k(x)= $$\int\limits_{\cos \alpha}^{\sin ^2x^{16}} 1/(3r+r^2)$$ and alpha is constant. Use the fundamental theorem of calculus to find dk/dx

Question

k(x)= $$\int\limits_{\cos \alpha}^{\sin ^2x^{16}} 1/(3r+r^2)$$     and  alpha is constant. Use the fundamental theorem of calculus to find dk/dx

anonymous · Answer

$$\frac{1}{3\sin^2x^{16}+\sin^4x^{16}} - \frac{1}{3\cos\alpha + \cos^2\alpha}$$Probably.

anonymous · Answer

no its not the one, i dont have the option in this multiple choice question

anonymous · Answer

no its not the one, i dont have this option in this multiple choice question

turingtest · Answer

Ishaan forgot to square the other x in the denom on the first term
should be x^32
is that a choice ?

turingtest · Answer

... or how about is this integral even with respect to dr ?
or something else?

anonymous · Answer

$$\frac{2\sin x^{16}\cos x^{16} \times 16x^{15}}{3\sin^2x^{16} + \sin^4x^{16}}$$Sorry, for the earlier mistake.

anonymous · Answer

Which can be simplified to
$$\frac{32 x^{15} \cot
   \left(x^{16}\right)}{\sin
   ^2\left(x^{16}\right)+3}
$$