Exercise 11.5 - Chapter 11 Integral Calculus 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex 11.5
Integrate the following functions with respect to $\mathrm{x}$

Question 1.
$\frac{x^3+4 x^2-3 x+2}{x^2}$
Solution:
$
\begin{aligned}
\int \frac{x^3+4 x^2-3 x+2}{x^2} d x & =\int\left(x+4-\frac{3}{x}+\frac{2}{x^2}\right) d x \\
& =\frac{x^2}{2}+4 x-3 \log |x|-\frac{2}{x}+c
\end{aligned}
$
Question 2.
$
\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2
$
$\begin{aligned} \int\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2 d x & =\int\left[(\sqrt{x})^2+\left(\frac{1}{\sqrt{x}}\right)^2+2 \sqrt{x} \cdot \frac{1}{\sqrt{x}}\right] d x \\ & =\int\left(x+\frac{1}{x}+2\right) d x=\frac{x^2}{2}+\log |x|+2 x+c\end{aligned}$
Question 3.
$
(2 \mathrm{x}-5)(36+4 \mathrm{x})
$
Solution:
$
\begin{aligned}
\int(2 x-5)(36+4 x) d x & =\int\left(72 x+8 x^2-180-20 x\right) d x \\
& =\int\left(8 x^2+52 x-180 x\right) d x \\
& =8 \frac{x^3}{3}+\frac{52 x^2}{2}-180 x+c \\
& =\frac{8 x^3}{3}+26 x^2-180 x+c
\end{aligned}
$
Question 4.
$
\cot ^2 x+\tan ^2 x
$

Solution:
$
\begin{aligned}
\int\left(\cot ^2 x+\tan ^2 x\right) d x & =\int \frac{\cos ^2 x}{\sin ^2 x} d x+\int \frac{\sin ^2 x}{\cos ^2 x} d x \\
& =\int \frac{1-\sin ^2 x}{\sin ^2 x} d x+\int \frac{1-\cos ^2 x}{\cos ^2 x} d x \\
& =\int\left(\frac{1}{\sin ^2 x}-\frac{\sin ^2 x}{\sin ^2 x}\right) d x+\int\left(\frac{1}{\cos ^2 x}-\frac{\cos ^2 x}{\cos ^2 x}\right) d x \\
& =\int(\operatorname{cosec} x-1) d x+\int\left(\sec ^2 x-1\right) d x \\
& =-\cot x-x+\tan x-x+c=\tan x-\cot x-2 x+c
\end{aligned}
$
Question 5.
$\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}$
Solution:
$
\begin{aligned}
\int \frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha} d x & =\int \frac{\left(2 \cos ^2 x-1\right)-\left(2 \cos ^2 \alpha-1\right)}{\cos x-\cos \alpha} d x \\
& =\int \frac{2\left[\cos ^2 x-\cos ^2 \alpha\right]}{\cos x-\cos \alpha} d x=2 \int \frac{(\cos x+\cos \alpha)(\cos x-\cos \alpha)}{(\cos x-\cos \alpha)} d x \\
& =2[\sin x+x \cos \alpha]+c
\end{aligned}
$
Question 6.
$
\frac{\cos 2 x}{\sin ^2 x \cos ^2 x}
$
Solution:
$
\begin{aligned}
\int \frac{\cos 2 x}{\sin ^2 x \cos ^2 x} d x & =\int \frac{\cos 2 x}{\left(\frac{\sin 2 x}{2}\right)^2} d x \quad \sin x \cos x=\frac{\sin 2 x}{2} \\
& =4 \int \frac{\cos 2 x}{\sin 2 x} \cdot \frac{1}{\sin 2 x} d x=4 \int \cot 2 x \operatorname{cosec} 2 x d x \\
& =4\left[-\frac{\operatorname{cosec} 2 x}{2}\right]+c=-2 \operatorname{cosec} 2 x+c
\end{aligned}
$
Question 7.
$\frac{3+4 \cos x}{\sin ^2 x}$

Solution:
$
\begin{aligned}
\int \frac{3+4 \cos x}{\sin ^2 x} d x & =\int\left(\frac{3}{\sin ^2 x}+\frac{4 \operatorname{co}}{\sin ^2}\right. \\
& =3 \int \operatorname{cosec}^2 x d x+
\end{aligned}
$

