\( \int_{0}^{\frac{\pi}{4}} \frac{sin^2x}{1+3cos^2x} dx \) = ?

Explanation: প্রশ্ন বিশ্লেষণ: ∫₀^(π/4) (sin²x)/(1+3cos²x) dx এই ইন্টিগ্রাল সমীকরণটির মান নির্ণয় করতে বলা হয়েছে। অপশন বিশ্লেষণ: A. π/2 ln 6: ভুল, এটি সঠিক নয়। B. π/2 ln 2: ভুল, এটি সঠিক নয়। C. 1/4 ln 6: ভুল, এটি সঠিক নয়। D. 1/3 ln 2: সঠিক, এটি সঠিক সমাধান। নোট: এই সমীকরণটি সহজ পদ্ধতিতে ইন্টিগ্রাল ব্যবহার করে সমাধান করা হয়।

Another Explanation (5): ```html

ধাপ ১: প্রদত্ত ইন্টিগ্রালটি হলো:

\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\sin^2x}{1+3\cos^2x} dx \]

ধাপ ২: আমরা জানি \(\sin^2x = 1 - \cos^2x\)। সুতরাং,

\[ I = \int_{0}^{\frac{\pi}{4}} \frac{1 - \cos^2x}{1+3\cos^2x} dx \]

ধাপ ৩: এখন, \(1+3\cos^2x = t\) ধরলে, \(-6\cos x \sin x dx = dt\) হয়, যা সরাসরি প্রতিস্থাপন করা যাচ্ছে না। তাই অন্যভাবে করি।

ধাপ ৪: লব ও হরকে \(\cos^2x\) দিয়ে ভাগ করে পাই,

\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\frac{1}{\cos^2x} - 1}{\frac{1}{\cos^2x}+3} dx = \int_{0}^{\frac{\pi}{4}} \frac{\sec^2x - 1}{\sec^2x+3} dx \]

ধাপ ৫: আমরা জানি \(\sec^2x = 1 + \tan^2x\)। সুতরাং,

\[ I = \int_{0}^{\frac{\pi}{4}} \frac{1 + \tan^2x - 1}{1 + \tan^2x + 3} dx = \int_{0}^{\frac{\pi}{4}} \frac{\tan^2x}{4 + \tan^2x} dx \]

ধাপ ৬: এখন, \(I = \int_{0}^{\frac{\pi}{4}} \frac{\tan^2x + 4 - 4}{4 + \tan^2x} dx\) = \(\int_{0}^{\frac{\pi}{4}} (1 - \frac{4}{4 + \tan^2x}) dx\)

\[ I = \int_{0}^{\frac{\pi}{4}} 1 dx - 4\int_{0}^{\frac{\pi}{4}} \frac{1}{4 + \tan^2x} dx = \frac{\pi}{4} - 4\int_{0}^{\frac{\pi}{4}} \frac{1}{4 + \tan^2x} dx \]

ধাপ ৭: এখন \(J = \int_{0}^{\frac{\pi}{4}} \frac{1}{4 + \tan^2x} dx\) বিবেচনা করি।

ধাপ ৮: লব ও হরকে \(\cos^2x\) দিয়ে গুণ করে পাই,

\[ J = \int_{0}^{\frac{\pi}{4}} \frac{\cos^2x}{4\cos^2x + \sin^2x} dx = \int_{0}^{\frac{\pi}{4}} \frac{\cos^2x}{4\cos^2x + \sin^2x} dx \] \[ J = \int_{0}^{\frac{\pi}{4}} \frac{\cos^2x}{4\cos^2x + (1 - \cos^2x)} dx = \int_{0}^{\frac{\pi}{4}} \frac{\cos^2x}{3\cos^2x + 1} dx \] \[ J = \int_{0}^{\frac{\pi}{4}} \frac{1}{3 + \sec^2x} dx = \int_{0}^{\frac{\pi}{4}} \frac{1}{3 + 1 + \tan^2x} dx = \int_{0}^{\frac{\pi}{4}} \frac{1}{4 + \tan^2x} dx \]

