Question

Qustion no 61
61.$\int$ sec³x dx

Answer 1

$$\begin{gathered} I=\int se{c}^{3}xdx \\ =\int \frac{se{c}^{2}x\left(secx\tan x\right)}{\tan x}dx \\ =\int \frac{se{c}^{2}x\left(secx\tan x\right)}{\sqrt{se{c}^{2}x-1}}dx \\ Letsecx=t \\ dt=secx\tan xdx \\ I=\int \frac{t^2}{\sqrt{t^2-1}}dt=\int \frac{t^2-1+1}{\sqrt{t^2-1}}dt=\int \frac{t^2-1}{\sqrt{t^2-1}}dt+\int \frac{1}{\sqrt{t^2-1}}dt=\int \sqrt{t^2-1}dt+\int \frac{1}{\sqrt{t^2-1}}dt \\ \int \sqrt{x^2-a^2}dx=\frac{1}{2}\left\{x\sqrt{x^2-a^2}-a^2\ln \left|x+\sqrt{x^2-a^2}\right|\right\}+cand\int \frac{1}{\sqrt{x^2-a^2}}dx=\ln \left|x+\sqrt{x^2-a^2}\right|+c \\ I=\frac{1}{2}\left\{t\sqrt{t^2-1}-\ln \left|t+\sqrt{t^2-1}\right|\right\}+\ln \left|t+\sqrt{t^2-1}\right|+c \\ I=\frac{t}{2}\sqrt{t^2-1}+\frac{1}{2}\ln \left|t+\sqrt{t^2-1}\right|+c \\ I=\frac{secx}{2}\sqrt{se{c}^{2}x-1}+\frac{1}{2}\ln \left|secx+\sqrt{se{c}^{2}x-1}\right|+c \end{gathered}$$