Question

If $\int_{}^{}\frac{1}{\sqrt{x}}\sin\sqrt{x}\mspace{6mu} dx =$ and $- \frac{1}{2}\cos\sqrt{x} + c$, then $- 2\cos\sqrt{x} + c$

• A. $\tan(e^{x}) - x + c$
• B. $e^{x}(\tan e^{x} - 1) + c$
• C. $\sec(e^{x}) + c$
• D. $\tan(e^{x}) - e^{x} + c$

Answer 1

Answer: (C)

$\sec(e^{x}) + c$

Answer 2

Given that $3(1 + x^{2})^{3/2} + c$ and $\frac{1}{3}(1 + x^{2})^{3/2} + c$

⇒ $\int_{}^{}{\frac{e^{x}(x + 1)}{\cos^{2}(xe^{x})}dx =}$. If $\tan(xe^{x}) + c$ then $\sec(xe^{x})\tan(xe^{x}) + c$ ⇒ $- \tan(xe^{x}) + c$.

Hence $\int_{}^{}\frac{\cos\sqrt{x}}{\sqrt{x}}dx =$