Question

\( \int \frac{d z}{\left[a^{2}+(a-z)^{2}\right]3 / 2} \)

\(\int\frac{dz}{[a^2+(a-z)^2]^{3/2}}\)

Answer 1

integration dt|{a^2+(a-z)^2}^3/2 

let a-z=t ,-dz=dt 

integration -dt/(a^2+t^2)^3/2

integration -dt/t^3(a^2/t^2+1)^3/2

let a^2/t^2 =k , -2a^2)t^3 ×dt =dk 

dt/t^3 =dk/ -2a^2 

1/2a^2 integration dk/(k+1)^3/2

1/2a^2 { (k+1)^-3/2+1/-3/2+1}

1/2a^2 {(k+1)^-1/2 /-1/2 } +C 

1/2a^2 ×-2{1/(k+1)^1/2 } +C 

-1/a^2 {1/{a^2/(a-z)^2+1}1/2}+C 

-1/a^2{1/{a^2+(a-z)^2/(a-z)^2}1/2}

-1/a^2{ a-z/(a^2+(a-z)^2}1/2 }  +C answr 

Answer 2

Let I = \(\int\frac{dz}{[a^2+(a-z)^2]^{3/2}}\)

Put a - z = a tan θ

⇒ dz = -asec² θ dθ

\(\therefore\) I = \(-\int\frac{asec^2\theta d\theta}{(a^2+a^2tan^2\theta)^{3/2}}\) 

 = \(-\int\frac{asec^2\theta d\theta}{a^3(1+tan^2\theta)^{3/2}}\) 

 = \(-\frac1{a^2}\int\frac{sec^2\theta d\theta}{sec^3\theta}\)

( ∵ 1 + tan² \(\theta\) = sec² \(\theta\)) 

= -\(\frac1{a^2}\int\frac1{sec\theta}d\theta\)

= - \(\frac1{a^2}\int cos\theta d\theta\) 

= \(-\frac{sin\theta}{a^2}\) ( ∵ \(\int cos\theta=sin\theta\))

 = \(-\frac1{a^2}\frac{tan\theta}{\sqrt{1+tan^2\theta}}\) (∵ sin \(\theta\) = \(\frac{tan\theta}{\sqrt{1+tan^2\theta}}\))

 =  \(-\frac1{a^2}\cfrac{\frac{a-z}a}{\sqrt{1+\frac{(a-z)^2}{a^2}}}\) \((\because tan\theta=\frac{a-z}a)\)

\(=-\frac1{a^2}\frac{a-z}{\sqrt{a^2+(a-z)^2}}\)

Hence, \(\int\frac{dz}{[a^2+(a-z)^2]^{3/2}}\) = \(-\frac1{a^2}\frac{a-z}{\sqrt{a^2+(a-z)^2}}\)