Evaluate ∫ (cos 2x)^(1/2) / sin x dx

MathematicsIntegration by SubstitutionJEE Advanced 1987Moderate

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

Visualized Solution (English)

Initial Integral and Double Angle Formula

Given integral: $I = \int \frac{( c o s 2 x ) ^{\frac{1}{2}}}{s i n x} d x$
Using the double angle identity: $cos 2 x = cos^{2} x - sin^{2} x$
Substituting this into the integral: $I = \int \frac{c o s ^{2} x - s i n ^{2} x}{s i n x} d x$

Simplifying the Root and Transforming to Cotangent

Bring $sin x$ inside the square root: $I = \int \frac{c o s ^{2} x - s i n ^{2} x}{s i n ^{2} x} d x$
Simplify the fraction inside the root: $I = \int \frac{c o s ^{2} x}{s i n ^{2} x} - \frac{s i n ^{2} x}{s i n ^{2} x} d x$
Resulting form: $I = \int cot^{2} x - 1 d x$

Choosing a Strategic Substitution

To eliminate the square root, use the identity: $sec^{2} θ - 1 = tan^{2} θ$
Let: $cot x = sec θ$
Then: $cot^{2} x - 1 = sec^{2} θ - 1 = tan θ$

Finding the Differential .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d x

Differentiate $cot x = sec θ$ : $- csc^{2} x d x = sec θ tan θ d θ$
Isolate $d x$ : $d x = - \frac{s e c θ t a n θ}{c s c ^{2} x} d θ$
Substitute $csc^{2} x = 1 + cot^{2} x = 1 + sec^{2} θ$ :
Final differential: $d x = - \frac{s e c θ t a n θ}{1 + s e c ^{2} θ} d θ$

Substituting into the Integral

Substitute back: $I = \int (tan θ) (- \frac{s e c θ t a n θ}{1 + s e c ^{2} θ}) d θ$
Combine terms: $I = - \int \frac{s e c θ t a n ^{2} θ}{1 + s e c ^{2} θ} d θ$

Simplifying the Integrand

Convert to sine and cosine: $I = - \int \frac{\frac{1}{c o s θ} \cdot \frac{s i n ^{2} θ}{c o s ^{2} θ}}{1 + \frac{1}{c o s ^{2} θ}} d θ$
Simplify the denominator: $1 + \frac{1}{c o s ^{2} θ} = \frac{c o s ^{2} θ + 1}{c o s ^{2} θ}$
Resulting expression: $I = - \int \frac{s i n ^{2} θ}{c o s θ ( 1 + c o s ^{2} θ )} d θ$

Preparing for Substitution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u = sin θ

Use $cos^{2} θ = 1 - sin^{2} θ$ : $I = - \int \frac{s i n ^{2} θ}{c o s θ ( 1 + 1 - s i n ^{2} θ )} d θ$
Simplify: $I = - \int \frac{s i n ^{2} θ}{c o s θ ( 2 - s i n ^{2} θ )} d θ$
Multiply numerator and denominator by $cos θ$ : $I = - \int \frac{s i n ^{2} θ c o s θ}{c o s ^{2} θ ( 2 - s i n ^{2} θ )} d θ$
Substitute $cos^{2} θ = 1 - sin^{2} θ$ again: $I = - \int \frac{s i n ^{2} θ c o s θ}{( 1 - s i n ^{2} θ ) ( 2 - s i n ^{2} θ )} d θ$

Substitution and Partial Fractions

Let $u = sin θ$ , then $d u = cos θ d θ$
Integral becomes: $I = - \int \frac{u ^{2}}{( 1 - u ^{2} ) ( 2 - u ^{2} )} d u$
Using partial fractions: $\frac{u ^{2}}{( 1 - u ^{2} ) ( 2 - u ^{2} )} = \frac{1}{1 - u ^{2}} - \frac{2}{2 - u ^{2}}$
So: $I = \int (\frac{2}{2 - u ^{2}} - \frac{1}{1 - u ^{2}}) d u$

Integrating the Terms

Integrate $\int \frac{2}{2 - u ^{2}} d u = \frac{1}{2} lo g \frac{2 + u}{2 - u}$
Integrate $\int \frac{1}{1 - u ^{2}} d u = \frac{1}{2} lo g \frac{1 + u}{1 - u} = lo g ∣ sec θ + tan θ ∣$
Combined result: $I = \frac{1}{2} lo g \frac{2 + s i n θ}{2 - s i n θ} - lo g ∣ sec θ + tan θ ∣ + C$

Final Result in Terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Substitute $sin θ = 1 - cos^{2} θ = 1 - tan^{2} x$
Substitute $sec θ = cot x$ and $tan θ = cot^{2} x - 1$
Final Answer: $I = \frac{1}{2} lo g \frac{2 + 1 - t a n ^{2} x}{2 - 1 - t a n ^{2} x} - lo g (cot x + cot^{2} x - 1) + C$

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Conceptually Similar Problems

MathematicsIntegration by PartsJEE Advanced 1994Moderate

View in:English Hindi

Find the indefinite integral $\int cos 2 θ ln (\frac{c o s θ + s i n θ}{c o s θ - s i n θ}) d θ$

Answer:

\frac{s i n 2 θ}{2} ln (\frac{c o s θ + s i n θ}{c o s θ - s i n θ}) - \frac{1}{2} ln sec 2 θ + C

