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Question Number 86313 by john santu last updated on 28/Mar/20

$$\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{dx}{\sqrt{1+\tan x}}$$

Commented by john santu last updated on 28/Mar/20

$$\frac{\sqrt{\sin x}}{\sqrt{\sin x+\cos x}}+\frac{\sqrt{\cos x}}{\sqrt{\sin x+\cos x}}=1$$$$how?$$

Commented by john santu last updated on 28/Mar/20

$$\sqrt{\cos x}+\sqrt{\sin x}=\sqrt{\sin x+\cos x}?$$

Answered by TANMAY PANACEA. last updated on 28/Mar/20

$$ithink...$$$${\int }_0^{\frac{\pi }{2}}\frac{dx}{1+\frac{\sqrt{sinx}}{\sqrt{cosx}}}={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{cosx}}{\sqrt{sinx}+\sqrt{cosx}}dx$$$$using{\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$$$$I={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{cos(\frac{\pi }{2}-x)}}{\sqrt{sin(\frac{\pi }{2}-x)}+\sqrt{cos(\frac{\pi }{2}-x)}}dx$$$$={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{sinx}}{\sqrt{cosx}+\sqrt{sinx}}dx$$$$2I={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{cosx}}{\sqrt{sinx}+\sqrt{cosx}}+\frac{\sqrt{sinx}}{\sqrt{cosx}+\sqrt{sinx}}dx$$$$2I={\int }_0^{\frac{\pi }{2}}dx$$$$I=\frac{\pi }{2}\times \frac{1}{2}=\frac{\pi }{4}$$

Commented by jagoll last updated on 28/Mar/20

$$\mathrm{king}\mathrm{formula}$$

Commented by TANMAY PANACEA. last updated on 28/Mar/20

$$thankyousir$$

Commented by john santu last updated on 28/Mar/20

$$\mathrm{it}\mathrm{does}\mathrm{mean}\mathrm{the}\mathrm{equation}\mathrm{wrong}\mathrm{sir}?$$

Commented by TawaTawa1 last updated on 28/Mar/20

$$\mathrm{Sir}\mathrm{tanmay},\mathrm{please}\mathrm{help}\mathrm{with}\mathrm{Q}86324.$$

Answered by MJS last updated on 28/Mar/20

$$\int \frac{dx}{\sqrt{1+\tan x}}=$$$$[t=\sqrt{1+\tan x}\to dx={2\cos }^{2}x\sqrt{1+\tan x}dt]$$$$=2\int \frac{dt}{t^4-2t^2+2}$$$$\mathrm{and}\mathrm{now}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}$$

Commented by jagoll last updated on 28/Mar/20

$$=2\int \frac{\mathrm{dt}}{{({\mathrm{t}}^2-1)}^2+1}$$$$\mathrm{let}{\mathrm{t}}^2-1=\tan \mathrm{w}\Rightarrow 2\mathrm{t}\mathrm{dt}=\sec {}^2\mathrm{w}\mathrm{dw}$$$$=\int \frac{\sec {}^2\mathrm{w}\mathrm{dw}}{\sec {}^2\mathrm{w}}=\mathrm{w}+\mathrm{c}$$$$={\tan }^{-1}({\mathrm{t}}^2-1)+\mathrm{c}$$$$={\tan }^{-1}\left(\tan \left(\mathrm{x}\right)\right)+\mathrm{c}$$$$\mathrm{it}\mathrm{correct}\mathrm{sir}?$$

Commented by MJS last updated on 28/Mar/20

$$\mathrm{no}$$$$w=\arctan (t^2-1)\to dt=\frac{t^4-2t^3+2}{2t}dw$$$$\Rightarrow \int \frac{dw}{\sqrt{1+\tan w}}\mathrm{so}{\mathrm{you}}^{\prime }\mathrm{re}\mathrm{back}\mathrm{at}\mathrm{the}\mathrm{start}$$

Commented by MJS last updated on 28/Mar/20

$$\mathrm{we}\mathrm{must}\mathrm{decompose}$$$$2\int \frac{dt}{t^4-2t^2+2}=$$$$=\int \frac{at+b}{t^2-\sqrt{2+2\sqrt{2}}t+\sqrt{2}}dt+\int \frac{ct+d}{t^2+\sqrt{2+2\sqrt{2}t+\sqrt{2}}}dt=$$$$=-\frac{\sqrt{-1+\sqrt{2}}}{2}\int \frac{t-\sqrt{2+2\sqrt{2}}}{t^2-\sqrt{2+2\sqrt{2}}t+\sqrt{2}}dt+\frac{\sqrt{-1+\sqrt{2}}}{2}\int \frac{t+\sqrt{2+2\sqrt{2}}}{t^2+\sqrt{2+\sqrt{2}}t+\sqrt{2}}dt$$$$\mathrm{now}\mathrm{use}\mathrm{formula}$$

Commented by jagoll last updated on 28/Mar/20

$$\mathrm{waw}...\mathrm{yes}\mathrm{sir}.\mathrm{i}\mathrm{try}\mathrm{solve}\mathrm{it}$$

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