Question 8 .
$
\frac{\sin ^2 x}{1+\cos x}
$
Solution:
$
\begin{aligned}
\int \frac{\sin ^2 x}{1+\cos x} d x & =\int \frac{1-\cos ^2 x}{1+\cos x} d x \\
& =\int \frac{(1+\cos x)(1-\cos x)}{1+\cos x} d x=\int(1-\cos x) d x \\
& =x-\sin x+c
\end{aligned}
$

Question 9.
$\frac{\sin 4 x}{\sin x}$
Solution:
$
\begin{aligned}
\int \frac{\sin 4 x}{\sin x} d x & =\int \frac{2 \sin 2 x \cos 2 x}{\sin x} d x \\
& =\int \frac{2(2 \sin x \cos x) \cos 2 x}{\sin x} d x=4 \int \cos 2 x \cos x d x \\
& =4 \times \frac{1}{2} \int[\cos (2 x+x)+\cos (2 x-x)] d x=2 \int(\cos 3 x+\cos x) d x \\
& =2\left[\frac{\sin 3 x}{3}+\sin x\right]+c
\end{aligned}
$

Question 10.
$\cos 3 \mathrm{x} \cos 2 \mathrm{x}$
Solution:
$
\begin{aligned}
\int \cos 3 x \cos 2 x d x & =\frac{1}{2} \int[\cos 5 x+\cos x] d x \quad\left(\cos \mathrm{A} \cos \mathrm{B}=\frac{1}{2}[\cos (\mathrm{A}+\mathrm{B})+\cos (\mathrm{A}-\mathrm{B})]\right) \\
& =\frac{1}{2}\left[\frac{\sin 5 x}{5}+\sin x\right]+c
\end{aligned}
$

Question 11.
$\sin ^2 5 x$
Solution:
$
\int \sin ^2 5 x d x=\int \frac{1-\cos 10 x}{2} d x=\frac{1}{2}\left[x-\frac{\sin 10 x}{10}\right]+c
$
Question 12 .
$\frac{1+\cos 4 x}{\cot x-\tan x}$
Solution:
$
\begin{aligned}
& \int \frac{1+\cos 4 x}{\cot x-\tan x} d x=\int \frac{2 \cos ^2 2 x}{\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}} d x .=\int \frac{2 \cos ^2 2 x}{\frac{\cos ^2 x-\sin ^2 x}{\sin x \cos x}} \\
& =\int 2 \cos ^2 2 x \times \frac{\sin x \cos x}{\cos ^2 x-\sin ^2 x} d x \quad\left(\cos ^2 x-\sin ^2 x=\cos 2 x\right) \\
& =\int 2 \cos ^2 2 x \times \frac{\sin x \cos x}{\cos ^2 x} d x \quad\left(\sin x \cos x=\frac{\sin 2 x}{2}\right) \\
& =\int 2 \cos 2 x\left(\frac{\sin 2 x}{2}\right) d x \\
& =\frac{1}{2} \int \sin 4 x d x=-\frac{1}{2}\left(\frac{\cos 4 x}{4}\right)+c \\
& =-\frac{1}{8} \cos 4 x+c \\
&
\end{aligned}
$
Question 13 .
$\mathrm{e}^{\mathrm{x} \log a} \mathrm{e}^{\mathrm{x}}$
Solution:
$
\begin{array}{rlr}
\int e^{x \log a} e^x d x & =\int a^x e^x d x & e^{x \log a}=e^{\log a^x}=a^x \\
& =\int(a e)^x d x=\frac{(a e)^x}{\log (a e)}+c
\end{array}
$
Question 14 .
$(3 \mathrm{x}+4) \sqrt{3 x+7}$
Solution:

$
\begin{aligned}
& \int(3 x+4) \sqrt{3 x+7} d x=\int(3 x+4+3-3) \sqrt{3 x+7} d x \\
&=\int(3 x+7-3) \sqrt{3 x+7} d x=\int(3 x+7) \sqrt{3 x+7} d x-3 \int \sqrt{3 x+7} d x \\
&=\int(3 x+7)^{3 / 2} d x-3 \int(3 x+7)^{3 / 2} d x \\
&=\frac{(3 x+7)^{5 / 2}}{\frac{5}{2} \times 3}-3 \frac{(3 x+7)^{3 / 2}}{\frac{3}{2} \times 3}+c \\
&=\frac{2}{15}(3 x+7)^{5 / 2}-\frac{2}{3}(3 x+7)^{3 / 2}+c
\end{aligned}
$
Question 15.
$\frac{8^{1+x}+4^{1-z}}{2^x}$
Solution
$
\begin{aligned}
& \int \frac{8^{1+x}+4^{1-x}}{2^x} d x \\
& \text { Consider } \frac{8^{1+x}+4^{1-x}}{2^x}=\frac{\left(2^3\right)^{1+x}+\left(2^2\right)^{1-x}}{2 x} \\
&=\left[2^{3+3 x}+2^{2-2 x}\right] 2^{-x}=2^{3+2 x}+2^{2-3 x} \\
&=\int\left(2^{3+2 x}+2^{2-3 x}\right) d x \\
&=\frac{2^{3+2 x}}{2 \log 2}-\frac{2^{2-3 x}}{3 \log 2}+c \\
&=\frac{2^{2+2 x}}{\log 2}-\frac{2^{2-3 x}}{3 \log 2}+c
\end{aligned}
$
Question 16.
$\frac{1}{\sqrt{x+3}-\sqrt{x-4}}$

Solution:
$
\begin{aligned}
& \int \frac{1}{\sqrt{x+3}-\sqrt{x-4}} d x=\int \frac{1}{\sqrt{x+3}-\sqrt{x-4}} \times \frac{\sqrt{x+3}+\sqrt{x-4}}{\sqrt{x+3}+\sqrt{x-4}} d x \\
&=\int \frac{\sqrt{x+3}+\sqrt{x-4}}{(x+3)-(x-4)} d x \\
&=\frac{1}{7} \int\left[(x+3)^{1 / 2}+(x-4)^{1 / 2}\right] d x \\
&=\frac{1}{7}\left[\frac{(x+3)^{3 / 2}}{\frac{3}{2}}+\frac{(x-4)^{3 / 2}}{\frac{3}{2}}\right]+c \\
&=\frac{2}{21}\left[(x+3)^{3 / 2}+(x-4)^{3 / 2}\right]+c
\end{aligned}
$

Question 17.
$
\frac{x+1}{(x+2)(x+3)}
$
Solution:
$
\begin{aligned}
& \int \frac{x+1}{(x+2)(x+3)} d x \\
& \text { Consider } \frac{x+1}{(x+2)(x+3)}=\frac{\mathrm{A}}{x+2}+\frac{\mathrm{B}}{x+3} \\
& \mathrm{x}+1=\mathrm{A}(\mathrm{x}+3)+\mathrm{B}(\mathrm{x}+2) \ldots+(\mathrm{i}) \\
& \text { Put } \mathrm{x}=-3 \text { in (i) } \\
& -2=\mathrm{B}(-1) \\
& \mathrm{B}=2 \\
& \text { Put } \mathrm{x}=-2 \text { in (i) } \\
& -1=\mathrm{A}(1) \\
& \mathrm{A}=-1 \\
& \int \frac{x+1}{(x+2)(x+3)} d x=\int \frac{-1}{x+2} d x+\int \frac{2}{x+3} d x \\
& .
\end{aligned}
$
Question 18 .
$
\frac{1}{(x-1)(x+2)^2}
$

Solution:
$
\begin{aligned}
& \int \frac{1}{(x-1)(x+2)^2} d x \\
& \text { Consider } \quad \frac{1}{(x-1)(x+2)^2}=\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{x+2}+\frac{\mathrm{C}}{(x+2)^2} \\
& 1=\mathrm{A}(x+2)^2+\mathrm{B}(x-1)(x+2)+\mathrm{C}(x-1) \\
& \text { put } x=1 \text { in }(i) \\
& 1=9 \mathrm{~A} \Rightarrow \mathrm{A}=1 / 9 \\
& \text { put } x=-2 \text { in }(i) \\
& 1=-3 \mathrm{C} \Rightarrow \mathrm{C}=-1 / 3 \\
&
\end{aligned}
$
Equating $x^2$ coefficients on $b / s$ in $(i)$
$
\begin{aligned}
\mathrm{A}+\mathrm{B} & =0 \\
\frac{1}{9}+\mathrm{B} & =0 \Rightarrow \mathrm{B}=-1 / 9 \\
\int \frac{1}{(x-1)(x+2)^2} d x & =\int \frac{1 / 9}{x-1} \int \frac{\frac{-1}{9}}{x+2} d x+\int \frac{\frac{-1}{3}}{(x+2)^2} d x \\
& =\frac{1}{9} \log |x-1|-\frac{1}{9} \log |x+2|+\frac{1}{3(x+2)}+c .
\end{aligned}
$

