Find the indefinite integral ∫ cos 2θ...

MathematicsIntegration by PartsJEE Advanced 1994Moderate

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

Visualized Solution (Hindi)

Simplify the Logarithmic Argument

Let the integral be $I = \int cos 2 θ ln (\frac{c o s θ + s i n θ}{c o s θ - s i n θ}) d θ$
Simplify the argument: $\frac{c o s θ + s i n θ}{c o s θ - s i n θ} = \frac{1 + t a n θ}{1 - t a n θ}$
Using the identity $tan (A + B) = \frac{t a n A + t a n B}{1 - t a n A t a n B}$ , we get:
$\frac{1 + t a n θ}{1 - t a n θ} = tan (\frac{π}{4} + θ)$

Set up Integration by Parts

Substitute the simplified argument back into the integral:
$I = \int cos 2 θ ln tan (θ + \frac{π}{4}) d θ$
Using Integration by Parts: $\int u d v = uv - \int v d u$
Let $u = ln tan (θ + \frac{π}{4})$
Let $d v = cos 2 θ d θ$

Differentiate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u using Chain Rule

Differentiate $u$ : $\frac{d u}{d θ} = \frac{1}{t a n ( θ + \frac{π}{4} )} \cdot sec^{2} (θ + \frac{π}{4})$
Simplify using $sin$ and $cos$ : $\frac{d u}{d θ} = \frac{c o s ( θ + \frac{π}{4} )}{s i n ( θ + \frac{π}{4} )} \cdot \frac{1}{c o s ^{2} ( θ + \frac{π}{4} )}$
$\frac{d u}{d θ} = \frac{1}{s i n ( θ + \frac{π}{4} ) c o s ( θ + \frac{π}{4} )}$
Multiply by $\frac{2}{2}$ : $\frac{d u}{d θ} = \frac{2}{s i n ( 2 θ + \frac{π}{2} )}$
Since $sin (\frac{π}{2} + 2 θ) = cos 2 θ$ :
$d u = 2 sec 2 θ d θ$

Integrate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d v to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v

Integrate $d v$ : $v = \int cos 2 θ d θ$
Using the rule $\int cos a x d x = \frac{s i n a x}{a}$ :
$v = \frac{s i n 2 θ}{2}$

Apply the IBP Formula

Substitute $u, v, d u$ into $uv - \int v d u$ :
$I = \frac{s i n 2 θ}{2} ln tan (θ + \frac{π}{4}) - \int \frac{s i n 2 θ}{2} (2 sec 2 θ) d θ$
Simplify the integral term: $I = \frac{s i n 2 θ}{2} ln tan (θ + \frac{π}{4}) - \int sin 2 θ sec 2 θ d θ$
Since $sec 2 θ = \frac{1}{c o s 2 θ}$ :
$I = \frac{s i n 2 θ}{2} ln tan (θ + \frac{π}{4}) - \int tan 2 θ d θ$

Final Integration and Result

Integrate $tan 2 θ$ : $\int tan 2 θ d θ = \frac{1}{2} ln ∣ sec 2 θ ∣$
Substitute back into the expression for $I$ :
$I = \frac{s i n 2 θ}{2} ln tan (θ + \frac{π}{4}) - \frac{1}{2} ln ∣ sec 2 θ ∣ + C$
Replace $tan (θ + \frac{π}{4})$ with the original fraction:
$I = \frac{s i n 2 θ}{2} ln (\frac{c o s θ + s i n θ}{c o s θ - s i n θ}) - \frac{1}{2} ln ∣ sec 2 θ ∣ + C$

Summary and Key Takeaways

Key Takeaway 1: Simplify logarithmic arguments using trigonometric identities before integrating.
Key Takeaway 2: Use the I L A T E rule to choose $u$ and $d v$ effectively.
Next Challenge: Try solving $\int sin 2 θ ln (tan θ) d θ$ using a similar approach.

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Conceptually Similar Problems

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

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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:

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

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

Answer:

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

Visualized Solution (Hindi)

Simplify the Logarithmic Argument

Set up Integration by Parts

Differentiate $u$ using Chain Rule

Integrate $d v$ to find $v$

Apply the IBP Formula

Final Integration and Result

Summary and Key Takeaways

Conceptually Similar Problems

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

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

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

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

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Integrate $\int_{0}^{π} \frac{e ^{c o s x}}{e ^{c o s x} + e ^{- c o s x}} d x$ .

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

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

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

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

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

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

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

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Integrate $\int_{0}^{π} \frac{e ^{c o s x}}{e ^{c o s x} + e ^{- c o s x}} d x$ .

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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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the indefinite integral ∫cos2θln(cosθ−sinθcosθ+sinθ​)dθ

Visualized Solution (Hindi)

Simplify the Logarithmic Argument

Set up Integration by Parts

Differentiate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u using Chain Rule

Integrate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dv to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v

Apply the IBP Formula

Final Integration and Result

Summary and Key Takeaways

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