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

Updated On May 15, 2024

By SaraNextGen

Additional Problems
I. One Mark Questions
Choose the correct answer.
Question 1.
The integral of $\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)$ equals
(a) $\frac{1}{3} x^{\frac{1}{3}}+2 x^{\frac{1}{2}}+c$
(b) $\frac{2}{3} x^{\frac{2}{3}}+\frac{1}{2} x^2+c$
(c) $\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+c$
(d) $\frac{3}{2} x^{\frac{3}{2}}+\frac{1}{2} x^{\frac{1}{2}}+c$
Answer:
(c) $\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+c$
Question 2.
If $\frac{d}{d x} f(x)=4 x^3-\frac{3}{x^4}$ such that $f(2)=0$. Then $f(x)$ is
(a) $x^4+\frac{1}{x^3}-\frac{129}{8}$
(b) $x^3+\frac{1}{x^4}+\frac{129}{8}$
(c) $x^4-\frac{1}{x^3}+\frac{129}{8}$
(d) $x^3+\frac{1}{x^4}-\frac{129}{8}$
Answer:
(a) $x^4+\frac{1}{x^3}-\frac{129}{8}$
Hint:
$
\begin{aligned}
f(x)= & \int\left(4 x^3-\frac{3}{x^4}\right) d x \\
= & 4 \int x^3 d x-3 \int x^{-4} d x \\
= & x^4-\frac{3 x^{-3}}{-3}+c=x^4+\frac{1}{x^3}+c \\
& f(2)=0 \Rightarrow 16+\frac{1}{8}+c=0 \Rightarrow c=\frac{-129}{8}
\end{aligned}
$

Question 3.
$\int \frac{\sin ^2 x-\cos ^2 x}{\sin ^2 x \cos ^2 x} d x$ is equal to
(a) $\tan x+\cot x+c$
(b) $\tan x+\operatorname{cosec} x+c$
(c) $-\tan x+\cot x+c$
(d) $\tan x-\sec x+c$
Answer:
(a) $\tan x+\cot x+c$
Hint:
$
\int \frac{\sin ^2 x d x}{\sin ^2 x \cos ^2 x}-\int \frac{\cos ^2 x}{\sin ^2 x \cos ^2 x} d x=\int \sec ^2 x d x-\int \operatorname{cosec}^2 x d x=\tan x+\cot x+c
$
Question 4.
$
\int_1^{\sqrt{3}} \frac{d x}{1+x^2} \cdots \cdots \cdots \ldots
$
(a) $\frac{\pi}{3}$
(b) $\frac{2 \pi}{3}$
(c) $\frac{\pi}{6}$
(d) $\frac{\pi}{12}$
Answer:
(d) $\frac{\pi}{12}$
Hint:
$
\int_1^{\sqrt{3}} \frac{d x}{1+x^2}=\left[\tan ^{-1} x\right]_1^{\sqrt{3}}=\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}
$
Question 5.
The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(x^3+x \cos x+\tan ^5 x\right) d x$ is
(a) 0
(b) 2
(c) $\pi$
(d) 1
Answer:
(a) 0
Hint:
Let $f(x)=x^3+x \cos x+\tan ^5 \mathrm{x}$
$f(-x)=-x^3-x \cos x-\tan ^5 x=-f(x)$
So $\mathrm{f}(\mathrm{x})$ is odd function.
Integral is 0.

Question 6.
Fill in the blanks.
(a) $\int_0^{\frac{\pi}{2}} \cos ^3 x d x$ is equal to
(b) $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^{31} x d x$ is equal to
Answer:
(a) $\frac{2}{3}$
(b) 0
Hint:
$
\begin{aligned}
\int_0^{\frac{\pi}{2}} \cos ^3 x d x & =\int_0^{\frac{\pi}{2}} \cos ^2 x \cos x d x \\
& =\int_0^{\frac{\pi}{2}}\left(1-\sin ^2 x\right) \cos x d x
\end{aligned}
$
Let $y=\sin x$. Then $d y=\cos x d x$
When $x=0, y=0$
$
\begin{gathered}
x=\frac{\pi}{2}, y=1 \\
\int_0^1\left(1-y^2\right) d y=\left[y-\frac{y^3}{3}\right]_0^1=1-\frac{1}{3}=\frac{2}{3} \\
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^{31} x=0 \text { since } \sin ^{31} x \text { is odd function. }
\end{gathered}
$
Question 7.
Match the following.

