——— 

Example2: Find ∫ \/ 4 + x²  dx

Let : x = 2 sinshθ Then dx = 2 coshθ dθ   OR  dx = 2 coshθ dθ , x = 2 sinshθ .... (1)

note: θ = sinsh^–1(x/2) .... (2)

4 + x² = 4 + 4 sinsh²θ = 4(1+sinsh²θ) = 4 cosh²θ

   ——— 

\/ 4 + x²  = 2 coshθ   ..... (3)

   ——— 

\/ 4 + x²   dx = 2coshθ.2coshθ = 4 cosh²θ = 4× 0.5(1+cosh2θ) = 2(1+cosh2θ) = 2 + 2 cosh2θ

     ——— 

∫ \/ 4 + x²  dx = ∫(2 + 2 cosh2θ) .dθ

                   = 2 θ + sinsh2θ + c   Such as: sinsh2θ = 2 sinshθ coshθ Then

                   = 2 θ + 2 sinshθ coshθ + c 

From (1), (2), (3)

                                                                ———

                                                          1         

                   = 2 sinsh^–1(x/2) + x — \/4 + x²  + c 

                                                          2

                                                                ———

                                                      1         

                   = 2 sinsh^–1(x/2) + — x \/4 + x²  + c 

                                                      2