Miscellaneous Problems - Chapter 2 Integral Calculus I 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Miscellaneous Problems
Evaluate the following integrals:
Question 1.
$\int \frac{1}{\sqrt{x+2}-\sqrt{x+3}} d x$
Solution:
$
\begin{aligned}
& \int \frac{1}{\sqrt{x+2}-\sqrt{x+3}} d x \\
& \text { Consider } \frac{1}{\sqrt{x+2}-\sqrt{x+3}} \times \frac{\sqrt{x+2}+\sqrt{x+3}}{\sqrt{x+2}+\sqrt{x+3}} \\
& =\frac{\sqrt{x+2}+\sqrt{x+3}}{(\sqrt{x+2})^2-(\sqrt{x+3})^2}=\frac{\sqrt{x+2}+\sqrt{x+3}}{x+2-x-3} \\
& =-\sqrt{x+2}-\sqrt{x+3} \\
&
\end{aligned}
$
So the integral becomes $\int-\sqrt{x+2} d x+\int-\sqrt{x+3} d x$
$
\begin{aligned}
& =\frac{-(x+2)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{(x+3)^{\frac{3}{2}}}{\frac{3}{2}}+c \\
& =\frac{-2}{3}\left[(x+2)^{\frac{3}{2}}+(x+3)^{\frac{3}{2}}\right]+c
\end{aligned}
$

Question 2.
$\int \frac{d x}{2-3 x-2 x^2}$
Solution:
$
\begin{aligned}
& \int \frac{d x}{2-3 x-2 x^2} \\
& \text { Consider } 2-3 x-2 x^2=2\left[1-\frac{3}{2} x-x^2\right] \\
&=2\left[1-\left(x^2+\frac{3}{2} x\right)\right]=2\left[1-\left(\left(x+\frac{3}{4}\right)^2-\frac{9}{16}\right)\right]
\end{aligned}
$
So integral becomes,
$
\begin{aligned}
& =2\left[1+\frac{9}{16}-\left(x+\frac{3}{4}\right)^2\right]=2\left[\frac{25}{16}-\left(x+\frac{3}{4}\right)^2\right] \\
& =2\left[\left(\frac{5}{4}\right)^2-\left(x+\frac{3}{4}\right)^2\right]
\end{aligned}
$
$
\begin{aligned}
\frac{1}{2} \int \frac{d x}{\left(\frac{5}{4}\right)^2-\left(x+\frac{3}{4}\right)^2} & =\frac{1}{2} \frac{1}{2\left(\frac{5}{4}\right)} \log \left|\frac{\frac{5}{4}+x+\frac{3}{4}}{\frac{5}{4}-x-\frac{3}{4}}\right| \\
& =\frac{1}{5} \log \left|\frac{2+x}{1-2 x}\right|+c
\end{aligned}
$

Question 3.
$
\int \frac{d x}{e^x+6+5 e^{-x}}
$
Solution:
$
\begin{aligned}
\int \frac{d x}{e^x+6+5 e^{-x}} & =\int \frac{d x}{e^x+6+\frac{5}{e^x}}=\int \frac{e^x d x}{e^{2 x}+6 e^x+5} \\
\text { Let } e^x=t . \text { Then } e^x d x & =d t
\end{aligned}
$
So integral becomes $=\int \frac{d t}{t^2+6 t+5}=\int \frac{d t}{(t+5)(t+1)}$
$
\text { Now }=\frac{1}{(t+5)(t+1)}=\frac{\mathrm{A}}{t+5}+\frac{\mathrm{B}}{t+1}
$
$
1=\mathrm{A}(t+1)+\mathrm{B}(t+5)
$
Put $t=-1 \Rightarrow 4 \mathrm{~B}=1, \mathrm{~B}=\frac{1}{4} ;$ Put $t=-5 \Rightarrow-4 \mathrm{~A}=1, \mathrm{~A}=-\frac{1}{4}$
So we have
$
\begin{aligned}
\frac{1}{4} \int\left(\frac{-1}{t+5}+\frac{1}{t+1}\right) d t & =\frac{1}{4}[-\log (t+5)+\log (t+1)]+c \\
& =\frac{1}{4} \log \left|\frac{(t+1)}{(t+5)}\right|+c \\
& =\frac{1}{4} \log \left|\frac{e^x+1}{e^x+5}\right|+c
\end{aligned}
$