Answer:
(a) - (iii)
(b) - (iv)
(c) - (v)
(d) - (ii)
(e) - (i)
Question 8.
State True or False.

(a)

(b)

(c)

(b)

(e)

(f)

Answer:
(a) True
(b) False
(c) True
(d) False
(e) False
(f) True
Question 9.
Which of the following is not equal to $\int \tan x \sec ^2 \mathrm{xdx}$?
(a) $\frac{1}{2} \tan ^2 x$
(b) $\frac{1}{2} \sec ^2 x$
(c) $\frac{1}{2 \cos ^2 x}$
(d) None of these
Answer:
(d) None of these
Hint:

Hint:
$
\begin{aligned}
& \int \tan x \sec ^2 x d x=\int \tan x \mathrm{~d}(\tan x)=\frac{\tan ^2 x}{2} \\
& \text { Also } \int \tan x \sec ^2 x d x=\sec x d(\sec x) \\
& =\frac{\sec ^2 x}{2} \\
& =\frac{1}{2 \cos ^2 x} \\
&
\end{aligned}
$
Question 10.
$\int e^x(\cos x-\sin x) d x$ is equal to
(a) $\mathrm{e}^{\mathrm{x}} \sin \mathrm{x}+\mathrm{c}$
(b) $\mathrm{e}^{\mathrm{x}} \cos \mathrm{x}+\mathrm{c}$
(c) $-\mathrm{e}^{\mathrm{x}} \cos \mathrm{x}+\mathrm{c}$
(d) $-\mathrm{e}^{\mathrm{x}} \sin \mathrm{x}+\mathrm{c}$
Answer:
(b) $\mathrm{e}^{\mathrm{x}} \cos \mathrm{x}+\mathrm{c}$
Hint:
Let $f(x)=\cos x$
$f^{\prime}(x)=-\sin x$
$\int e^x\left[f(x)+f^{\prime}(x)\right] d x=e^x f(x)+c$
Question 11.
$\int \frac{1-\cos 2 x}{1+\cos 2 x} d x$ is
(a) $\tan x-x+c$
(b) $x+\tan x+c$
(c) $x-\tan x+c$
(d) $-x-\cot x+c$
Answer:
(a) $\tan x-x+c$

Hint:
$
\begin{aligned}
\int \frac{1-\cos 2 x}{1+\cos 2 x} d x & =\int \frac{2 \sin ^2 x}{2 \cos ^2 x} d x=\int \tan ^2 x d x \\
& =\int\left(\sec ^2 x-1\right) d x=\tan x-x+c
\end{aligned}
$
Question 12.
$\int_0^{\frac{\pi}{2}} \cos x e^{\sin x} d x$ is equal to
(a) $\mathrm{e}=1$
(b) $1-\mathrm{e}$
(c) $e^{\frac{\pi}{2}}-1$
(d) $1-e^{\frac{\pi}{2}}$
Answer:
(a) $\mathrm{e}=1$
Hint:
$
\int_0^{\frac{\pi}{2}} \cos x e^{\sin x} d x=\left[e^{\sin x}\right]_0^{\frac{\pi}{2}}=e-1
$
Question 13.
Which of the following is an even function?
(a) $\sin \mathrm{x}$
(b) $e^x-e^{-x}$
(c) $x \cos x$
(d) $\cos x$
Answer:
(d) $\cos x$
Question 14.
Which of the following is neither odd nor even functional?
(a) $x \sin x$
(b) $x^2$
(c) $e^{-x}$
(d) $x \cos x$
Answer:
(c) $e^{-x}$

Question 15.
$\int \sec ^2(7-4 x) d x$ equal to
(a) $\tan (7-4 x)$
(b) $-\tan (7-4 x)$
(c) $-\frac{1}{4} \tan (7-4 x)$
(d) $\frac{1}{4} \tan (7-4 x)$
Answer:
(c) $-\frac{1}{4} \tan (7-4 x)$
II. 2 Mark Questions.
Question 1.
$
\int \frac{1}{\sqrt{(x+1)(x+5)}}=\int \frac{1}{\sqrt{x^2+6 x+5}} d x
$
Solution:
$
\int \frac{1}{\sqrt{(x+3)^2-2^2}} d x=\log \left|(x+3)+\sqrt{x^2+6 x+5}\right|+c
$
Question 2.
If $f^{\prime}(x)=x^2-\frac{1}{x^2}$ and $f(1)=\frac{1}{3}$, find $f(x)$
Answer:
$
\text { Given } \begin{aligned}
f^{\prime}(x) & =x^2-\frac{1}{x^2} \\
\Rightarrow f(x) & =\int\left(x^2-\frac{1}{x^2}\right) d x \\
f(x) & =\frac{x^3}{3}+\frac{1}{x}+c
\end{aligned}
$

