Question

If $\int \frac{x\cos \alpha +1}{(x^2+2x\cos \alpha +1{)}^{3/2}}dx=\frac{f\left(x\right)}{\sqrt{g\left(x\right)+1}}+C$, then $g\left(x\right)-2\cos \alpha f\left(x\right)$ is
(where $C$ is constant of integration)

• A. $x^2$
• B. $x$
• C. $2x^2$
• D. $2x$

Answer 1

The correct option is A $x^2$

Let $I=\int \frac{x\cos \alpha +1}{(x^2+2x\cos \alpha +1{)}^{3/2}}dx$
Now,
$\Rightarrow I=\int \frac{x\cos \alpha +1}{x^3{(1+\frac{2}{x}\cos \alpha +\frac{1}{x^2})}^{3/2}}dx\Rightarrow I=\int \frac{\frac{\cos \alpha }{x^2}+\frac{1}{x^3}}{{(1+\frac{2}{x}\cos \alpha +\frac{1}{x^2})}^{3/2}}dx$

Taking $1+\frac{2}{x}\cos \alpha +\frac{1}{x^2}=t^2$
$\Rightarrow -\frac{2\cos \alpha }{x^2}-\frac{2}{x^3}=2tdt\Rightarrow \frac{\cos \alpha }{x^2}+\frac{1}{x^3}=-tdt$

Now,
$I=-\int \frac{t}{t^3}dt\Rightarrow I=-\int \frac{1}{t^2}dt\Rightarrow I=\frac{1}{t}+C\Rightarrow I=\frac{1}{\sqrt{1+\frac{2}{x}\cos \alpha +\frac{1}{x^2}}}+C\Rightarrow I=\frac{x}{\sqrt{x^2+2x\cos \alpha +1}}+C\Rightarrow f\left(x\right)=x,g\left(x\right)=x^2+2x\cos \alpha \therefore g\left(x\right)-2\cos \alpha f\left(x\right)=x^2$