Answer
(a) Zero
(b) Zero
Work Step by Step
(a) We are required to find the integral: $\int_{-1}^{1} x \sqrt{\cos \left(x^{2}\right)} d x$ Splitting the integral into two parts and replacing $(x) \ with \ (-t)$ in one of them:
Reversing the limits of the first integral:
\[
\begin{aligned}
\int_{-1}^{1} x \sqrt{\cos \left(x^{2}\right)} d x &=\int_{0}^{1} x \sqrt{\cos \left(x^{2}\right)} d x+\int_{-1}^{0} x \sqrt{\cos \left(x^{2}\right)} d x \\
&=\int_{0}^{1} x \sqrt{\cos \left(x^{2}\right)} d x+\int_{1}^{0}(-t) \sqrt{\cos \left((-t)^{2}\right)}(-d t) \quad\{-t=x \Rightarrow d x=-d t\} \\
&=\int_{0}^{1} x \sqrt{\cos \left(x^{2}\right)} d x+\int_{1}^{0} t \sqrt{\cos \left(t^{2}\right)} d t
\end{aligned}
\]
(b) To simplify the integral, substitute $-(\pi / 2)+x=u \Rightarrow d x=d u$
$-\int_{b}^{a} f(x) d x=\int_{a}^{b} f(x) d x$
Note that the Integrand is an odd function, so the integral is $ 0 .$ More formally
\[
\begin{array}{l}
=\int_{0}^{1} x \sqrt{\cos \left(x^{2}\right)} d x-\int_{0}^{1} t \sqrt{\cos \left(t^{2}\right)} d t \\
=0
\end{array}
\]
Replacing $(w) \ with\ (-u) $ in the second integral
\[
\begin{array}{l}
\int_{0}^{\pi} \cos ^{5} x \sin ^{8} x d x=\int_{-\pi / 2}^{\pi / 2} \cos ^{5}(\pi / 2+u) \sin ^{8}(\pi / 2+u) d u \\
\int_{0}^{\pi} \cos ^{5} x \sin ^{8} x d x=\int_{-\pi / 2}^{\pi / 2}-\cos ^{8} u \sin ^{5} u d u
\end{array}
\]
$\int_{-\pi / 2}^{\pi / 2}-\sin ^{5} u \cos ^{8} u d u=\int_{0}^{\pi / 2}-\sin ^{5} u
\cos ^{8} u d u+\int_{-\pi / 2}^{0}-\sin ^{5} u \cos ^{8} u d u$
\[
\begin{array}{l}
=\int_{0}^{\pi / 2}-\sin ^{5} u \cos ^{8} u d u+\int_{\pi / 2}^{0}-\sin ^{5} w \cos ^{8} w d w \\
=\int_{0}^{\pi / 2}-\sin ^{5} u \cos ^{8} u d u-\int_{0}^{\pi / 2}-\sin ^{5} w \cos ^{8} w d w
\end{array}
\]
$=0 \quad\{\text { As the two integrations are exactly the same }\}$