$
f(1)=\frac{1}{3} \Rightarrow 1+\frac{1}{3}+c=\frac{1}{3} \Rightarrow \mathrm{C}=-1
$
So $\quad f(x)=\frac{x^3}{3}+\frac{1}{x}-1$
Question 3.
$
\int\left(e^x+e^{-x}\right)^2 d x
$
Solution:
$
\int\left(e^{2 x}+e^{-2 x}+2\right) d x=\frac{e^{2 x}}{2}+\frac{e^{-2 x}}{-2}+2 x+c
$
Question 4.
Find $\int x^5 \sqrt{3+5 x^6} d x$
Solution:
Let $t=3+5 x^6 \quad$ Then $d t=30 x^5 d x$
So integral becomes $\begin{aligned} \frac{1}{30} \int \sqrt{t} d t & =\frac{1}{30} \frac{\left(3+5 x^6\right)^{\frac{3}{2}}}{\frac{3}{2}}+c \\ & =\frac{1}{45}\left(3+5 x^6\right)^{\frac{3}{2}}+c\end{aligned}$

Question 5.
$
\int \frac{(x+1)(x+\log x)^2}{x} d x
$
Solution:
Let $x+\log x=t$ Then $\left(1+\frac{1}{x}\right) d x=d t$
$
\begin{aligned}
& =\int(x+\log x)^2 d(x+\log x) \\
= & \frac{(x+\log x)^3}{3}+c
\end{aligned}
$
Question 6.
$
\int \frac{e^{2 x}-1}{e^{2 x}+1} d x
$
Solution:
Dividing numerator and denominator by ex, we get $\int \frac{e^x-e^{-x}}{e^x+e^{-x}} d x$
Now $\frac{d}{d x}\left(e^x+e^{-x}\right)=e^x-e^{-x}=\log \left(e^x+e^{-x}\right)+c$
Question 7.
$
\int \frac{1}{x+\sqrt{x}} d x
$
Solution:
Let $t=\sqrt{x} \quad$ Then $x=t^2 \Rightarrow d x=2 t d t$
$
\begin{aligned}
& =\int \frac{2 t d t}{t^2+t}=\int \frac{2 d t}{t+1}=2 \log |t+1|+c \\
& =2 \log (\sqrt{x}+1)+c
\end{aligned}
$

III. 3 and 5 Mark Questions.
Question 1.
Find $\int \frac{e^x(x+1)}{(x+3)^3} d x$
Solution:
$
\begin{aligned}
& \frac{x+1}{(x+3)^3}=\frac{x+3-2}{(x+3)^3}=\frac{1}{(x+3)^2}-\frac{2}{(x+3)^3} \\
& \text { Now } \frac{d}{d x}\left[\frac{1}{(x+3)^3}\right]=\frac{-2(x+3)}{(x+3)^4}=\frac{-2}{(x+3)^3}
\end{aligned}
$
So this is of the form, $\int e^x\left[f(x)+f^{\prime}(x)\right] d x=\frac{e^x}{(x+3)^2}+c$
Question 2.
$
\int \frac{1}{2 x^2-x-1} d x
$
Solution:
$
\begin{aligned}
2 x^2-x-1 & =\frac{1}{2}\left(x^2-\frac{x}{2}-\frac{1}{2}\right) \\
& =\frac{1}{2}\left[\left(x-\frac{1}{4}\right)^2-\frac{1}{16}-\frac{1}{2}\right] \\
& =\frac{1}{2}\left[\left(x-\frac{1}{4}\right)^2-\frac{9}{16}\right] \\
& =\frac{1}{2}\left[\left(x-\frac{1}{4}\right)^2-\left(\frac{3}{4}\right)^2\right]
\end{aligned}
$

