Question

Evaluate $\int \frac{1}{\sqrt{{\csc }^{2}x+{\cot }^{2}x}}dx$.

• A. $-\log (t+\sqrt{1+t^2})$ where $t=\cos x$
• B. $\log (t+\sqrt{1+t^2})$ eher $t=\cos x$
• C. ${\tan }^{-1}\left(\cos x\right)+c$
• D. $Noneofthese$

Answer 1

The correct option is A $-\log (t+\sqrt{1+t^2})$ where $t=\cos x$

$\int \frac{dx}{\sqrt{cose{c}^2+co{t}^{2}x}}$

$=\int \frac{dx}{\sqrt{\frac{1}{si{n}^{2}x}+\frac{co{s}^{2}x}{si{n}^{2}x}}}$

$=\int \frac{sinx.dx}{\sqrt{1+co{s}^{2}x}}$

Put cosx = t

Differentiating.

-sinx.dx = dt

$=\int \frac{-dt}{\sqrt{1+t^2}}$ $\therefore \int \frac{1}{\sqrt{x^2+a^2}}=log(x+\sqrt{x^2+a^2})+c$

$=-log(t+\sqrt{1+t^2})+c$

where t = cosx