Question 5.
$\int \sqrt{9 x^2+12 x+3} d x$
Solution:
$
\begin{aligned}
\int \sqrt{9 x^2+12 x+3} & d x \\
& =\int \sqrt{9\left(x^2+\frac{12}{9} x+\frac{3}{9}\right)} d x \\
& =3 \int \sqrt{x^2+\frac{4}{3} x+\frac{1}{3}} d x \\
& =3 \int \sqrt{\left(x+\frac{4}{6}\right)^2-\frac{16}{36}+\frac{1}{3}} d x
\end{aligned}
$

$
\begin{aligned}
& =3 \int \sqrt{\left(x+\frac{2}{3}\right)^2-\frac{1}{9}} d x \\
& =3 \int \sqrt{\left(x+\frac{2}{3}\right)^2-\left(\frac{1}{3}\right)^2} d x \\
& =3\left[\frac{3 x+2}{6} \sqrt{\frac{(3 x+2)^2}{9}-\frac{1}{9}}-\frac{1}{18} \log \left|\frac{(3 x+2)}{2}+\sqrt{\left(\frac{3 x+2}{9}\right)^2-\frac{1}{9}}\right|\right]+c \\
& =3\left[\frac{(3 x+2)}{6} \frac{\sqrt{9 x^2+12 x+3}}{3}-\frac{1}{18} \log \left|\frac{(3 x+2)}{3}+\frac{\sqrt{9 x^2+12 x+3}}{3}\right|\right]+c \\
& =\frac{(3 x-2)}{6} \sqrt{9 x^2+12 x+3}-\frac{1}{6} \log \left|(3 x+2)+\sqrt{9 x^2+12 x+3}\right|+k \\
&
\end{aligned}
$
Note: The constant $\frac{1}{6} \log 3$ can be merged with the constant ' $c$ '.
Question 6.
$
\int(\mathrm{x}+1)^2 \log \mathrm{x} \mathrm{dx}
$
Solution:
$
\int(x+1)^2 \log x d x
$

We use the integration-by-parts method.
Let $u=\log x \Rightarrow d u=\frac{1}{x} d x \quad d v=(x+1)^2 d x$ So $v=\frac{(x+1)^3}{3}$
We have $\int(x+1)^2 \log x d x=\frac{(x+1)^3}{3} \log x-\int \frac{(x+1)^3}{3}\left(\frac{1}{x}\right) d x$
$
=\frac{(x+1)^3}{3} \log x-\frac{1}{3} \int \frac{\left(x^3+3 x^2+3 x+1\right)}{x} d x
$
$=\frac{(x+1)^3}{3} \log x-\frac{1}{3} \int\left(x^2+3 x+3+\frac{1}{x}\right) d x$
$
=\frac{1}{3}\left[(x+1)^3 \log x-\frac{x^3}{3}-\frac{3 x^2}{2}-3 x-\log |x|\right]+c
$

Question 7.
$\int \log \left(x-\sqrt{x^2}-1\right) d x$
Solution:
$
\int \log \left(x-\sqrt{x^2}-1\right) d x
$
Let $u=\log \left(x-\sqrt{x^2-1}\right)$
$
\begin{aligned}
d u & =\frac{1-\frac{2 x}{2 \sqrt{x^2-1}}}{x-\sqrt{x^2-1}} \\
d u & =\frac{2 \sqrt{x^2-1}-2 x}{2 \sqrt{x^2-1}\left(x-\sqrt{x^2-1}\right)} \\
d u & =\frac{-2\left(x-\sqrt{x^2-1}\right)}{2 \sqrt{x^2-1}\left(x-\sqrt{x^2-1}\right)} \\
d u & =\frac{-1}{\sqrt{x^2-1}}
\end{aligned}
$

So integral becomes
$
\begin{aligned}
x \log \left(x-\sqrt{x^2-1}\right)+\int \frac{x}{\sqrt{x^2-1}} d x & =x \log \left(x-\sqrt{x^2-1}\right)+\frac{1}{2} \int \frac{2 x}{\sqrt{x^2-1}} d x \\
& =x \log \left(x-\sqrt{x^2-1}\right)+\frac{1}{2} \frac{\left(x^2-1\right)^{\frac{1}{2}}}{\frac{1}{2}}+c \\
& =x \log \left(x-\sqrt{x^2-1}\right)+\sqrt{x^2-1}+c
\end{aligned}
$
Question 8.
$\int_0^1 \sqrt{x(x-1)} d x$
Solution:
$
\begin{aligned}
& \int_0^1 \sqrt{x(x-1)} d x=\int_0^1 \sqrt{x^2-x} d x \\
& =\int_0^1 \sqrt{\left(x-\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2} \\
& \text { This is of the form } \int_0^1 \sqrt{x^2-a^2} \\
& \left.=\left[\frac{\left(x-\frac{1}{2}\right)}{2} \sqrt{\left(x-\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2}-\frac{1}{(4)(2)} \log \mid\left(x-\frac{1}{2}\right)+\sqrt{x(x-1)}\right)\right]_0^1 \\
& =\frac{\left(1-\frac{1}{2}\right)}{2} \sqrt{\left(1-\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2}-\frac{1}{8} \log \left|\left(1-\frac{1}{2}\right)+\sqrt{1(1-1)}\right| \\
& \frac{-\left(\frac{-1}{2}\right)}{2} \sqrt{\left(\frac{-1}{2}\right)^2-\left(\frac{1}{2}\right)^2}+\frac{1}{8} \log \left|\frac{-1}{2}+\sqrt{0(0-1)}\right| \\
& =\frac{1}{4}(0)-\frac{1}{8} \log \frac{1}{2}+\frac{1}{4}(0)-\frac{1}{8} \log \frac{1}{2}=0 \\
&
\end{aligned}
$

