Q1257: Evaluate: $\int_{0}^{\pi/4}\frac{1}{sin~x+cos~x}dx$...

Class CBSE Class 12 Mathematics Definite Integrals Q #1257

KNOWLEDGE BASED

UNDERSTAND

3 Marks 2024 AISSCE(Board Exam) SA

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

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AI Tutor Explanation

Detailed Solution

Step 1: Rewrite the integrand

We can rewrite the integrand using the identity $sin~x + cos~x = \sqrt{2}sin(x + \frac{\pi}{4})$. Therefore, the integral becomes: $\int_{0}^{\pi/4}\frac{1}{sin~x+cos~x}dx = \int_{0}^{\pi/4}\frac{1}{\sqrt{2}sin(x + \frac{\pi}{4})}dx = \frac{1}{\sqrt{2}}\int_{0}^{\pi/4}\frac{1}{sin(x + \frac{\pi}{4})}dx$

Step 2: Convert sine to cosecant

We know that $\frac{1}{sin~\theta} = cosec~\theta$. So, the integral becomes: $\frac{1}{\sqrt{2}}\int_{0}^{\pi/4}cosec(x + \frac{\pi}{4})dx$

Step 3: Integrate cosecant

The integral of $cosec~x$ is $ln|tan(\frac{x}{2})|$. Therefore, the integral of $cosec(x + \frac{\pi}{4})$ is $ln|tan(\frac{x}{2} + \frac{\pi}{8})|$. So, we have: $\frac{1}{\sqrt{2}}\int_{0}^{\pi/4}cosec(x + \frac{\pi}{4})dx = \frac{1}{\sqrt{2}}[ln|tan(\frac{x}{2} + \frac{\pi}{8})|]_{0}^{\pi/4}$

Step 4: Evaluate the limits

Now, we substitute the limits: $\frac{1}{\sqrt{2}}[ln|tan(\frac{\pi}{8} + \frac{\pi}{8})| - ln|tan(\frac{\pi}{8})|] = \frac{1}{\sqrt{2}}[ln|tan(\frac{\pi}{4})| - ln|tan(\frac{\pi}{8})|]$ Since $tan(\frac{\pi}{4}) = 1$, $ln(1) = 0$. So, we have: $\frac{1}{\sqrt{2}}[0 - ln|tan(\frac{\pi}{8})|] = -\frac{1}{\sqrt{2}}ln|tan(\frac{\pi}{8})|$

Step 5: Simplify using the half-angle formula

We know that $tan(\frac{\theta}{2}) = \frac{1 - cos~\theta}{sin~\theta}$. Therefore, $tan(\frac{\pi}{8}) = \frac{1 - cos(\frac{\pi}{4})}{sin(\frac{\pi}{4})} = \frac{1 - \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = \sqrt{2} - 1$. So, the integral becomes: $-\frac{1}{\sqrt{2}}ln(\sqrt{2} - 1)$ Since $ln(\frac{1}{x}) = -ln(x)$, we can rewrite this as: $\frac{1}{\sqrt{2}}ln(\frac{1}{\sqrt{2} - 1}) = \frac{1}{\sqrt{2}}ln(\frac{\sqrt{2} + 1}{(\sqrt{2} - 1)(\sqrt{2} + 1)}) = \frac{1}{\sqrt{2}}ln(\sqrt{2} + 1)$

Final Answer: $\frac{1}{\sqrt{2}}ln(\sqrt{2} + 1)$

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Pedagogical Audit

Bloom's Analysis: This is an UNDERSTAND question because the student needs to understand trigonometric identities and integration techniques to solve the problem. They must recognize the form of the integrand and apply appropriate substitutions and formulas.

Knowledge Dimension: PROCEDURAL

Justification: The question requires the student to apply specific procedures for integration, including trigonometric substitutions and the evaluation of definite integrals. The student must know the steps involved in integrating trigonometric functions and applying the limits of integration.<\/span>

Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's knowledge of integration techniques and trigonometric identities, which are core concepts covered in the syllabus. The question is a standard application of these concepts.