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Question Number 54248 by rahul 19 last updated on 01/Feb/19

Commented by rahul 19 last updated on 01/Feb/19

$$Ans.for3)\to 2{\pi }^2.$$$$for2)\to putx=\tan \theta andthenuse$$$$thepropertyofreplacingxbya+b-x.$$$$henceans={\tan }^{-1}2.ln\sqrt{5}.$$

Answered by rahul 19 last updated on 01/Feb/19

Commented by Meritguide1234 last updated on 01/Feb/19

$$\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{name}\mathrm{of}\mathrm{book}?$$

Commented by rahul 19 last updated on 01/Feb/19

grb problems in calculus for JEE.

Answered by Prithwish sen last updated on 01/Feb/19

$$1){\int }_0^1\frac{\mathrm{dx}}{\sqrt{1+\mathrm{x}}+\sqrt{1-\mathrm{x}}+2}$$$$\mathrm{Put},\mathrm{x}=\sin 2\theta \Rightarrow \mathrm{dx}=2\cos 2\theta$$$$0\ll \mathrm{x}\ll 1\Rightarrow 0\ll \theta \ll \frac{\pi }{4}$$$${\int }_0^{\int \frac{\pi }{4}}\frac{2\cos 2\theta \mathrm{d}\theta }{2\sin \theta +2}$$$$={\int }_0^{\frac{\pi }{4}}\frac{\cos 2\theta (1-\sin \theta )\mathrm{d}\theta }{{\cos }^2\theta }$$$$={\int }_0^{\frac{\pi }{4}}\frac{({2\cos }^2\theta -1)(1-\sin \theta )\mathrm{d}\theta }{{\cos }^2\theta }$$$$=2\theta +2\cos \theta -\tan \theta +\sec \theta {\mid }_0^{\frac{\pi }{4}}$$$$=(\frac{\pi }{2}+\sqrt{2}-1+\sqrt{2})-(2-1)$$$$=\frac{\pi }{2}+2\sqrt{2}-2$$$$$$

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