Tutorial note 4

Problem 2. (c) By completing square we have \[

3+2x-x^2=3-(x^2-2x)=4-(x-1)^2,

\] hence \[


\int\frac{x\,dx}{\sqrt{3+2x-x^2}}\,dx=\int\frac{x\,dx}{\sqrt{4-(x-1)^2}} = \int\frac{u\,du}{\sqrt{4-u^2}} +\int\frac{du}{\sqrt{4-u^2}},

\] where we let $u=x-1$. To evaluate the first integral, we note that $u\,du=-\frac{1}{2}d(4-u^2)$. To evaluate the second integral, we substitute $u=2\sin\theta$. So we get \[

-\sqrt{4-(x-1)^2}+\sin^{-1}\frac{x-1}{2}+C

\] for some constant $C$.

(d) To get rid of the square root, recall that $\sin^2\theta - 1 = \tan^2\theta$, hence we let $x=\sec\theta$ to get \[

\int\frac{1}{1+\cos\theta}\,d\theta.

\] Note that $2\cos^2\frac{\theta}{2} = 1+\cos\theta$, the integral becomes\begin{align*}

\int \frac{1}{2 \cos^2\frac{\theta}{2}}\,d\theta& = \int \sec^2\frac{\theta}{2} \,d\brac{\frac{\theta}{2}}
=\tan\brac{\frac{\theta}{2}}+C\\
&=\tan\brac{\frac{1}{2}\cos^{-1}\frac{1}{x}} +C.

\end{align*}

(e) As above, we let $x=\sec\theta$, then the integral becomes \[
\int \tan^2\theta\,d\theta,
\] from that we easily obtain \[

\tan\theta-\theta+C=\sqrt{x^2-1}-\cos^{-1}\frac{1}{x}+C.

\]

Problem 3. We substitute $x=\tan\theta$. Note that \[


1=(\sin^2\theta+\cos^2\theta)^2=\sin^4\theta+\cos^4\theta+2\sin^2\theta\cos^2\theta

\] and thus we simplify the integral to 
\[

\sqrt{2}\int \frac{\sin^2\theta - \cos^2\theta\,d\theta}{\sqrt{2-\sin^22\theta}}.

\] Note that $\sin^2\theta=\frac{1-\cos2\theta}{2}$ and $\cos^2\theta= \frac{1+\cos2\theta}{2}$, hence our integral becomes \[

-\frac{1}{\sqrt{2}}\int\frac{d(\sin 2\theta)}{\sqrt{2-\sin^22\theta}}.

\] Now the rest is routine, let $y=\sin 2\theta$, the integral becomes $\dis  -\frac{1}{\sqrt{2}}\int\frac{dy}{\sqrt{2-y^2}}$. This is easily computed by letting $y=\sqrt{2}\sin\phi$, so finally our answer is \[

-\frac{1}{\sqrt{2}}\sin^{-1}\brac{\sqrt{2}\frac{x}{1+x^2}}+C,

\] for some constant $C$.