Q1318: Evaluate $\int_{0}^{\frac{\pi}{4}}\frac{x~dx}{1+cos~2x+sin~2...

Class CBSE Class 12 Mathematics Definite Integrals Q #1318

COMPETENCY BASED

APPLY

3 Marks 2024 AISSCE(Board Exam) SA

Evaluate $\int_{0}^{\frac{\pi}{4}}\frac{x~dx}{1+cos~2x+sin~2x}$

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AI Tutor Explanation

Detailed Solution

Step 1: Simplify the denominator

We have $1 + \cos 2x + \sin 2x$ in the denominator. We can rewrite this using trigonometric identities:\r\n\r\n$1 + \cos 2x = 2 \cos^2 x$\r\n$\sin 2x = 2 \sin x \cos x$\r\n\r\nSo, $1 + \cos 2x + \sin 2x = 2 \cos^2 x + 2 \sin x \cos x = 2 \cos x (\cos x + \sin x)$

Step 2: Rewrite the integral

The integral becomes:\r\n\r\n$\int_{0}^{\frac{\pi}{4}} \frac{x}{2 \cos x (\cos x + \sin x)} dx = \frac{1}{2} \int_{0}^{\frac{\pi}{4}} \frac{x}{\cos x (\cos x + \sin x)} dx$\r\n\r\nDivide numerator and denominator by $\cos^2 x$:\r\n\r\n$\frac{1}{2} \int_{0}^{\frac{\pi}{4}} \frac{x \sec^2 x}{1 + \tan x} dx$

Step 3: Use integration by parts

Let $u = x$ and $dv = \frac{\sec^2 x}{1 + \tan x} dx$. Then $du = dx$ and to find $v$, we integrate $dv$:\r\n\r\n$v = \int \frac{\sec^2 x}{1 + \tan x} dx$\r\n\r\nLet $t = 1 + \tan x$, so $dt = \sec^2 x dx$. Then:\r\n\r\n$v = \int \frac{dt}{t} = \ln |t| = \ln |1 + \tan x|$\r\n\r\nUsing integration by parts, $\int u dv = uv - \int v du$:\r\n\r\n$\frac{1}{2} \int_{0}^{\frac{\pi}{4}} \frac{x \sec^2 x}{1 + \tan x} dx = \frac{1}{2} \left[ x \ln (1 + \tan x) \right]_{0}^{\frac{\pi}{4}} - \frac{1}{2} \int_{0}^{\frac{\pi}{4}} \ln (1 + \tan x) dx$

Step 4: Evaluate the first term

$\frac{1}{2} \left[ x \ln (1 + \tan x) \right]_{0}^{\frac{\pi}{4}} = \frac{1}{2} \left[ \frac{\pi}{4} \ln (1 + \tan \frac{\pi}{4}) - 0 \cdot \ln (1 + \tan 0) \right] = \frac{1}{2} \left[ \frac{\pi}{4} \ln (1 + 1) - 0 \right] = \frac{\pi}{8} \ln 2$

Step 5: Evaluate the remaining integral

Let $I = \int_{0}^{\frac{\pi}{4}} \ln (1 + \tan x) dx$. Use the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx$:\r\n\r\n$I = \int_{0}^{\frac{\pi}{4}} \ln \left(1 + \tan \left(\frac{\pi}{4} - x\right)\right) dx = \int_{0}^{\frac{\pi}{4}} \ln \left(1 + \frac{\tan \frac{\pi}{4} - \tan x}{1 + \tan \frac{\pi}{4} \tan x}\right) dx = \int_{0}^{\frac{\pi}{4}} \ln \left(1 + \frac{1 - \tan x}{1 + \tan x}\right) dx$\r\n\r\n$I = \int_{0}^{\frac{\pi}{4}} \ln \left(\frac{1 + \tan x + 1 - \tan x}{1 + \tan x}\right) dx = \int_{0}^{\frac{\pi}{4}} \ln \left(\frac{2}{1 + \tan x}\right) dx = \int_{0}^{\frac{\pi}{4}} (\ln 2 - \ln (1 + \tan x)) dx$\r\n\r\n$I = \int_{0}^{\frac{\pi}{4}} \ln 2 dx - \int_{0}^{\frac{\pi}{4}} \ln (1 + \tan x) dx = \ln 2 \int_{0}^{\frac{\pi}{4}} dx - I$\r\n\r\n$I = (\ln 2) \left[ x \right]_{0}^{\frac{\pi}{4}} - I = \frac{\pi}{4} \ln 2 - I$\r\n\r\n$2I = \frac{\pi}{4} \ln 2$, so $I = \frac{\pi}{8} \ln 2$

Step 6: Combine the results

The original integral is:\r\n\r\n$\frac{1}{2} \left[ x \ln (1 + \tan x) \right]_{0}^{\frac{\pi}{4}} - \frac{1}{2} \int_{0}^{\frac{\pi}{4}} \ln (1 + \tan x) dx = \frac{\pi}{8} \ln 2 - \frac{1}{2} \left( \frac{\pi}{8} \ln 2 \right) = \frac{\pi}{8} \ln 2 - \frac{\pi}{16} \ln 2 = \frac{\pi}{16} \ln 2$

Final Answer: $\frac{\pi}{16} \ln 2$

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Pedagogical Audit

Bloom's Analysis: This is an APPLY question because the student needs to apply integration techniques such as trigonometric identities, integration by parts, and properties of definite integrals to solve the problem.

Knowledge Dimension: PROCEDURAL

Justification: The question requires the student to execute a sequence of steps involving trigonometric simplification, integration by parts, and application of definite integral properties.

Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. The question requires application of multiple concepts and techniques to arrive at the solution, testing the student's problem-solving abilities.