 s i n ^{2} θ + 6} \frac{d θ}{1 - c o s 2 θ}$ is (where $c$ is a constant of integration)

If $\int \frac{s i n ^{3/2} x + c o s ^{3/2} x}{s i n ^{3} x c o s ^{3} x s i n ( x - θ )} d x = A cos θ tan x - sin θ + B cos θ - sin θ cot x + C$ where C is the integration constant, then AB is equal to

The integral $\int sec^{2/3} x csc^{4/3} x d x$ is equal to (Hence C is a constant of integration)

If $\int \frac{1}{5 ( x - 1 ) ^{4} ( x + 3 ) ^{6}} d x = A (\frac{α x - 1}{β x + 3})^{B} + C$ , where $C$ is the constant of integration, then the value of $α + β + 20 A B$ is __________

If $\int sec 2 x - 1 d x = α lo g_{e} cos 2 x + β + cos 2 x (1 + cos \frac{1}{β} x) + constant$ , then $β - α$ is equal to ______.

The value of the integral $\int \frac{c o s ^{3} x + c o s ^{5} x}{s i n ^{2} x + s i n ^{4} x} d x$ is

If $\int \frac{d θ}{c o s ^{2} θ ( t a n 2 θ + s e c 2 θ )} = λ tan θ + 2 lo g_{e} ∣ f (θ) ∣ + C$ where $C$ is a constant of integration, then the ordered pair $(λ, f (θ))$ is equal to:

If $f (\frac{x - 4}{x + 2}) = 2 x + 1, (x \in R - {1, - 2})$ , then $\int f (x) d x$ is equal to : (where C is a constant of integration)