Chủ đề này có 1 trả lời

[tritanngo99]

Tính: $I=\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{cot(x)}{1+sin^9(x)}dx$

[L Lawliet]

Tính: $I=\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{cot(x)}{1+sin^9(x)}dx$

Lời giải.

$$I=\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\dfrac{\cot x}{1+\sin ^{9}x}dx=\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\dfrac{\cos x}{1+\sin ^{9}x}dx$$

Đặt $\sin x=t\Rightarrow \cos xdx=dt$.

\begin{align*} I &=\int_{\frac{1}{2}}^{1}\dfrac{1}{t+t^{10}}dt \\ &=\int_{\frac{1}{2}}^{1}\dfrac{\left ( 1+t^{9} \right )-t^{9}}{t\left ( 1+t^{9} \right )}dt \\ &=\int_{\frac{1}{2}}^{1}\dfrac{1}{t}dt-\int_{\frac{1}{2}}^{1}\dfrac{t^{8}}{1+t^{9}}dt \\ &=\ln \left | t \right |\bigg|_{\frac{1}{2}}^{1}-\dfrac{1}{9}\ln \left | 1+t^{9} \right |\bigg|_{\frac{1}{2}}^{1} \\ \end{align*}