Evaluate the following ∫ √((1 - √x)/(1...

MathematicsIntegration by SubstitutionJEE Advanced 1985Moderate

Evaluate the following $\int \frac{1 - x}{1 + x} d x$

Answer:

- 2 1 - x + cos^{- 1} x + x 1 - x + C

Visualized Solution (English)

Analyze the Integrand

Given integral: $I = \int \frac{1 - x}{1 + x} d x$
Observe the structure: $\frac{1 - u}{1 + u}$ where $u = x$ .
The domain of the integrand is $x \in [0, 1]$ .

Choosing the Substitution

To simplify the radical, we use the substitution: $x = cos^{2} θ$
This implies $x = cos θ$ .
Note: $θ \in [0, π /2]$ ensures $cos θ$ is positive.

Differentiating the Substitution

Differentiate $x = cos^{2} θ$ with respect to $θ$ :
$\frac{d x}{d θ} = 2 cos θ (- sin θ)$
$d x = - 2 sin θ cos θ d θ$

Substituting into the Integral

Substitute $x$ and $d x$ into the integral:
$I = \int \frac{1 - c o s θ}{1 + c o s θ} (- 2 sin θ cos θ) d θ$

Simplifying the Radical

Use half-angle identities:
$1 - cos θ = 2 sin^{2} (\frac{θ}{2})$ and $1 + cos θ = 2 cos^{2} (\frac{θ}{2})$
The radical becomes: $\frac{2 s i n ^{2} ( θ /2 )}{2 c o s ^{2} ( θ /2 )} = tan (\frac{θ}{2})$
Integral: $I = \int tan (\frac{θ}{2}) (- 2 sin θ cos θ) d θ$

Expanding the Sine Term

Substitute $tan (\frac{θ}{2}) = \frac{s i n ( θ /2 )}{c o s ( θ /2 )}$ and $sin θ = 2 sin (\frac{θ}{2}) cos (\frac{θ}{2})$ :
$I = \int \frac{s i n ( θ /2 )}{c o s ( θ /2 )} (- 2 \cdot 2 sin (\frac{θ}{2}) cos (\frac{θ}{2}) cos θ) d θ$
Cancel $cos (\frac{θ}{2})$ : $I = - 4 \int sin^{2} (\frac{θ}{2}) cos θ d θ$

Linearizing the Squared Sine

Use the identity $2 sin^{2} (\frac{θ}{2}) = 1 - cos θ$ :
$I = - 2 \int (1 - cos θ) cos θ d θ$

Distributing the Cosine

Distribute $cos θ$ :
$I = - 2 \int (cos θ - cos^{2} θ) d θ$
Use the identity $cos^{2} θ = \frac{1 + c o s 2 θ}{2}$ :
$I = - 2 \int (cos θ - \frac{1 + c o s 2 θ}{2}) d θ$

Performing the Integration

Integrate term by term:
$I = - 2 [sin θ - \frac{1}{2} (θ + \frac{s i n 2 θ}{2})] + C$
Simplify: $I = - 2 sin θ + θ + \frac{s i n 2 θ}{2} + C$
Using $sin 2 θ = 2 sin θ cos θ$ :
$I = - 2 sin θ + θ + sin θ cos θ + C$

Back Substitution

Substitute back using $x = cos^{2} θ$ :
$cos θ = x \Rightarrow θ = cos^{- 1} x$
$sin θ = 1 - cos^{2} θ = 1 - x$
Final Answer: $I = - 2 1 - x + cos^{- 1} x + x 1 - x + C$

Key Takeaways

Key Takeaway: Trigonometric substitution is powerful for simplifying radical expressions.
Pattern Recognition: $\frac{1 - u}{1 + u}$ suggests $u = cos θ$ .
Challenge: Try evaluating $\int \frac{1 - x}{1 + x} d x$ using $x = cos θ$ and compare the steps.

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Conceptually Similar Problems

MathematicsIntegration by SubstitutionJEE Advanced 1989Moderate

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Evaluate $\int (tan x + cot x) d x$

Answer:

2 tan^{- 1} (\frac{t a n x - c o t x}{2}) + c

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Evaluate the following $\int_{0}^{1} \frac{x s i n ^{- 1} x}{1 - x ^{2}} d x$ .

Answer:1

MathematicsFundamental Theorem & Properties of Definite IntegralsJEE Advanced 1997Moderate

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Determine the value of $\int_{- π}^{π} \frac{2 x ( 1 + s i n x )}{1 + c o s ^{2} x} d x$ .

Answer:

π^{2}

MathematicsFundamental Theorem & Properties of Definite IntegralsJEE Advanced 1986Moderate

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Evaluate: $\int_{0}^{π} \frac{x d x}{1 + c o s α s i n x}, 0 < α < π$

Answer:

\frac{π α}{s i n α}

MathematicsFundamental Theorem & Properties of Definite IntegralsJEE Advanced 1988Moderate

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Evaluate $\int_{0}^{1} lo g [1 - x + 1 + x] d x$

Answer:

\frac{1}{2} [lo g 2 + \frac{π}{2} - 1]

MathematicsIntegration by SubstitutionJEE Advanced 1987Moderate

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Evaluate $\int \frac{( c o s 2 x ) ^{1/2}}{s i n x} d x$

Answer:

\frac{1}{2} lo g \frac{2 + 1 - t a n ^{2} x}{2 - 1 - t a n ^{2} x} - lo g (cot x + cot^{2} x - 1) + C

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Evaluate $\int \frac{s i n x}{s i n x - c o s x} d x$

Answer:

\frac{1}{2} lo g ∣ sin x - cos x ∣ + \frac{x}{2} + C

MathematicsFundamental Theorem & Properties of Definite IntegralsJEE Advanced 1985Moderate

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Evaluate the following: $\int_{0}^{π /2} \frac{x s i n x c o s x}{c o s ^{4} x + s i n ^{4} x} d x$

Answer:

\frac{π ^{2}}{16}

MathematicsIntegration by SubstitutionJEE Advanced 2001Moderate

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Evaluate $\int sin^{- 1} (\frac{2 x + 2}{4 x ^{2} + 8 x + 13}) d x$

Answer:

(x + 1) tan^{- 1} (\frac{2 x + 2}{3}) - \frac{3}{4} lo g (4 x^{2} + 8 x + 13) + C

MathematicsFundamental Theorem & Properties of Definite IntegralsJEE Advanced 2004Easy

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The value of the integral $\int_{0}^{1} \frac{1 - x}{1 + x} d x$ is

(A)

\frac{π}{2} + 1

(B)

\frac{π}{2} - 1

(C)

-1

(D)

Answer:B

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluate the following ∫1+x​1−x​​​dx

Visualized Solution (English)

Analyze the Integrand

Choosing the Substitution

Differentiating the Substitution

Substituting into the Integral

Simplifying the Radical

Expanding the Sine Term

Linearizing the Squared Sine

Distributing the Cosine

Performing the Integration

Back Substitution

Key Takeaways

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