Q1418: Evaluate: $\int_{0}^{\pi}\frac{dx}{a^{2}\cos^{2}x+b^{2}\sin^...

Class CBSE Class 12 Mathematics Definite Integrals Q #1418

KNOWLEDGE BASED

UNDERSTAND

5 Marks 2025 AISSCE(Board Exam) LA

Evaluate: $\int_{0}^{\pi}\frac{dx}{a^{2}\cos^{2}x+b^{2}\sin^{2}x}$.

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AI Tutor Explanation

Detailed Solution

Step 1: Use the property of definite integrals

We use the property $\int_{0}^{a}f(x)dx = \int_{0}^{a}f(a-x)dx$. Applying this to the given integral, we have $$I = \int_{0}^{\pi}\frac{dx}{a^{2}\cos^{2}x+b^{2}\sin^{2}x} = \int_{0}^{\pi}\frac{dx}{a^{2}\cos^{2}(\pi-x)+b^{2}\sin^{2}(\pi-x)}$$ Since $\cos(\pi-x) = -\cos x$ and $\sin(\pi-x) = \sin x$, we get $$I = \int_{0}^{\pi}\frac{dx}{a^{2}(-\cos x)^{2}+b^{2}(\sin x)^{2}} = \int_{0}^{\pi}\frac{dx}{a^{2}\cos^{2}x+b^{2}\sin^{2}x}$$ This doesn't simplify the integral, so we proceed with a different approach.

Step 2: Divide numerator and denominator by $\cos^2 x$

Divide both the numerator and the denominator by $\cos^{2}x$: $$I = \int_{0}^{\pi}\frac{\sec^{2}x}{a^{2}+b^{2}\tan^{2}x}dx$$

Step 3: Split the integral at $\pi/2$

Since $\tan x$ is not defined at $x = \frac{\pi}{2}$, we split the integral at this point: $$I = \int_{0}^{\pi/2}\frac{\sec^{2}x}{a^{2}+b^{2}\tan^{2}x}dx + \int_{\pi/2}^{\pi}\frac{\sec^{2}x}{a^{2}+b^{2}\tan^{2}x}dx$$

Step 4: Substitute $t = \tan x$ in the first integral

For the first integral, let $t = \tan x$, so $dt = \sec^{2}x dx$. When $x = 0$, $t = 0$. When $x = \frac{\pi}{2}$, $t \to \infty$. $$I_{1} = \int_{0}^{\infty}\frac{dt}{a^{2}+b^{2}t^{2}} = \frac{1}{b^{2}}\int_{0}^{\infty}\frac{dt}{\frac{a^{2}}{b^{2}}+t^{2}} = \frac{1}{b^{2}}\left[\frac{b}{a}\arctan\left(\frac{bt}{a}\right)\right]_{0}^{\infty} = \frac{1}{ab}\left[\arctan\left(\frac{bt}{a}\right)\right]_{0}^{\infty} = \frac{1}{ab}\left(\frac{\pi}{2}-0\right) = \frac{\pi}{2ab}$$

Step 5: Substitute $t = \tan x$ in the second integral

For the second integral, let $t = \tan x$, so $dt = \sec^{2}x dx$. When $x = \frac{\pi}{2}$, $t \to -\infty$. When $x = \pi$, $t = 0$. $$I_{2} = \int_{-\infty}^{0}\frac{dt}{a^{2}+b^{2}t^{2}} = \frac{1}{b^{2}}\int_{-\infty}^{0}\frac{dt}{\frac{a^{2}}{b^{2}}+t^{2}} = \frac{1}{b^{2}}\left[\frac{b}{a}\arctan\left(\frac{bt}{a}\right)\right]_{-\infty}^{0} = \frac{1}{ab}\left[\arctan\left(\frac{bt}{a}\right)\right]_{-\infty}^{0} = \frac{1}{ab}\left(0-\left(-\frac{\pi}{2}\right)\right) = \frac{\pi}{2ab}$$

Step 6: Combine the two integrals

Therefore, $$I = I_{1} + I_{2} = \frac{\pi}{2ab} + \frac{\pi}{2ab} = \frac{\pi}{ab}$$

Final Answer: $\frac{\pi}{ab}$

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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 understand how to handle singularities within the integration limits.

Knowledge Dimension: PROCEDURAL

Justification: The question requires the student to apply specific procedures such as using properties of definite integrals, trigonometric identities, and substitution to evaluate the integral. The student must know the steps to solve definite integrals.

Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's knowledge of definite integrals and their properties, which is a standard topic in the syllabus. The question is a direct application of textbook knowledge.