(a)মান নির্ণয় কর: int_0^pi(dx/(a^2cos^2x+b^2sin^2x) ) (b) মান নির্ণয় কর: int_0^1(In(x^2+1)dx 

Explanation:

Another Explanation (5):

সমাধান:

(a) \( \int_0^\pi \frac{dx}{a^2\cos^2x + b^2\sin^2x} \)

আমরা জানি, \( \int_0^{2a} f(x) dx = 2\int_0^a f(x) dx \) যদি \( f(2a-x) = f(x) \) হয়। এখানে, \( f(x) = \frac{1}{a^2\cos^2x + b^2\sin^2x} \) তাহলে, \( f(\pi - x) = \frac{1}{a^2\cos^2(\pi - x) + b^2\sin^2(\pi - x)} = \frac{1}{a^2\cos^2x + b^2\sin^2x} = f(x) \) সুতরাং, \( \int_0^\pi \frac{dx}{a^2\cos^2x + b^2\sin^2x} = 2\int_0^{\pi/2} \frac{dx}{a^2\cos^2x + b^2\sin^2x} \) এখন, লব ও হরকে \( \cos^2x \) দিয়ে ভাগ করে পাই, \( 2\int_0^{\pi/2} \frac{\sec^2x}{a^2 + b^2\tan^2x} dx \) ধরি, \( \tan x = z \), সুতরাং \( \sec^2x dx = dz \) যখন \( x = 0 \), \( z = 0 \) এবং যখন \( x = \pi/2 \), \( z = \infty \) তাহলে, \( 2\int_0^{\infty} \frac{dz}{a^2 + b^2z^2} = \frac{2}{b^2}\int_0^{\infty} \frac{dz}{(a/b)^2 + z^2} \) \( = \frac{2}{b^2} \cdot \frac{1}{a/b} \left[ \tan^{-1}\left(\frac{z}{a/b}\right) \right]_0^{\infty} \) \( = \frac{2}{ab} \left[ \tan^{-1}\left(\frac{bz}{a}\right) \right]_0^{\infty} = \frac{2}{ab} \left[ \frac{\pi}{2} - 0 \right] = \frac{\pi}{ab} \) 🎉 সুতরাং, \( \int_0^\pi \frac{dx}{a^2\cos^2x + b^2\sin^2x} = \frac{\pi}{ab} \)

(b) \( \int_0^1 \ln(x^2+1) dx \)

আমরা parts এর মাধ্যমে ইন্টিগ্রেট করি: \( \int u dv = uv - \int v du \) ধরি, \( u = \ln(x^2+1) \) এবং \( dv = dx \) তাহলে, \( du = \frac{2x}{x^2+1} dx \) এবং \( v = x \) সুতরাং, \( \int_0^1 \ln(x^2+1) dx = \left[ x\ln(x^2+1) \right]_0^1 - \int_0^1 \frac{2x^2}{x^2+1} dx \) \( = \left[ 1\cdot\ln(1^2+1) - 0\cdot\ln(0^2+1) \right] - 2\int_0^1 \frac{x^2}{x^2+1} dx \) \( = \ln 2 - 2\int_0^1 \frac{x^2+1-1}{x^2+1} dx = \ln 2 - 2\int_0^1 \left(1 - \frac{1}{x^2+1}\right) dx \) \( = \ln 2 - 2\left[ x - \tan^{-1}(x) \right]_0^1 = \ln 2 - 2\left[ (1 - \tan^{-1}(1)) - (0 - \tan^{-1}(0)) \right] \) \( = \ln 2 - 2\left[ 1 - \frac{\pi}{4} \right] = \ln 2 - 2 + \frac{\pi}{2} \) ✨ সুতরাং, \( \int_0^1 \ln(x^2+1) dx = \ln 2 - 2 + \frac{\pi}{2} \) 🚀