Question 9.
$\int_{-1}^1 x^2 e^{-2 x} d x$
Solution:
$\int_{-1}^1 x^2 e^{-2 x} d x \quad$ We use integration by parts
Let $u=x^2$, then $d u=2 x d x$
Let $d \nu=e^{-2 x} d x, \quad v=\frac{e^{-2 x}}{-2}$
$
\begin{aligned}
& =\left[\frac{x^2 e^{-2 x}}{-2}\right]_{-1}^1-\int_{-1}^1(2 x) \frac{e^{-2 x}}{-2} d x \\
& =\frac{e^{-2}}{-2}-\frac{e^2}{-2}+\int_{-1}^1 x e^{-2 x} d x \\
& =\frac{e^2-e^{-2}}{2}+\left[\frac{x e^{-2 x}}{-2}\right]_{-1}^1-\int_{-1}^1 \frac{e^{-2 x}}{-2} d x
\end{aligned}
$
\{We have used integration by parts again \}
$
\begin{aligned}
& =\frac{e^2-e^{-2}}{2}-\frac{e^{-2}}{2}-\frac{e^2}{2}+\frac{1}{2}\left[\frac{e^{-2 x}}{-2}\right]_{-1}^1 \\
& =-e^{-2}+\frac{1}{2}\left[\frac{e^{-2}}{-2}+\frac{e^2}{2}\right] \\
& =\frac{-1}{e^2}-\frac{1}{4 e^2}+\frac{1}{4} e^2
\end{aligned}
$

$=\frac{e^2}{4}-\frac{5}{4 e^2}$

Question 10.
$
\int_0^3 \frac{x d x}{\sqrt{x+1}+\sqrt{5 x+1}}
$

Solution:
$
\begin{aligned}
& \int_0^3 \frac{x d x}{\sqrt{x+1}+\sqrt{5 x+1}} \\
& \frac{x}{\sqrt{x+1}+\sqrt{5 x+1}}=\frac{x}{\sqrt{x+1}+\sqrt{5 x+1}} \times \frac{\sqrt{x+1}-\sqrt{5 x+1}}{\sqrt{x+1}-\sqrt{5 x+1}} \\
&=\frac{x[\sqrt{x+1}-\sqrt{5 x+1}]}{(x+1)-(5 x+1)} \\
&=\frac{x(\sqrt{x+1}-\sqrt{5 x+1})}{} \\
&=-\frac{1}{4}(\sqrt{x+1}-\sqrt{5 x+1})
\end{aligned}
$
So integral becomes
$
\begin{aligned}
-\frac{1}{4} \int_0^3(\sqrt{x+1}-\sqrt{5 x+1}) d x & =-\frac{1}{4}\left[\frac{(x+1)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{(5 x+1)^{\frac{3}{2}}}{\left(\frac{3}{2}\right)^5}\right]_0^3 \\
& =-\frac{1}{4}\left[\frac{4^{\frac{3}{2}}}{\frac{3}{2}}-\frac{(16)^{\frac{3}{2}}}{\frac{15}{2}}\right]+\frac{1}{4}\left[\frac{(1)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{-(1)^{\frac{3}{2}}}{\frac{15}{2}}\right] \\
& =-\frac{1}{4}\left(\frac{2}{3}(8)-\frac{2}{15}(64)\right)+\frac{1}{4}\left[\frac{2}{3}-\frac{2}{15}\right] \\
& =-\frac{1}{4}\left(\frac{16}{3}-\frac{128}{15}-\frac{2}{3}+\frac{2}{15}\right) \\
& =-\frac{1}{4}\left[\frac{14}{3}-\frac{126}{15}\right]=-\frac{1}{4}\left(\frac{70-126}{15}\right)=\frac{14}{15}
\end{aligned}
$