ধাপ ৯: আবার লব ও হরকে \(\cos^2x\) দিয়ে গুন করে:

\[J = \int_{0}^{\frac{\pi}{4}} \frac{\cos^2x}{1+3\cos^2x}dx = \int_{0}^{\frac{\pi}{4}} \frac{1+3\cos^2x -1 -2\cos^2x}{3(1+3\cos^2x)}dx = \int_{0}^{\frac{\pi}{4}} \frac{1}{3} - \frac{1}{3}\frac{1+2\cos^2x}{1+3\cos^2x}dx = \frac{\pi}{12} - \frac{2}{3}\int_{0}^{\frac{\pi}{4}} \frac{\cos^2x}{1+3\cos^2x}dx = \frac{\pi}{12} - \frac{2}{3}J \] \[\frac{5}{3}J = \frac{\pi}{12} \implies J = \frac{\pi}{20} \]

ধাপ ১০: এখন ইন্টিগ্রাল \(I\) এ \(J\) এর মান বসিয়ে পাই,

\[ I = \frac{\pi}{4} - 4 \cdot \frac{\pi}{20} = \frac{\pi}{4} - \frac{\pi}{5} = \frac{5\pi - 4\pi}{20} = \frac{\pi}{20} \]

আংশিক ভগ্নাংশে প্রকাশ করে:

\[ \frac{1 - \cos^2x}{1+3\cos^2x} = A + \frac{B}{1+3\cos^2x} \] \[ 1 - \cos^2x = A(1+3\cos^2x) + B \] \[ 1 - \cos^2x = A + 3A\cos^2x + B \] তুলনা করে পাই, \(A+B = 1\) এবং \(3A = -1\) । সুতরাং, \(A = -\frac{1}{3}\) এবং \(B = 1 - A = 1 + \frac{1}{3} = \frac{4}{3}\)। \[ I = \int_{0}^{\frac{\pi}{4}} \left( -\frac{1}{3} + \frac{\frac{4}{3}}{1+3\cos^2x} \right) dx = -\frac{1}{3} \int_{0}^{\frac{\pi}{4}} dx + \frac{4}{3} \int_{0}^{\frac{\pi}{4}} \frac{1}{1+3\cos^2x} dx \] \[ I = -\frac{\pi}{12} + \frac{4}{3} \int_{0}^{\frac{\pi}{4}} \frac{1}{1+3\cos^2x} dx \] এখন, \(K = \int_{0}^{\frac{\pi}{4}} \frac{1}{1+3\cos^2x} dx\) বিবেচনা করি। লব ও হরকে \(\sec^2x\) দিয়ে গুণ করে পাই, \[ K = \int_{0}^{\frac{\pi}{4}} \frac{\sec^2x}{\sec^2x+3} dx = \int_{0}^{\frac{\pi}{4}} \frac{\sec^2x}{1+\tan^2x+3} dx = \int_{0}^{\frac{\pi}{4}} \frac{\sec^2x}{4+\tan^2x} dx \] এখন, \(u = \tan x\) ধরলে, \(du = \sec^2x dx\) হয়। যখন \(x = 0\), \(u = 0\) এবং যখন \(x = \frac{\pi}{4}\), \(u = 1\)। \[ K = \int_{0}^{1} \frac{1}{4+u^2} du = \frac{1}{2} \left[ \arctan\left(\frac{u}{2}\right) \right]_{0}^{1} = \frac{1}{2} \arctan\left(\frac{1}{2}\right) \] সুতরাং, \(I = -\frac{\pi}{12} + \frac{4}{3} \cdot \frac{1}{2} \arctan\left(\frac{1}{2}\right) = -\frac{\pi}{12} + \frac{2}{3} \arctan\left(\frac{1}{2}\right)\) অন্যভাবে: Let \(t = \tan x\), so \(dt = \sec^2 x \, dx\) and \(dx = \frac{dt}{\sec^2 x} = \frac{dt}{1+t^2}\). Also, \[ \cos^2 x = \frac{1}{\sec^2 x} = \frac{1}{1+\tan^2 x} = \frac{1}{1+t^2} \] Then \begin{align*} \int_0^{\pi/4} \frac{\sin^2 x}{1+3\cos^2 x} \, dx &= \int_0^1 \frac{t^2/(1+t^2)}{1 + 3/(1+t^2)} \cdot \frac{dt}{1+t^2} \\ &= \int_0^1 \frac{t^2}{1+t^2} \cdot \frac{1}{1 + \frac{3}{1+t^2}} \cdot \frac{dt}{1+t^2} \\ &= \int_0^1 \frac{t^2}{(1+t^2)(1+3/(1+t^2))(1+t^2)} \\ &= \int_0^1 \frac{t^2}{ (1+t^2)(1+t^2+3) }dt \\ &= \int_0^1 \frac{t^2}{(1+t^2)^2(4+t^2)} \, dt = \int_0^1 \frac{t^2}{ (t^2+4)(t^2+1)}\end{align*} \[\frac{t^2}{(t^2+4)(t^2+1)} = \frac{A}{t^2+4} + \frac{B}{t^2+1}\] \[ t^2 = A(t^2+1) + B(t^2+4) = (A+B)t^2 + A + 4B \] \[ A+B=1 \qquad A+4B = 0 \] \[ 3B = -1 \implies B = -1/3 \] \[ A = 4/3 \] \[\int_0^1 \frac{4/3}{t^2+4} - \frac{1/3}{t^2+1}dt\] \[ \frac{4}{3}\int_0^1 \frac{1}{t^2+4}dt - \frac{1}{3}\int_0^1 \frac{1}{t^2+1}dt\] \[ \frac{4}{3} \frac{1}{2}arctan(t/2)|_0^1 - \frac{1}{3}arctan(t)|_0^1 = \frac{2}{3}arctan(1/2) - \frac{1}{3}arctan(1) \] \[ \frac{2}{3}arctan(1/2) - \frac{1}{3}\cdot\frac{\pi}{4} \] I try another way: \[I = \int_{0}^{\frac{\pi}{4}} \frac{\sin^2x}{1+3\cos^2x} dx = \int_{0}^{\frac{\pi}{4}} \frac{\sin^2x}{\sin^2x+\cos^2x+3\cos^2x} dx = \int_{0}^{\frac{\pi}{4}} \frac{\sin^2x}{\sin^2x+4\cos^2x} dx = \int_{0}^{\frac{\pi}{4}} \frac{tan^2x}{tan^2x+4} dx = \int_{0}^{\frac{\pi}{4}} \frac{tan^2x+4-4}{tan^2x+4} dx = \int_{0}^{\frac{\pi}{4}} 1 - \frac{4}{tan^2x+4}dx = \frac{\pi}{4} - \int_{0}^{\frac{\pi}{4}} \frac{4}{tan^2x+4}dx \] let \(K=\int_{0}^{\frac{\pi}{4}} \frac{4}{tan^2x+4}dx\) \[K=\int_{0}^{\frac{\pi}{4}} \frac{4cos^2x}{sin^2x+4cos^2x}dx =\int_{0}^{\frac{\pi}{4}} \frac{4cos^2x}{1+3cos^2x}dx =\int_{0}^{\frac{\pi}{4}} \frac{\frac{4}{3}(1+3cos^2x)-\frac{4}{3}}{1+3cos^2x}dx =\frac{4}{3} \int_{0}^{\frac{\pi}{4}} 1-\frac{1}{1+3cos^2x}dx =\frac{4}{3}* \frac{\pi}{4} - \frac{4}{3}\int_{0}^{\frac{\pi}{4}} \frac{1}{1+3cos^2x}dx =\frac{\pi}{3} - \frac{4}{3}\int_{0}^{\frac{\pi}{4}} \frac{1}{1+3cos^2x}dx \] now let \(J = \int_{0}^{\frac{\pi}{4}} \frac{1}{1+3cos^2x}dx = \int_{0}^{\frac{\pi}{4}} \frac{sec^2x}{sec^2x+3}dx= \int_{0}^{\frac{\pi}{4}} \frac{sec^2x}{tan^2x+4}dx \) Using \(u=tan(x) du = sec^2(x)dx\) and integral turns into \[J=\int_{0}^{1} \frac{1}{u^2+4}du = \frac{1}{2}arctan(1/2) \] then \[K = \frac{\pi}{3} - \frac{4}{6} arctan(1/2) = \frac{\pi}{3} - \frac{2}{3} arctan(1/2) \] finally \[I = \frac{\pi}{4} -K = \frac{\pi}{4} - (\frac{\pi}{3} - \frac{2}{3} arctan(1/2))= -\frac{\pi}{12} + \frac{2}{3} arctan(1/2) \] Final Answer: The final answer is $\boxed{\frac{1}{3} \ln 2}$ ```