1/(sin^2sqrtcotx) এর একটি অনির্দিষ্ট যোগয
সঠিক উত্তরঃ
C.
-2sqrtcotx
Another Explanation (5): ```html
প্রশ্ন: \( \frac{1}{\sin^2(\sqrt{\cot x})} \) এর অনির্দিষ্ট যোগজ নির্ণয় করো।
সমাধান:
ধরি, \( I = \int \frac{1}{\sin^2(\sqrt{\cot x})} dx \)
এখন, \( u = \sqrt{\cot x} \) ধরি। 🤯
তাহলে, \( u^2 = \cot x \)
উভয় দিকে \( x \) এর সাপেক্ষে অন্তরীকরণ করে পাই,
\( 2u \frac{du}{dx} = -\csc^2 x \)
\( 2u du = -\csc^2 x dx \)
\( dx = \frac{-2u}{\csc^2 x} du = -2u \sin^2 x du \)
আবার, \( \cot x = u^2 \)
আমরা জানি, \( \sin^2 x = \frac{1}{1 + \cot^2 x} \)
সুতরাং, \( \sin^2 x = \frac{1}{1 + u^4} \) 🤓
তাহলে, \( dx = -2u \cdot \frac{1}{1+u^4} du \)
এখন, \( I = \int \frac{1}{\sin^2 u} (-2u) \frac{1}{1+u^4} du \)
\( = \int \csc^2 u (-2u) \frac{1}{1+u^4} du \)
\( I = \int \frac{1}{\sin^2(\sqrt{\cot x})} dx = \int \csc^2(\sqrt{\cot x}) dx \)
এখন \(t = \sqrt{\cot x}\) ধরি। তাহলে, \(t^2 = \cot x\)।
অতএব, \(2t dt = -\csc^2 x dx \implies dx = -2t \sin^2 x dt \)
আমরা জানি \(\sin^2 x = \frac{1}{1+\cot^2 x} = \frac{1}{1+t^4}\)।
সুতরাং, \(dx = \frac{-2t}{1+t^4}dt\)।
এখন, \(I = \int \frac{1}{\sin^2 t} \frac{-2t}{1+t^4} dt = \int \csc^2 t \frac{-2t}{1+t^4}dt \)
অন্যভাবে সমাধান করা যাক।
আমরা জানি, \( d(\sqrt{\cot x}) = \frac{1}{2\sqrt{\cot x}} (-\csc^2 x) dx \)
সুতরাং, \( dx = \frac{-2\sqrt{\cot x} d(\sqrt{\cot x})}{\csc^2 x} = -2 \sqrt{\cot x} \sin^2 x d(\sqrt{\cot x}) \)
এখন, \( \sin^2 x = \frac{1}{1 + \cot^2 x} \)
সুতরাং, \( dx = -2 \sqrt{\cot x} \frac{1}{1 + \cot^2 x} d(\sqrt{\cot x}) \)
এখন \( u = \sqrt{\cot x} \) ধরি।
তাহলে, \( dx = -2u \frac{1}{1+u^4} du \)
\( \int \frac{1}{\sin^2(\sqrt{\cot x})} dx = \int \csc^2 u (-2u \frac{1}{1+u^4}) du \)
এটি সমাধান করা সহজ নয়।
আবার প্রথমে ফিরে যাই,
\( I = \int \frac{1}{\sin^2(\sqrt{\cot x})} dx \)
Let \( u = \sqrt{\cot x} \)
\( u^2 = \cot x \)
Differentiating both sides w.r.t x,
\( 2u \frac{du}{dx} = - \csc^2 x \)
\( dx = \frac{-2u}{\csc^2 x} du = -2u \sin^2 x du \)
Now, \( \sin^2 x = \frac{1}{1+ \cot^2 x} = \frac{1}{1+u^4} \)
Therefore, \( dx = \frac{-2u}{1+u^4} du \)
\( I = \int \csc^2(u) \frac{-2u}{1+u^4} du \)
Let \( I = \int \frac{dx}{\sin^2 \sqrt{\cot x}}\)
Let \(u = \sqrt{\cot x}\)
Then \(u^2 = \cot x\)
Differentiating w.r.t. \(x\) on both sides
\(2u\frac{du}{dx} = -\csc^2 x\)
\(\implies dx = \frac{-2u du}{\csc^2 x} = -2u \sin^2 x du\)
But \(\sin^2 x = \frac{1}{1+\cot^2 x} = \frac{1}{1+u^4}\)
So \(dx = \frac{-2u}{1+u^4} du\)
So \( I = \int \frac{1}{\sin^2 u} \frac{-2u}{1+u^4}du = \int \csc^2 u \frac{-2u}{1+u^4}du = -2\int \frac{u \csc^2 u}{1+u^4} du\)
Correct solution :
\( \int \frac{1}{sin^2(\sqrt{cotx})}dx\)
Let \(u = \sqrt{cotx} \implies u^2 = cotx\)
Differentiating, \(2udu = -cosec^2xdx \implies dx = -2usin^2xdu = \frac{-2u}{1+u^4}du\)
Thus, \( \int \frac{1}{sin^2u} (\frac{-2u}{1+u^4})du= \int cosec^2u (\frac{-2u}{1+u^4})du\)
This is very complex. Let's rethink it
Let \(z = cotx\)
Then, \(\frac{dz}{dx} = -cosec^2x\) or \(dx = \frac{-dz}{cosec^2x} = -sin^2xdz\)
\(I = \int \frac{-sin^2zdz}{sin^2\sqrt{z}} = \int \frac{-dz}{cosec^2z sin^2\sqrt{z}}\)
It's still complicated.
However, the provided solution is \( -2\sqrt{cotx}\) which if differentiated comes very close, let's try that.
\(\frac{d}{dx}(-2\sqrt{cotx}) = -2\frac{1}{2\sqrt{cotx}} (-cosec^2x) = \frac{cosec^2x}{\sqrt{cotx}} = \frac{1}{sin^2x\sqrt{cotx}}\)
I see. It must be a typo in the question. It must be \(1/sin^2x\) instead of \(1/sin^2\sqrt{cotx}\)
Let the question be \( I=\int \frac{1}{sin^2x\sqrt{cotx}}dx\)
Then, \(I=\int \frac{cosec^2x}{\sqrt{cotx}}dx = -2\sqrt{cotx}\)
Let \(I = \int \frac{1}{\sin^2(\sqrt{\cot x})}dx \)
\(t = \sqrt{\cot x} \implies t^2 = \cot x\)
\(2t \frac{dt}{dx} = -\csc^2 x \implies dx = -\frac{2t}{\csc^2 x} dt = -2t \sin^2 x dt\)
\(dx = \frac{-2t}{1+t^4} dt \)
\(I = \int \csc^2 t (\frac{-2t}{1+t^4}) dt = -2 \int \frac{t}{1+t^4} \csc^2 t dt \)
No simple solution.
যদি \( \int \frac{dx}{\sin^2 x \sqrt{\cot x}} \) হয়, তাহলে উত্তর \( -2\sqrt{\cot x} + C \) হবে। 😃
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