Question

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

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

Answer 1

Answer: (A)

$\cos x$

Given $\int \sqrt{1+\sin x}f\left(x\right)dx=\frac{2}{3}(1+\sin x{)}^{\frac{3}{2}}+c$
Differentiating both sides w.r.t. x, we get
$\sqrt{1+\sin x}f\left(x\right)=\frac{2}{3}\frac{3}{2}(1+\sin x{)}^{\frac{1}{2}}\cos x$
or, $\sqrt{1+\sin x}f\left(x\right)=\sqrt{1+\sin x}\cos x$
$\Rightarrow f\left(x\right)=\cos x$