Question

integration of [1+2tanx(tanx+secx)]^1/2 dx

Answer 1

Hope this help.............

Answer 2

$\int\sqrt{1+2tanx(tanx+secx)} = - log cosx +log (secx+tanx) + C$

Step-by-step explanation:

Given:

We have,

$\int\sqrt{1+2tanx(tanx+secx)}$

= $\int\sqrt{1+2tan^2x+2tanxsecx}dx$

= $\int \sqrt{1 +2\frac{sin^2x}{cos^2x}+2 \times\frac{sinx}{cosx} \times \frac{1}{cosx}} dx$       $[tanx=\frac{sinx}{cosx}, secx=\frac{1}{cosx} ]$

= $\int \sqrt{1 +2\frac{sin^2x}{cos^2x}+2 \frac{sinx}{cos^2x}} dx$    

= $\int \sqrt{\frac{cos^2x+sin^2x+sin^2x+2sinx}{cos^2x} }dx$

Using the trigonometric identity $cos^2x+sin^2x=1$

= $\int \sqrt{\frac{1+sin^2x+2sinx}{cos^2x} }dx$

= $\int \sqrt{\frac{sinx+1}{cos^2x} }dx$      $[(sinx+1)^2=1+sin^2x+2sinx]$

= $\int(\frac{sinx+1}{cosx})dx$

= $\int \frac{sinx}{cosx}+ \frac{1}{cosx} dx$

= $\int(tanx+secx)dx$

= - log cosx +log (secx+tanx) + C

Therefore,

$\int\sqrt{1+2tanx(tanx+secx)} = - log cosx +log (secx+tanx) + C$