MathematicsEvaluation of Special Integral FormsJEE Advanced 1978Easy

View in:English Hindi

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 1991Moderate

View in:English Hindi

Evaluate $\int_{0}^{π} \frac{x s i n 2 x s i n ( \frac{π}{2} c o s x )}{2 x - π} d x$

Answer:

\frac{8}{π ^{2}}

MathematicsIntegration by SubstitutionJEE Advanced 1985Moderate

View in:English Hindi

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

Answer:

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

MathematicsIntegration by PartsJEE Advanced 1981Easy

View in:English Hindi

Evaluate $\int (e^{l o g x} + sin x) cos x d x$

Answer:

x sin x + cos x - \frac{1}{4} cos 2 x + C

MathematicsFundamental Theorem & Properties of Definite IntegralsJEE Advanced 2005Moderate

View in:English Hindi

Evaluate $\int_{0}^{π} e^{∣ c o s x ∣} (2 sin (\frac{1}{2} cos x) + 3 cos (\frac{1}{2} cos x)) sin x d x$ .

Answer:

24 [e cos (\frac{1}{2}) + \frac{1}{2} e sin (\frac{1}{2}) - 1]

MathematicsIntegration by SubstitutionJEE Main 2008Easy

View in:English Hindi

The value of $2 \int \frac{s i n x d x}{s i n ( x - \frac{π}{4} )}$ is

(A)

x + lo g cos (x - \frac{π}{4}) + c

(B)

x - lo g sin (x - \frac{π}{4}) + c

(C)

x + lo g sin (x - \frac{π}{4}) + c

(D)

x - lo g cos (x - \frac{π}{4}) + c

Answer:C

MathematicsEvaluation of Special Integral FormsJEE Main 2004Moderate

View in:English Hindi

$\int \frac{d x}{c o s x - s i n x}$ is equal to

(A)

\frac{1}{2} lo g tan (\frac{x}{2} + \frac{3 π}{8}) + C

(B)

\frac{1}{2} lo g cot (\frac{x}{2}) + C

(C)

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

(D)

\frac{1}{2} lo g tan (\frac{x}{2} - \frac{π}{8}) + C

Answer:A

MathematicsFundamental Theorem & Properties of Definite IntegralsJEE Advanced 1985Moderate

View in:English Hindi

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}

MathematicsFundamental Theorem & Properties of Definite IntegralsJEE Advanced 1997Moderate

View in:English Hindi

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}

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

Visualized Solution (English)

Initial Integral and Double Angle Formula

Simplifying the Root and Transforming to Cotangent

Choosing a Strategic Substitution

Finding the Differential $d x$

Substituting into the Integral

Simplifying the Integrand

Preparing for Substitution $u = sin θ$

Substitution and Partial Fractions

Integrating the Terms

Final Result in Terms of $x$

Conceptually Similar Problems

Find the indefinite integral $\int cos 2 θ ln (\frac{c o s θ + s i n θ}{c o s θ - s i n θ}) d θ$

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

Evaluate $\int_{0}^{π} \frac{x s i n 2 x s i n ( \frac{π}{2} c o s x )}{2 x - π} d x$

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

Evaluate $\int (e^{l o g x} + sin x) cos x d x$

Evaluate $\int_{0}^{π} e^{∣ c o s x ∣} (2 sin (\frac{1}{2} cos x) + 3 cos (\frac{1}{2} cos x)) sin x d x$ .

The value of $2 \int \frac{s i n x d x}{s i n ( x - \frac{π}{4} )}$ is

$\int \frac{d x}{c o s x - s i n x}$ is equal to

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

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

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

Find the indefinite integral $\int cos 2 θ ln (\frac{c o s θ + s i n θ}{c o s θ - s i n θ}) d θ$

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

Evaluate $\int_{0}^{π} \frac{x s i n 2 x s i n ( \frac{π}{2} c o s x )}{2 x - π} d x$

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

Evaluate $\int (e^{l o g x} + sin x) cos x d x$

Evaluate $\int_{0}^{π} e^{∣ c o s x ∣} (2 sin (\frac{1}{2} cos x) + 3 cos (\frac{1}{2} cos x)) sin x d x$ .

The value of $2 \int \frac{s i n x d x}{s i n ( x - \frac{π}{4} )}$ is

$\int \frac{d x}{c o s x - s i n x}$ is equal to

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluate ∫sinx(cos2x)1/2​dx

Visualized Solution (English)

Initial Integral and Double Angle Formula

Simplifying the Root and Transforming to Cotangent

Choosing a Strategic Substitution

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

Substituting into the Integral

Simplifying the Integrand

Preparing for Substitution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u=sinθ

Substitution and Partial Fractions

Integrating the Terms

Final Result in Terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

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Preparing for Substitution $u = sin θ$

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

Evaluate $\int_{0}^{π} \frac{x s i n 2 x s i n ( \frac{π}{2} c o s x )}{2 x - π} d x$

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

Evaluate $\int (e^{l o g x} + sin x) cos x d x$

Evaluate $\int_{0}^{π} e^{∣ c o s x ∣} (2 sin (\frac{1}{2} cos x) + 3 cos (\frac{1}{2} cos x)) sin x d x$ .

The value of $2 \int \frac{s i n x d x}{s i n ( x - \frac{π}{4} )}$ is

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

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

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The value of $2 \int \frac{s i n x d x}{s i n ( x - \frac{π}{4} )}$ is

$\int \frac{d x}{c o s x - s i n x}$ is equal to

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

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