Evaluating $\int_{-5}^{\sqrt{x}}(\frac{\cos t}{t^{10}})dt$ 
Evaluate $y=\int_{-5}^{\sqrt{x}}(\frac{\cos t}{t^{10}})dt$


I've tried differentiating both sides of the fraction until the denominator was 1, and then integrating that by parts, but this was marked wrong. I know I can't just integrate by parts right off the bat because differentiating $t^{-b}$ will always end in another varied reciprocal, it'll just make things worse and worse.
How can I solve this?

First note that this integral is improper because $-5 < 0<\sqrt x.$ Split off a bad piece near 0 by writing $c=\min(5,\sqrt x,\pi/3)$ so $$\int_{-5}^\sqrt x \frac{\cos t}{t^{10}}dt=\int_{-5}^{-c} \frac{\cos t}{t^{10}}dt+\int_{-c}^c \frac{\cos t}{t^{10}}dt+\int_c^\sqrt x \frac{\cos t}{t^{10}}dt$$. Of the three integrals on the right side of the equation above, the first and third can be expressed in terms of the special functions Si and Ci, called sine-integral and cosine-integral but we shall see that there is no need to do so. For now, note that in both the first and third integrals, we have a bounded function integrated over a finite interval, so both the first and third integrals are definite real numbers. Now look at the second of the three inegrals on the  rirght side. That inegral is improper with a discontinuity at 0. By definition, $$\int_{-c}^c \frac{\cos t}{t^{10}}dt=\lim_{u,v \to 0}(\int_{-c}^u \frac{\cos t}{t^{10}}dt+\int_v^c \frac{\cos t}{t^{10}}dt)$$ where $u<0,v>0.$ The function $\frac{\cos t}{t^{10}}$ is even so it is enough to look at the integral $$\int_v^c \frac{\cos t}{t^{10}}dt).$$ (Alternatively, it will turn out that all we have to do is note that $\frac{\cos t}{t^{10}}$ is positive on $[-c,u].)$ On the interval $[v,c], \cos t \ge 1/2 $ so $$\int_v^c \frac{\cos t}{t^{10}}dt \ge \frac{1}{2}\int_v^c \frac{1}{t^{10}}dt$$ $$\lim_{v \to 0}\int_v^c\frac{1}{t^{10}}dt=\lim_{v \to 0}\frac{1}{9}(\frac{1}{v^9}-\frac{1}{c^9})=\infty.$$ Thus $$\int_v^c \frac{\cos t}{t^{10}}dt=\infty.$$ We conclude that 
$$\int_{-5}^\sqrt x \frac{\cos t}{t^{10}}dt=\infty.$$
