<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\parindent=0pt
\begin{document}

{\bf Question}

\begin{description}
\item[(i)]
$\ds\int_0^{\frac{\pi}{2}} \sin x \,dx$

\item[(ii)]
$\ds\int_0^{\pi} \sin x \,dx$

\item[(iii)]
$\ds\int_0^{2\pi} \sin x \,dx$

\item[(iv)]
$\ds\int \sin^2 x \,dx$

\item[(v)]
$\ds\int \cos^2 x \,dx$

\item[(vi)]
$\ds\int_0^{2\pi} \sin^2 x \,dx$
\end{description}
\medskip

{\bf Answer}

(i)

$\ds\int_0^{\frac{\pi}{2}} \sin x \,dx=[-\cos
x]_0^{\frac{\pi}{2}}=-\cos\frac{\pi}{2}+\cos 0 = -0+1=\un{1}$

(ii)

$\ds\int_0^{\pi} \sin x \,dx=[-\cos x]_0^{\pi}=-\cos{\pi}+\cos 0 =
-(-1)+1=\un{2}$

(iii)

$\ds\int_0^{2\pi} \sin x \,dx=[-\cos x]_0^{2\pi}=-\cos{2\pi}+\cos
0 = -1+1=\un{0}$


Pictures of (i), (ii) and (iii): PICTURE \vspace{2in}

(iv)

$\ds\int \sin^2 x \,dx$: Now, $1-2\sin^2 x=\cos 2x \Rightarrow
\sin^2 x=\ds\frac{1}{2}(1-\cos 2x)$

Therefore \begin{eqnarray*} \ds\int \sin^2 x \,dx &amp; = &amp;
\ds\frac{1}{2}(1+\cos 2x) \,dx\\ &amp; = &amp; \ds\frac{1}{2}\ds\int
\,dx+\ds\frac{1}{2}\ds\int\cos 2x \,dx\\ &amp; = &amp; \un{\ds\frac{x}{2}
-\ds\frac{1}{4} \sin 2x +c}
\end{eqnarray*}

(standard power and standard multiple sine/cosine)

(v)

$\ds\int \cos^2 x \,dx$:

Now, $2\cos^2x-1=\cos 2x \Rightarrow \cos^2
x=\ds\frac{1}{2}(1+\cos 2x)$

Therefore \begin{eqnarray*} \ds\int \cos^2 x \,dx &amp; = &amp;
\ds\frac{1}{2}\ds\int \,dx +\ds\frac{1}{2}\ds\int \cos 2x \,dx\\ &amp;
= &amp; \un{\ds\frac{x}{2}+\ds\frac{1}{4} \sin 2x +c} \end{eqnarray*}

(compare with (iv) above)

(vi)

$\begin{array} {rcl} \ds\int_0^{2\pi} \sin^2 x \,dx &amp; = &amp;
\left[\ds\frac{x}{2}-\ds\frac{1}{4} \sin 2x\right]_0^{2\pi}\
[{\rm{from}}\ x]\\ &amp; = &amp;
\left(\ds\frac{2\pi}{2}-\ds\frac{1}{4}\sin 4\pi\right)-(0-0)=\pi
\end{array}$

NB $\ds\int_0^{\pi} \sin^2 x \,dx=0$, but $\ds\int_0^{2\pi} \sin^2
x \,dx=\pi \ne 0$

Why?

Plot graph of $\sin^2 x$: PICTURE \vspace{1in}

\end{document}
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