Evaluate the following definite integrals \int_{0}^{1}(x^{4}+7e^{x}-3)dx

How to compute the integral of ${\displaystyle \int \sin (1-x^2)dx}$
I tried substitution, ${\displaystyle u=1-x^2,du=-2xdx,dx=-\frac{du}{2x}}$, yielding:
${\displaystyle \int \sin u(-\frac{du}{2})}$
${\displaystyle =-\frac{1}{2}\int \sin \left(u\right)du}$
${\displaystyle =-\frac{1}{2}\cos u}$
${\displaystyle =-\frac{1}{2}\cos (1-x^2)}$
WolframAlpha is showing something completely different, however. What's wrong with my solution?

Evaluate the integral.
${\displaystyle {\int }_1^4\frac{t^2+1}{\sqrt{t}}dt}$

Compute $\int \frac{4t}{5\sqrt[3]{2t+3}}dt$.

Calculate the iterated integral.
${\displaystyle {\int }_0^4{\int }_0^{12}2e^{x+3y}dxdy}$

Write the integral in terms of u and du. Then evaluate.
${\displaystyle \int \frac{{\left(\ln x\right)}^{2}dx}{x},u=\ln x}$

Evaluate the integral.
${\displaystyle {\int }_{-4}^4\left|x\right|dx}$

Evaluate the integral of ${\displaystyle x{e}^{x^2}dx}$