Question

The value of the integral $\int \cos \sqrt{x}dx=$

• A. $2[\sqrt{x}\sin \sqrt{x}+\cos \sqrt{x}]+c$
• B. $2[\sqrt{x}\sin \sqrt{x}-\cos \sqrt{x}]+c$
• C. $2[\cos \sqrt{x}-\sqrt{x}\sin \sqrt{x}]+c$
• D. $-2[\sqrt{x}\sin \sqrt{x}+\cos \sqrt{x}]+c$

Answer 1

The correct option is A $2[\sqrt{x}\sin \sqrt{x}+\cos \sqrt{x}]+c$

Put $\sqrt{x}=t$
$\Rightarrow \frac{1}{2\sqrt{x}}dx=dt\Rightarrow dx=2tdt,$
It then reduces to
$\int 2t\cdot \cos tdt=2[t\cdot \sin t-\int \sin tdt]$
$=2t\sin t+2\cos t+c=2[\sqrt{x}\sin \sqrt{x}+\cos \sqrt{x}]+c$