So integral becomes, $\frac{1}{2} \int \frac{d x}{\left(x-\frac{1}{4}\right)^2-\left(\frac{3}{4}\right)^2}$
$
\begin{aligned}
& =\frac{1}{2} \frac{1}{(2) \frac{3}{4}} \log \left|\frac{x-\frac{1}{4}-\frac{3}{4}}{x-\frac{1}{4}+\frac{3}{4}}\right|+c \\
& =\frac{1}{3} \log \left|\frac{x-1}{x+\frac{1}{2}}\right|+c=\frac{1}{3} \log \left|\frac{2(x-1)}{2 x+1}\right|+c
\end{aligned}
$
Question 3.
$\int \frac{1}{1-3 \sin ^2 x} d x$
Solution:
Divide the numerator and denominator by $\cos ^2 \mathrm{x}$
$
\int \frac{\sec ^2 x}{1-3 \tan ^2 x} d x
$
Let $t=\tan x$. Then $d t=\sec ^2 x d x$
$
\begin{aligned}
& =\int \frac{d t}{1-3 \mathrm{t}^2}=\frac{1}{3} \int \frac{d t}{\frac{1}{3}-\mathrm{t}^2}=\frac{1}{3} \int \frac{d t}{\left(\frac{1}{\sqrt{3}}\right)^2-t^2} \\
& =\frac{1}{3} \frac{1}{2\left(\frac{1}{\sqrt{3}}\right)} \log \left|\frac{\frac{1}{\sqrt{3}}+t}{\frac{1}{\sqrt{3}}-t}\right|+c \\
& =\frac{\sqrt{3}}{6} \log \left|\frac{1+\sqrt{3} \tan x}{1-\sqrt{3} \tan x}\right|+c
\end{aligned}
$

Question 4.
$\int \frac{d x}{e^x-e^{-x}}$ Solution:
$
e^x-e^{-x}=e^x-\frac{1}{e^x}=\frac{e^{2 x}-1}{e^x}
$
So $\int \frac{d x}{e^x-e^{-x}}=\int \frac{e^x d x}{e^{2 x}-1}$
Let $t=e^x$, Then $e^x d x=d t$
$
\begin{aligned}
& =\int \frac{d t}{t^2-1}=\frac{1}{2} \log \left|\frac{t-1}{t+1}\right|+c \\
& =\frac{1}{2} \log \left|\frac{e^x-1}{e^x+1}\right|+c
\end{aligned}
$
Question 5.
Evaluate $\int_{-1}^2(7 x-5) d x$ as the limit of a sum.

Solution:
Here $f(x)=7 x-5, \quad a=-1, b=2$
$
\begin{aligned}
& h=\frac{b-a}{n}=\frac{2-(-1)}{n}=\frac{3}{n} \\
& f(a+r h)=f\left(-1+\frac{3}{n} r\right)=7\left(-1+\frac{3}{n} r\right)-5 \\
&=-7+\frac{21}{n} r+5=-12+\frac{21}{n} r
\end{aligned}
$
$
\text { So } \begin{aligned}
\int_{-1}^2(7 x-5) d x & =\lim _{n \rightarrow \infty} \sum_{r=1}^n\left(-12+\frac{21}{n} r\right)\left(\frac{3}{n}\right) \\
& =\lim _{n \rightarrow \infty}\left[-12 \sum_{r=1}^n(1)+\frac{21}{n} \sum_{r=1}^n r\right]\left(\frac{3}{n}\right) \\
& =\lim _{n \rightarrow \infty}\left[\frac{-36 n}{n}+\frac{63}{n^2} \frac{n(n+1)}{2}\right] \\
& =-36+\frac{63}{2} \lim _{n \rightarrow \infty}\left[1+\frac{1}{n^2}\right] \\
& =-36+\frac{63}{2}=\frac{-9}{2}
\end{aligned}
$

Question 6.
Evaluate $\int_1^2\left(x^2-1\right) d x$ as the limit of a sum.
Solution:
Here $f(x)=x^2-1, a=1, b=2$
$
\begin{aligned}
& h=\frac{b-a}{n}=\frac{2-1}{n}=\frac{1}{n} \\
& f(a+r h)=f\left(1+\frac{r}{n}\right)=\left(1+\frac{r}{n}\right)^2-1=\frac{r^2}{n^2}+\frac{2 r}{n} \\
& \therefore \int_1^2\left(x^2-1\right) d x=\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{n}\left[\frac{r^2}{n^2}+\frac{2 r}{n}\right] \\
&=\lim _{n \rightarrow \infty}\left[\frac{1}{n^3} \sum_{r=1}^n r^2+\frac{2}{n^2} \sum_{r=1}^n r\right] \\
&=\lim _{n \rightarrow \infty}\left[\frac{1}{n^3} \frac{n(n+1)(2 n+1)}{6}+\frac{2}{n^2} \frac{n(n+1)}{2}\right]
\end{aligned}
$
$
\begin{aligned}
& =\lim _{n \rightarrow \infty}\left[\frac{\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}{6}+\left(1+\frac{1}{n^2}\right)\right] \\
& =\frac{2}{6}+1=\frac{8}{6}=\frac{4}{3}
\end{aligned}
$

