Помогите с решением интегралов,срочно

Решите задачу:

$\displaystyle \int sin^3x\sqrt[3]{cosx}dx=-3\int sin^2x*t^3dt=-3\int(1-t^6)t^3dt=\\-3\int(t^3-t^9)dt=-\frac{3t^4}{4}+\frac{3t^{10}}{10}+C=-\frac{3\sqrt[3]{cos^4x}}{4}+\frac{3\sqrt[3]{cos^{10}x}}{10}+C\\\\\\cosx=t^3\\-sinxdx=3t^2dt=\ \textgreater \ dx=-\frac{3t^2dt}{sinx}$

$\displaystyle \int\frac{dx}{\sqrt{x}+\sqrt[3]{x}}=6\int\frac{t^3dt}{t+1}=6\int(t^2-t+1-\frac{1}{t+1})dt=\\=2t^3-3t^2+6t-6ln|t+1|+C=\\\\=2\sqrt{x}-3\sqrt[3]{x}+6\sqrt[6]{x}-6ln|\sqrt[6]{x}+1|+C\\\\\\x=t^6=\ \textgreater \ dx=6t^5dt\\t^3:(t+1)=t^2-t+1-\frac{1}{t+1}$

$\displaystyle \int\frac{dx}{x\sqrt{7+x^2}}=\int\frac{dt}{t^2-7}=\frac{1}{2\sqrt7}ln|\frac{t-\sqrt7}{t+\sqrt7}|+C=\\=\frac{1}{2\sqrt7}ln|\frac{\sqrt{x^2+7}-\sqrt7}{\sqrt{x^2+7}+\sqrt7}|+C\\\\\\7+x^2=t^2\\x^2=t^2-7\\x=\sqrt{t^2-7}\\dx=\frac{tdt}{\sqrt{t^2-7}}$