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Question Number 94574 by student work last updated on 19/May/20

$$\int \frac{\sqrt{\mathrm{cosx}}}{\sqrt{\mathrm{sinx}}+\sqrt{\mathrm{cosx}}}\mathrm{dx}=?$$

Commented by student work last updated on 19/May/20

$$\mathrm{please}\mathrm{solve}\mathrm{who}\mathrm{can}?$$

Answered by MJS last updated on 19/May/20

$$\int \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx=$$$$[t=\sqrt{\tan x}\to dx={2\cos }^{2}x\sqrt{\tan x}dt]$$$$=2\int \frac{t}{(t^4+1)(t+1)}dt=$$$$=\int \frac{t^3-t^2+t+1}{t^4+1}dt+\int \frac{dt}{t+1}=$$$$=\frac{1-\sqrt{2}}{2}\int \frac{t-1-\sqrt{2}}{t^2-\sqrt{2}t+1}dt+\frac{1+\sqrt{2}}{2}\int \frac{t-1+\sqrt{2}}{t^2+\sqrt{2}t+1}dt-\int \frac{dt}{t+1}$$$$\mathrm{now}\mathrm{it}\mathrm{should}\mathrm{be}\mathrm{easy}$$

Commented by student work last updated on 20/May/20

$$\int \frac{2\mathrm{t}}{({\mathrm{t}}^4+1)(1+\mathrm{t})}\mathrm{dt}=?\mathrm{how}\mathrm{did}\mathrm{you}\mathrm{solve}?$$

Commented by MJS last updated on 20/May/20

$$\mathrm{just}\mathrm{decompose}$$

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