Answer to Question #139982 in Calculus for Moel Tariburu

Question

Evaluate the integral using a rational function ∫▒(cos x)/(√sin+ ∛(sin⁡x )) dx

Answer 1

$\displaystyle \textsf{Let}\, I =\int \frac{\cos{x}}{\sqrt{\sin{x}} + \sqrt[3]{\sin{x}}}\, \mathrm{d}x\\ \textsf{Substitute}\, u = \sin{x} \\ \begin{aligned} I &=\int \frac{\mathrm{d}(\sin{x})}{\sqrt{\sin{x}} + \sqrt[3]{\sin{x}}} \\ &=\int \frac{\mathrm{d}u}{\sqrt{u} + \sqrt[3]{u}} \end{aligned}\\ \textsf{Substitute}\, u = v^6 \\ \begin{aligned} I &=\int \frac{6v^5}{v^3 + v^2} \, \mathrm{d}v\\ &=\int \frac{6v^3}{v + 1} \, \mathrm{d}v\\ &=\int \left(6v^2 - 6v + 6 - \frac{6}{v + 1}\right) \mathrm{d}v\\ &= 2v^3 - 3v^2 + 6v - 6\ln(v + 1) + C\\ &= 2\sqrt{u} - 3\sqrt[3]{u} + 6\sqrt[6]{u} - 6\ln(\sqrt[6]{u} + 1) + C\\ &= 2\sqrt{\sin{x}} - 3\sqrt[3]{\sin{x}} + 6\sqrt[6]{\sin{x}} - 6\ln(\sqrt[6]{\sin{x}} + 1) + C \end{aligned}\\ \therefore \int \frac{\cos{x}}{\sqrt{\sin{x}} + \sqrt[3]{\sin{x}}}\, \mathrm{d}x \\ = 2\sqrt{\sin{x}} - 3\sqrt[3]{\sin{x}} + 6\sqrt[6]{\sin{x}} - 6\ln(\sqrt[6]{\sin{x}} + 1) + C$