Help with two integrals I need help calculating two integrals
1)
$$\int_1^2 \sqrt{4+ \frac{1}{x}}\mathrm{d}x$$
2)
$$\int_0^{\frac{\pi}{2}}x^n sin(x)\mathrm{d}x$$
So I think on the 1st one I will have to use substitution, but I don't know what to do to get something similar to something that's in the known basic integrals.
2nd one I think is per partes but the $n$ is confusing.
Any help would be appreciated.
Thank you in advance.
(If there happens to be any duplicates I am apologizing also, but didn't find any.)
 A: Hint:
For the first problem:
$$\int \sqrt{4+ \frac{1}{x}}\mathrm{d}x = \int \frac{\sqrt{4x+ 1 }}{\sqrt x}\mathrm{d}x$$
Let $u = \sqrt x$ , $x = u^2$, $2u du =dx $
$$ \int \frac{\sqrt{4u^2+ 1 }}u 2u \mathrm{d}u$$
Try some trigonometric substitution.
For the second problem:
Set $$I(n) = \int x^n sin(x)\mathrm{d}x$$
Let $u = x^{n},dv = \sin x$,$du = n x^{n-1}, v = -\cos x$  by integration by parts.
$$I(n) = x^n \times \cos x + n \int  x^{n-1} \cos x\mathrm{d}x  $$
Let $u = x^{n-1},dv = \cos x$,$du = (n-1) x^{n-2}, v = \sin x$  by integration by parts
$$I(n) = x^n \times \cos x + n (x^{n-1} \sin x - (n-1)\int x^{n-2} \sin x dx )  = x^n \times \cos x + n x^{n-1} \sin x - n (n-1)I(n-2) $$
Then use mathematic induction you will get the formula.
A: You can indeed work out the second integral by parts.
$$I_n:=\int_0^{\pi/2}x^n\sin x\,dx=-\left.x^n\cos x\right|_0^{\pi/2}+n\int_0^{\pi/2}x^{n-1}\cos x\,dx.$$
Repeat with the cosine integral,
$$J_n:=\int_0^{\pi/2}x^n\cos x\,dx=\left.x^n\sin x\right|_0^{\pi/2}-n\int_0^{\pi/2}x^{n-1}\sin x\,dx.$$
This gives you the recurrence relations
$$I_n=nJ_{n-1},\\J_n=\left(\frac\pi2\right)^n-nI_{n-1},$$
so that
$$I_n=n\left(\left(\frac\pi2\right)^{n-1}-(n-1)I_{n-2}\right).$$
When you decrease $n$, you will eventually reach $n=1$ or $n=0$, you need to explictly compute $I_1$ and $I_0$. It is an easy matter to establish
$$I_0=J_0=1,$$ then $$I_1=1.$$

To solve the recurrence, you can rewrite
$$\frac{I_n}{n!}=\frac1{(n-1)!}\left(\frac\pi2\right)^{n-1}-\frac{I_{n-2}}{(n-2)!}.$$
Then
$$\frac{I_{n-2}}{(n-2)!}=\frac1{(n-3)!}\left(\frac\pi2\right)^{n-3}-\frac{I_{n-4}}{(n-4)!}$$ and summming,
$$\frac{I_n}{n!}=\frac1{(n-1)!}\left(\frac\pi2\right)^{n-1}-\frac1{(n-3)!}\left(\frac\pi2\right)^{n-3}+\frac{I_{n-4}}{(n-4)!}$$ and finally
$$\frac{I_n}{n!}=\frac{I_{m}}{m!}+\sum_{k=m}^{n}\left(\frac1{(k-1)!}\left(\frac\pi2\right)^{k-1}-\frac1{(k-3)!}\left(\frac\pi2\right)^{k-3}\right)$$
where $m=n\bmod4$ and the summation must be performed with a step of $4$. You will recognize the truncated development of $\sin\pi/2$ or $\cos\pi/2$, depending on the parity of $m$.
A: HINT...for the first one, try substituting $$\frac 1x=4\tan^2\theta$$
For the second, try integrating by parts twice to obtain a reduction formula
A: Just as an interesting fact,  the general antiderivative  for the second integral can be written as 
$$\int x^n \sin(x)  \,  \mathrm{d}x = \frac 12 i x^{n+1} (\operatorname{E_{-n}} (-i x)-\operatorname{E_{-n}} (i x))+C$$
This can be proven directly through substitutions. Write the answer in terms of integrals,  combine, and substitute to get back to the original integral
