Question

If $\int \sqrt{1+\cos x}f\left(x\right)dx=\frac{2}{3}(1+\cos x{)}^{3/2}+c,$ then $f\left(x\right)$ equals

• A. $\cos x$
• B. $-\sin x$
• C. $\tan x$
• D. $1$

Answer 1

The correct option is B $-\sin x$

$\int \sqrt{1+\cos x}f\left(x\right)dx=\frac{2}{3}(1+\cos x{)}^{3/2}+c$

Differentiating both sides w.r.t. $x,$

$\Rightarrow$ $\frac{d}{dx}(\int \sqrt{1+\cos x}f(x\left)dx\right)=\frac{d}{dx}\left(\frac{2}{3}\right(1+\cos x{)}^{3/2}+c)$

$\Rightarrow$ $\sqrt{1+\cos x}f\left(x\right)=\frac{2}{3}\times \frac{3}{2}(1+\cos x{)}^{1/2}(-\sin x)+0$

$\Rightarrow$ $\sqrt{1+\cos x}f\left(x\right)=-\sin x\sqrt{1+\cos x}$

$\therefore$ $f\left(x\right)=-\sin x$