Evaluate the following indefinite integrals :
`(i) intsqrt(a^(2)-x^(2))dx` `(ii) int(1)/(sqrt(49+x^(2)))dx`
`(iii) int(1)/(xsqrt(x^(6)-1))dx` `(iv) intsqrt((1+x)/(1-x))dx`
`(v) intsqrt((x^(2009))/(2x^(2008)-x^(2009)))dx` `(vi) int(1)/(sqrt((x-4)(x-5)))dx`

Explanation

`(i)` Let us put `x=asintheta` so that `dx=acostheta d theta` and `a^(2)-x^(2)=a^(2)(1-sin^(2)theta)=a^(2)cos^(2)theta`
Thus `intsqrt(a^(2)-x^(2))dx=intacostheta*acostheta d theta=(a^(2))/(2)int(1+cos2theta)d theta`
`=(a^(2))/(2)[theta+(sin2theta)/(2)]+C=(a^(2))/(2)[sin^(-1).(x)/(a)+(x)/(a)*(sqrt(a^(2)-x^(2)))/(a)]+C`
`=(a^(2))/(2)[sin^(-1).(x)/(a)]+(1)/(2)[xsqrt(a^(2)-x^(2))]+C`, where `C` is a constant of integration.
`(ii)` Let us put `x=7tantheta` so that
`dx=7sec^(2)theta d theta` and `49+x^(2)=49sec^(2)theta`.
Thus `int(1)/(sqrt(49+x^(2)))dx=int(1)/(7sectheta)*7sec^(2)theta d theta=intsectheta d theta`
`=int(sectheta(sectheta+tantheta))/(sectheta+tantheta)d theta=int(d(sectheta+tan theta))/(sectheta+tantheta)`
`=ln|(sectheta+tantheta)|+C=ln|(x)/(7)+(sqrt(49+x^(2)))/(7)|+C`
`=ln|x+sqrt(49+x^(2))|+C-ln7`
`=ln|x+sqrt(49+x^(2))|+K`, `K` being a constant.
`(iii)` Let us put `x^(3)=sectheta`
`implies3x^(2)dx=secthetatanthetad theta`
and `x^(6)-1=sec^(2)theta-1=tan^(2)theta`
Thus `int(1)/(xsqrt(x^(6)-1))dx=(1)/(3)int(3x^(2))/(x^(3)sqrt((x^(3))^(2)-1))dx`
`=(1)/(3)int(secthetatantheta)/(secthetatantheta)d theta`
`=(1)/(3)theta+C=(1)/(3)sec^(-1)(x^(3))+C`
`(iv)` Let us substitute `x=cos2theta` so that
`(1+x)/(1-x)=(1+cos2theta)/(1-cos2theta)=(2cos^(2)theta)/(2sin^(2)theta)=(cos^(2)theta)/(sin^(2)theta)`
and `dx=-2sin2thetad theta=-4sinthetacostheta d theta`
under above substitution the given integral reduces to
`intsqrt((1+x)/(1-x))dx=int(costheta)/(sintheta)xx-4sinthetacosthetad theta`
`=-2int(1+cos2theta)d theta=-2[theta+(sin2theta)/(2)]+C`
`=-[cos^(-1)x+sqrt(1-x^(2))]+C`
`(v)` We have,
`intsqrt((x^(2009))/(2x^(2008)-x^(2009)))dx=intsqrt((x)/(2-x))dx`
Let us substitute `x=2sin^(2)theta`
`impliesdx=4sinthetacosthetad theta` and `2-x=2-2sin^(2)theta=2cos^(2)theta`
`:.intsqrt((x^(2009))/(2x^(2008)-x^(2009)))dx=intsqrt((x)/(2-x))dx`
`=4int(sintheta)/(costheta)*sinthetacostheta d theta=2int(1-cos2theta)d theta`
`=2[theta-(sin2theta)/(2)]+C=2[sin^(-1).sqrt((x)/(2))-sqrt((x)/(2))sqrt((2-x)/(2))]+C`
`=2sin^(-1).sqrt((x)/(2)-sqrt(2x-x^(2))+C`, `C` is a constant of integration.
`(vi)` Let us substitute `x-4=t^(2)`
so that `dx=2tdt`
and `x-5=t^(2)-1`
Thus `int(1)/(sqrt((x-4)(x-5)))dx=int(2t)/(sqrt(t^(2)(t^(2)-1)))dt=2int(1)/(sqrt(t^(2)-1))dt`
`=2ln|t+sqrt(t^(2)-1)|+C`
`=2ln|sqrt(x-4)+sqrt(x-5)|+C`, where `C` is a constant of integration.