int x sqrt{frac{2 sin left(x^2+1right)-sin 2l | vedclass

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Question

(A) ધારો કે $I = \int x \sqrt{\frac{2 \sin \left(x^2+1\right)-\sin 2\left(x^2+1\right)}{2 \sin \left(x^2+1\right)+\sin 2\left(x^2+1\right)}} \, dx$ $\

Vedclass · Accepted Answer

$\log \left(\sec \left(\frac{x^2+1}{2}\right)\right)+c$,જ્યાં $c$ એ સંકલનનો અચળાંક છે.

Vedclass · Answer

(A) ધારો કે $I = \int x \sqrt{\frac{2 \sin \left(x^2+1ight)-\sin 2\left(x^2+1ight)}{2 \sin \left(x^2+1ight)+\sin 2\left(x^2+1ight)}} \, dx$ $\sin 2	heta = 2 \sin 	heta \cos 	heta$ નો ઉપયોગ કરતા: $I = \int x \sqrt{\frac{2 \sin \left(x^2+1ight) - 2 \sin \left(x^2+1ight) \cos \left(x^2+1ight)}{2 \sin \left(x^2+1ight) + 2 \sin \left(x^2+1ight) \cos \left(x^2+1ight)}} \, dx$ $I = \int x \sqrt{\frac{1 - \cos \left(x^2+1ight)}{1 + \cos \left(x^2+1ight)}} \, dx$ $1 - \cos 	heta = 2 \sin^2(	heta/2)$ અને $1 + \cos 	heta = 2 \cos^2(	heta/2)$ નો ઉપયોગ કરતા: $I = \int x \sqrt{\frac{2 \sin^2 \left(\frac{x^2+1}{2}ight)}{2 \cos^2 \left(\frac{x^2+1}{2}ight)}} \, dx$ $I = \int x 	an \left(\frac{x^2+1}{2}ight) \, dx$ ધારો કે $t = \frac{x^2+1}{2}$,તેથી $dt = x \, dx$. $I = \int 	an t \, dt = \log |\sec t| + c$ $I = \log \left| \sec \left(\frac{x^2+1}{2}ight) ight| + c$

$\int x \sqrt{\frac{2 \sin \left(x^2+1\right)-\sin 2\left(x^2+1\right)}{2 \sin \left(x^2+1\right)+\sin 2\left(x^2+1\right)}} \, dx =$

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