Q.Evaluate $\int(3x^2+4x+5)\,dx$.

$\int\frac{\sin 2x}{2\cos x}\,dx=\int\sin x\,dx=-\cos x+c$.

Using the identity $\sin 2x=2\sin x\cos x$,

$\int\frac{\sin 2x}{2\cos x}\,dx=\int\frac{2\sin x\cos x}{2\cos x}\,dx=\int\sin x\,dx=-\cos x+c.$

Simplify the integrand with the double-angle identity $\sin 2x=2\sin x\cos x$; it collapses to $\int\cos x\,dx=\sin x+c$.

Start with the given integral:

$\int\frac{\sin 2x}{2\sin x}\,dx.$

Apply $\sin 2x=2\sin x\cos x$:

$\int\frac{2\sin x\cos x}{2\sin x}\,dx=\int\cos x\,dx.$

Integrate the standard result:

$\int\cos x\,dx=\sin x+c.$

Convert the surd to an exponent, $\sqrt{e^x}=e^{x/2}$, then use $\int e^{ax}dx=\frac{1}{a}e^{ax}$ with $a=\tfrac12$.

Rewrite the integrand:

$\int\sqrt{e^x}\,dx=\int e^{x/2}\,dx.$

Using $\displaystyle\int e^{ax}\,dx=\frac{1}{a}e^{ax}+c$ with $a=\tfrac12$:

$\int e^{x/2}\,dx=\frac{1}{1/2}\,e^{x/2}+c=2e^{x/2}+c=2\sqrt{e^x}+c.$