Question 19.
$
\frac{3 x-9}{(x-1)(x+2)\left(x^2+1\right)}
$
Solution:
$
\int \frac{3 x-9}{(x-1)(x+2)\left(x^2+1\right)} d x
$
Consider,
$
\begin{aligned}
& \begin{aligned}
& \frac{3 x-9}{(x-1)(x+2)\left(x^2+1\right)}=\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{x+2}+\frac{\mathrm{C} x+\mathrm{D}}{x^2+1} \\
& 3 x-9=\mathrm{A}(x+2)\left(x^2+1\right)+\mathrm{B}(x-1)\left(x^2+1\right)+(\mathrm{C} x+\mathrm{D})(x-1)(x+2) \ldots(i) \\
& \text { put } \mathrm{x}=1 \text { in }(\mathrm{i}) \\
& 3-9=\mathrm{A}(3)(2)+0+0 \\
&-6=6 \mathrm{~A}
\end{aligned} \\
& \mathrm{~A}=-1
\end{aligned}
$

$
\begin{aligned}
& \text { put } \mathrm{x}=-2 \text { in (i) } \\
& -6-9=0+\mathrm{B}(-3)(5)+0 \\
& -15=-15 \mathrm{~B} \\
& \mathrm{~B}=1 \\
& \text { put } \mathrm{x}=0 \text { in (i) } \\
& 0-9=\mathrm{A}(2)(1)+\mathrm{B}(-1)(1)+\mathrm{D}(-1)(2) \\
& -9=2 \mathrm{~A}-\mathrm{B}-2 \mathrm{D} \\
& 2 \mathrm{D}=2 \mathrm{~A}-\mathrm{B}+9 \\
& 2 \mathrm{D}=2(-1)-1+9 \\
& =-2-1+9 \\
& 2 \mathrm{D}=6 \\
& \mathrm{D}=3
\end{aligned}
$
Equating $\mathrm{x}^3$ coefficients on $\mathrm{b} / \mathrm{s}$ in (i)
$
\begin{aligned}
& \mathrm{A}+\mathrm{B}+\mathrm{C}=0 \\
&-1+1+\mathrm{C}=0 \\
& \mathrm{C}=0 \\
& \int \frac{3 x-9}{(x-1)(x+2)\left(x^2+1\right)} d x=\int \frac{-1}{x-1} d x+\int \frac{1}{x+2} d x+\int \frac{3}{x^2+1} d x \\
&=-\log |x-1|+\log |x+2|+3 \tan ^{-1} x+c \\
&=\log \left|\frac{x+2}{x-1}\right|+3 \tan ^{-1} x+c
\end{aligned}
$
Question 20.
$
\frac{x^3}{(x-1)(x-2)}
$
Solution:

$
\int \frac{x^3}{(x-1)(x-2)} d x
$
Consider, $\frac{x^3}{(x-1)(x-2)}=\frac{x^3}{x^3-3 x+2}$

$\begin{aligned}
& \frac{x^3}{x^2-3 x+2}=\frac{7 x-6}{x^2-3 x+2} \\
& \text { But } \quad \frac{7 x-6}{x^2-3 x+2}=\frac{7 x-6}{(x-1)(x-2)}=\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{x-2} \\
& \Rightarrow \quad 7 x-6=\mathrm{A}(x-2)+\mathrm{B}(x-1) \\
& \operatorname{sub} x=2 \text { in (ii) } \\
& 14-6=0+B \Rightarrow B=8 \\
& \text { sub } x=1 \text { in }(i i) \\
& 7-6=A(1-2) \\
& 1=-\mathrm{A} \\
& A=-1 \\
& \int \frac{x^3}{(x-1)(x-2)}=\int\left[x+3+\frac{-1}{x-1}+\frac{8}{x-2}\right] d x \\
& =\frac{x^2}{2}+3 x-\log |x-1|+8 \log |x-2|+c \\
&
\end{aligned}$