IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

ramkumar_november

3 problem

I = $\int_0^{\frac{\pi}{2}}\frac{\sqrt{sinx}.dx}{\sqrt{sinx}+\sqrt{cosx}}$ ..........(1)

apply property 4 we get

I = $\int_0^{\frac{\pi}{2}}\frac{\sqrt{cosx}.dx}{\sqrt{sinx}+\sqrt{cosx}}$ ...........(2)

adding (1) and (2) we get

2I = $\int_0^{\frac{\pi}{2}}\frac{\sqrt{cosx}+\sqrt{sinx}.dx}{\sqrt{sinx}+\sqrt{cosx}}$

2I = $\int_0^{\frac{\pi}{2}}dx$

I = $\frac{\pi}{4}$

4th problem is wrong the limits should be from 0 to 5........

I = $\int_0^5 \frac{\sqrt{x}.dx}{\sqrt{5-x}+\sqrt{x}}$ ....(1)

apply property 4 we get

I = $\int_0^5 \frac{\sqrt{5-x}.dx}{\sqrt{5-x}+\sqrt{x}}$ .............(2)

adding (1) and (2)..

I = 5/2

5th problem

I = $\int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\frac{\pi+4x^3}{2-cos(|x|+\frac{\pi}{3})}$

I = $\int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\frac{\pi}{2-cos(|x|+\frac{\pi}{3})}+\int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\frac{4x^3}{2-cos(|x|+\frac{\pi}{3})}$

the second integral becomes zero since it is an even function

I = $\int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\frac{\pi}{2-cos(|x|+\frac{\pi}{3})}$

which you can do easily......

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