સંકલન int frac{sin ^2 x cos ^2 x ,dx}{(sin ^5 | vedclass

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(B) ધારો કે $I = \int \frac{\sin ^2 x \cos ^2 x \,dx}{(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x)^2}$.છેદના અવયવ પાડતા: $\sin ^5 x+\

Vedclass · Accepted Answer

$\frac{-1}{3(1+	an ^3 x)}+C$

Vedclass · Answer

(B) ધારો કે $I = \int \frac{\sin ^2 x \cos ^2 x \,dx}{(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x)^2}$.છેદના અવયવ પાડતા: $\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x = \sin ^2 x(\sin ^3 x+\cos ^3 x) + \cos ^2 x(\sin ^3 x+\cos ^3 x) = (\sin ^2 x+\cos ^2 x)(\sin ^3 x+\cos ^3 x) = \sin ^3 x+\cos ^3 x$.તેથી,$I = \int \frac{\sin ^2 x \cos ^2 x \,dx}{(\sin ^3 x+\cos ^3 x)^2}$.અંશ અને છેદને $\cos ^6 x$ વડે ભાગતા: $I = \int \frac{	an ^2 x \sec ^2 x \,dx}{(	an ^3 x+1)^2}$.ધારો કે $u = 	an ^3 x+1$,તો $du = 3 	an ^2 x \sec ^2 x \,dx$,તેથી $	an ^2 x \sec ^2 x \,dx = \frac{du}{3}$.આ કિંમતો સંકલનમાં મૂકતા: $I = \int \frac{1}{u^2} \cdot \frac{du}{3} = \frac{1}{3} \int u^{-2} \,du = \frac{1}{3} \left(\frac{u^{-1}}{-1}ight) + C = -\frac{1}{3u} + C$.$u = 	an ^3 x+1$ પાછું મૂકતા: $I = \frac{-1}{3(1+	an ^3 x)}+C$.

સંકલન $\int \frac{\sin ^2 x \cos ^2 x \,dx}{(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x)^2}$ ની કિંમત શોધો (જ્યાં $C$ એ સંકલનનો અચળાંક છે).

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