Question 7.
Evaluate $\int_1^2 \frac{1}{x\left(x^4+1\right)} d x$
Solution:
Consider $\frac{1}{x\left(x^4+1\right)}=\frac{x^3}{\left.x^4\left(x^4+1\right)\right)}$
Let $x^4=t \Rightarrow 4 x^3 d x=d t$
When $x=1, t=1 ; \quad$ When $x=2, t=16$
$
\begin{aligned}
& =\int_1^{16} \frac{\frac{d t}{4}}{t(t+1)}=\frac{1}{4} \int_1^{16}\left(\frac{1}{t}-\frac{1}{t+1}\right) d t \text { (Using partial fraction) } \\
& =\frac{1}{4}[\log |t|-\log |1+t|]_1^{16} \\
& =\frac{1}{4}[\log 16-\log 17-\log 1+\log 2] \\
& =\frac{1}{4} \log \left(\frac{32}{17}\right)
\end{aligned}
$
Question 8.
Evaluate $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d x}{1+\sqrt{\tan x}}$
Solution:

Let $\quad I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d x}{1+\sqrt{\tan x}}=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x$
By using $\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$
$
\begin{aligned}
I & =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos \left(\frac{\pi}{6}+\frac{\pi}{3}-x\right)}}{\sqrt{\cos \left(\frac{\pi}{6}+\frac{\pi}{3}-x\right)}+\sqrt{\sin \left(\frac{\pi}{6}+\frac{\pi}{3}-x\right)}} d x \\
I & =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos \left(\frac{\pi}{2}-x\right)}}{\sqrt{\cos \left(\frac{\pi}{2}-x\right)}+\sqrt{\sin \left(\frac{\pi}{2}-x\right)}} d x \\
I & =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x
\end{aligned}
$
Adding (1) and (2) we get
$
\begin{aligned}
2 \mathrm{I} & =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} d x \\
2 \mathrm{I} & =\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6} \\
\Rightarrow \mathrm{I} & =\frac{\pi}{12}
\end{aligned}
$
Question 9.
Evaluate $\int_0^1 x(1-x)^5 d x$

Solution:
By using $\int_0^a f(x) d x=\int_0^a f(a-x) d x$, we get
$
\begin{aligned}
\int_0^1 x(1-x)^5 d x & =\int_0^1(1-x)(1-1+x)^5 d x \\
& =\int_0^1(1-x) x^5 d x=\int_0^1\left(x^5-x^6\right) d x \\
& =\left[\frac{x^6}{6}-\frac{x^7}{7}\right]_0^1=\frac{1}{6}-\frac{1}{7}=\frac{1}{42}
\end{aligned}
$
Question 10.
Evaluate $\int \frac{2 x+3}{\sqrt{x^2+x+1}} d x$
Solution:
Now $\frac{d}{d x}\left(x^2+x+1\right)=2 x+1$
So we write $2 x+3=2 x+1+2$
$
\begin{aligned}
& \text { Thus } \int \frac{2 x+3}{\sqrt{x^2+x+1}} d x=\int \frac{2 x+1}{\sqrt{x^2+x+1}} d x+2 \int \frac{d x}{\sqrt{x^2+x+1}} \\
& =2 \sqrt{x^2+x+1}+2 \int \frac{d x}{\sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}} \\
& =2 \sqrt{x^2+x+1}+2 \log \left|\left(x+\frac{1}{2}\right)+\sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\right|+c \\
& \text { Integral }=2 \sqrt{x^2+x+1}+2 \log \left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right|+c \\
&
\end{aligned}
$
Question 11.
$
\int\left(x^2+1\right) \log x d x
$

