Q1482: Evaluate: $\int_{0}^{\pi}\frac{x\tan x}{\sec x+\tan x}dx$...

Class CBSE Class 12 Mathematics Definite Integrals Q #1482

KNOWLEDGE BASED

UNDERSTAND

5 Marks 2025 AISSCE(Board Exam) LA

Evaluate: $\int_{0}^{\pi}\frac{x\tan x}{\sec x+\tan x}dx$

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Detailed Solution

Step 1: Define the integral

Let $I = \int_{0}^{\pi}\frac{x\tan x}{\sec x+\tan x}dx$

Step 2: Use the property $\int_{0}^{a}f(x)dx = \int_{0}^{a}f(a-x)dx$

Using the property, we have $I = \int_{0}^{\pi}\frac{(\pi-x)\tan (\pi-x)}{\sec (\pi-x)+\tan (\pi-x)}dx$ $I = \int_{0}^{\pi}\frac{(\pi-x)(-\tan x)}{-\sec x-\tan x}dx$ $I = \int_{0}^{\pi}\frac{(\pi-x)\tan x}{\sec x+\tan x}dx$ $I = \int_{0}^{\pi}\frac{\pi\tan x}{\sec x+\tan x}dx - \int_{0}^{\pi}\frac{x\tan x}{\sec x+\tan x}dx$ $I = \pi\int_{0}^{\pi}\frac{\tan x}{\sec x+\tan x}dx - I$

Step 3: Solve for I

From the above equation, we have $2I = \pi\int_{0}^{\pi}\frac{\tan x}{\sec x+\tan x}dx$ $I = \frac{\pi}{2}\int_{0}^{\pi}\frac{\tan x}{\sec x+\tan x}dx$

Step 4: Simplify the integrand

We can rewrite the integrand as follows: $\frac{\tan x}{\sec x+\tan x} = \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+\frac{\sin x}{\cos x}} = \frac{\sin x}{1+\sin x}$ So, $I = \frac{\pi}{2}\int_{0}^{\pi}\frac{\sin x}{1+\sin x}dx$

Step 5: Further simplification of the integrand

We can multiply and divide by $(1-\sin x)$ to get: $\frac{\sin x}{1+\sin x} = \frac{\sin x(1-\sin x)}{(1+\sin x)(1-\sin x)} = \frac{\sin x - \sin^2 x}{1-\sin^2 x} = \frac{\sin x - \sin^2 x}{\cos^2 x} = \frac{\sin x}{\cos^2 x} - \frac{\sin^2 x}{\cos^2 x} = \sec x \tan x - \tan^2 x$ Since $\tan^2 x = \sec^2 x - 1$, we have $\frac{\sin x}{1+\sin x} = \sec x \tan x - (\sec^2 x - 1) = \sec x \tan x - \sec^2 x + 1$ Therefore, $I = \frac{\pi}{2}\int_{0}^{\pi}(\sec x \tan x - \sec^2 x + 1)dx$

Step 6: Evaluate the integral

$I = \frac{\pi}{2}\left[\int_{0}^{\pi}\sec x \tan x dx - \int_{0}^{\pi}\sec^2 x dx + \int_{0}^{\pi}1 dx\right]$ $I = \frac{\pi}{2}\left[ \sec x - \tan x + x \right]_{0}^{\pi}$ However, $\sec x$ and $\tan x$ are not defined at $x = \frac{\pi}{2}$. So, we need to split the integral at $\frac{\pi}{2}$. $I = \frac{\pi}{2}\int_{0}^{\pi}\frac{\sin x}{1+\sin x}dx = \frac{\pi}{2}\int_{0}^{\pi}\frac{\sin x(1-\sin x)}{1-\sin^2 x}dx = \frac{\pi}{2}\int_{0}^{\pi}\frac{\sin x - \sin^2 x}{\cos^2 x}dx$ $I = \frac{\pi}{2}\int_{0}^{\pi}(\sec x \tan x - \tan^2 x)dx = \frac{\pi}{2}\int_{0}^{\pi}(\sec x \tan x - (\sec^2 x - 1))dx$ $I = \frac{\pi}{2}\int_{0}^{\pi}(\sec x \tan x - \sec^2 x + 1)dx = \frac{\pi}{2}[\sec x - \tan x + x]_{0}^{\pi}$ $I = \frac{\pi}{2}\left[ (\sec \pi - \tan \pi + \pi) - (\sec 0 - \tan 0 + 0) \right] = \frac{\pi}{2}[(-1 - 0 + \pi) - (1 - 0 + 0)] = \frac{\pi}{2}[-2 + \pi] = \frac{\pi}{2}(\pi - 2)$

Step 7: Final Answer

Therefore, $I = \frac{\pi}{2}(\pi - 2)$

Final Answer: $\frac{\pi}{2}(\pi - 2)$

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Pedagogical Audit

Bloom's Analysis: This is an UNDERSTAND question because the student needs to understand the properties of definite integrals and trigonometric functions to solve the problem. They must also manipulate the integrand to make it integrable.

Knowledge Dimension: PROCEDURAL

Justification: The question requires the student to apply specific procedures such as using the property of definite integrals $\int_{0}^{a}f(x)dx = \int_{0}^{a}f(a-x)dx$, simplifying trigonometric expressions, and evaluating the resulting integral.<\/span>

Syllabus Audit: In the context of CBSE Class 12, this is classified as APPLICATION. The question requires the application of definite integrals and trigonometric identities, which are part of the syllabus.