Solution:
Let $u=\log x$
$
\begin{aligned}
d v & =\left(x^2+1\right) d x \\
v & =\frac{x^3}{3}+x
\end{aligned}
$
$
d u=\frac{1}{x} d x \quad v=\frac{x^3}{3}+x
$
So
$
\begin{aligned}
\int\left(x^2+1\right) \log x d x & =\left(\frac{x^3}{3}+x\right) \log x-\int\left(\frac{x^3}{3}+x\right) \frac{1}{x} d x \\
& =\left(\frac{x^3}{3}+x\right) \log x-\int\left(\frac{x^3}{3}+1\right) d x \\
& =\left(\frac{x^3}{3}+x\right) \log x-\frac{x^3}{9}-x+c
\end{aligned}
$
Question 12.
$
\int \sqrt{x^2+4 x-5} d x
$
Solution:
$
\begin{aligned}
x^2+4 x-5 & =(x+2)^2-4-5 \\
& =(x+2)^2-(3)^2
\end{aligned}
$
So $\int \sqrt{x^2+4 x-5} d x=\int \sqrt{(x+2)^2-3^2} d x$.
$
\begin{aligned}
& =\frac{x+2}{2} \sqrt{(x+2)^2-3^2}-\frac{9}{2} \log \left|x+2+\sqrt{(x+2)^2-3^2}\right|+c \\
& =\frac{x+2}{2} \sqrt{x^2+4 x-5}-\frac{9}{2} \log \left|x+2+\sqrt{x^2+4 x-5}\right|+c
\end{aligned}
$
Question 13.

$
\int_0^{\frac{\pi}{4}}\left(2 \sec ^2 x+x^3+2\right) d x
$
Solution:

$\begin{aligned}
& \int_0^{\frac{\pi}{4}}\left(2 \sec ^2 x+x^3+2\right) d x=\left[2 \tan x+\frac{x^4}{4}+2 x\right]_0^{\frac{\pi}{4}} \\
& =\left[2 \tan \frac{\pi}{4}+\frac{\left(\frac{\pi}{4}\right)^4}{4}+2\left(\frac{\pi}{4}\right)\right]-\left[2 \tan 0+\frac{0}{4}+2(0)\right] \\
& =2+\frac{\pi^4}{1024}+\frac{\pi}{2}
\end{aligned}$
Question 14.
Evaluate $\int_0^1 \frac{2 x}{\left(x^2+1\right)\left(x^2+2\right)} d x$
Solution:
Let $x^2=t$, then $2 x d x=d t$
when $\mathrm{x}=0, \mathrm{t}=0$ and $\mathrm{x}=1, \mathrm{t}=1$
so integral becomes, $\int_0^1 \frac{d t}{(t+1)(t+2)}$
We use partial fractions to proceed further
$
\begin{array}{r}
\frac{1}{(t+1)(t+2)}=\frac{\mathrm{A}}{t+1}+\frac{\mathrm{B}}{t+2} \\
1=\mathrm{A}(t+2)+\mathrm{B}(t+1)
\end{array}
$
Let $t=-2 \quad$ Then $1=-\mathrm{B} \Rightarrow \mathrm{B}=-1$
Let $t=-1 \quad$ Then $1=A \Rightarrow A=1$
$
\text { Thus } \begin{aligned}
\int_0^1 \frac{d t}{(t+1)(t+2)} & =\int_0^1\left(\frac{1}{(t+1)}-\frac{1}{(t+2)}\right) d t \\
& =[\log (t+1)-\log (t+2)]_0^1 \\
& =\log 2-\log 3-\log 1+\log 2 \\
& =2 \log 2-\log 3 \\
& =\log 4-\log 3=\log \left(\frac{4}{3}\right)
\end{aligned}
$

Question 15.
$
\int_0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x
$
Solution:
We use $\int_0^a f(x) d x=\int_0^a f(a-x) d x$
So
$
\begin{aligned}
I & =\int_0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x \\
I & =\int_0^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{2}-x\right)-\cos \left(\frac{\pi}{2}-x\right)}{1+\sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)} \\
I & =\int_0^{\frac{\pi}{2}} \frac{\cos x-\sin x}{1+\sin x \cos x} d x
\end{aligned}
$
Adding (1) and (2)
$
\begin{aligned}
2 \mathrm{I} & =\int_0^{\frac{\pi}{2}} \frac{\sin x-\cos x+\cos x-\sin x}{1+\sin x \cos x} d x \\
I & =0
\end{aligned}
$