Maths M.C.Q (1 Marks)

3. $\frac{\pi}{2}$

Solution:

We have,

$\text{I}=\int\limits^1_0\frac{\text{d}}{\text{dx}}\Big\{\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\Big\}\text{dx}$

We know since $\int\text{f}'(\text{x})=\text{f}(\text{x})$

$\text{f}(\text{x})=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ and $\text{f}'(\text{x})=\frac{\text{d}}{\text{dx}}\Big\{\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\Big\}$

Therefore, $\text{I}=\Big[\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\Big]^1_0$

$=\sin^{-1}(1)-\sin^{-1}(0)$

$=\frac{\pi}{2}$

1. $\frac{\pi}{12}$

Solution:

$\int\limits^\sqrt{3}_1\frac{1}{1+\text{x}^2}\text{dx}$

$=\big[\tan^{-1}\text{x}\big]^\sqrt{3}_1$

$=\frac{\pi}{3}-\frac{\pi}{4}$

$=\frac{\pi}{12}$

4. $\frac{\pi}{8}$

Solution:

$\text{I}=\int\limits^1_0\sqrt{\text{x}(1-\text{x})}\text{ dx} $

$\text{I}= \int\limits^1_0\sqrt{\text{x}-\text{x}^2}\text{dx}$

$\text{I}=\int\limits^1_0\sqrt{\frac{1}{4}+\text{x}-\text{x}^2+\frac{1}{4}}\text{dx}$

$\text{I}=\int\limits^1_0\sqrt{\frac{1}{4}-\Big(\text{x}^2-\text{x}+\frac{1}{4}\Big)}\text{dx}$

$\text{I}=\int\limits^1_0\sqrt{\Big(\frac{1}{2}\Big)^2-\Big(\text{x}-\frac{1}{2}\Big)^2}\text{dx}$

$\text{I}=\Bigg[\frac{\text{x}-\frac{1}{2}}{2}\sqrt{\text{x}(1-\text{x})}+\frac{1}{2}\times\frac{1}{4}\sin^{-1}(2\text{x}-1)\Bigg]^1_0$

$\text{I}=0+\frac{1}{8}\big(\sin^{-1}(1)-\sin^{-1}(-1)\big)$

$\text{I}= \frac{1}{8}\Big(\frac{\pi}{2}-\Big(\frac{\pi}{2}\Big)\Big)$

$\text{I}= \frac{\pi}{8}$

1.  $1$

Solution:

$\int\limits^\text{e}_1\log\text{x}\text{ dx}$

$=\int\limits^\text{e}_1\log\text{x}\text{ x}^0\text{dx}$

$=\big[\text{x}\log\text{x}\big]^\text{e}_1-\int\limits^\text{e}_1\frac{1}{\text{x}}\text{dx}$

$=\big[\text{x}\log\text{x}\big]^\text{e}_1-\big[\text{x}\big]^\text{e}_1$

$=(\text{e}-0)-(\text{e}-1)$

$= \text{e}-\text{e}+1$

$=1$

1. $\frac{\pi}{12}+\log\big (2\sqrt{2}\big)$

Solution:

We have, 

$\text{I}=\int\limits^3_0\frac{3\text{x}+1}{\text{x}^2+9}\text{ dx}$

$\text{I}=\int\limits^3_0\frac{3\text{x}}{\text{x}^2+9}\text{ dx}+\int\limits^3_0\frac{1}{\text{x}^2+9}\text{ dx}$

$\text{I}_1=\int\limits^3_0\frac{3\text{x}}{\text{x}^2+9}\text{ dx}$ and $\text{I}_2=\int\limits^3_0\frac{1}{\text{x}^2+9}\text{ dx}$

Putting $\text{x}^2+9=\text{t}$ in I₁

$\Rightarrow 2\text{x}\text{ dx}=\text{dt}$

$\Rightarrow \text{x}\text{ dx}=\frac{\text{dt}}{2}$

when $\text{x}\rightarrow0;\text{t}\rightarrow9$

and $\text{x}\rightarrow3;\text{t}\rightarrow18$

$\therefore\text{I}=\int\limits^{18}_9\frac{3\text{ dt}}{2\text{ t}}+\int\limits^3_0\frac{1}{\text{x}^2+9}\text{ dx}$

$=\frac{3}{2}\int\limits^{18}_9\frac{\text{dt}}{\text{t}}+\int\limits^3_0\frac{1}{\text{x}^2+3^2}\text{ dx}$

$=\frac{3}{2}\big[\log(\text{t})\big]^{18}_9+\frac{1}{3}\Big[\tan^{-1}\Big(\frac{\pi}{3}\Big)\Big]^3_0$

$=\frac{3}{2}\big[\log18-\log9\big]+\frac{1}{3}\Big(\frac{\pi}{4}-0\Big)$

$=\frac{3}{2}\Big[\log\frac{18}{9}\Big]+\frac{\pi}{12}$

$=\frac{3}{2}\big[\log2\big]+\frac{\pi}{12}$

$=\log(\sqrt{8})+\frac{\pi}{12}$

$=\log(2\sqrt{2})+\frac{\pi}{12}$

$=\frac{\pi}{12}+\log(2\sqrt{2})$

2.  $\log_\text{e}\sqrt{3}$

Solution:

$\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\frac{1}{\sin2\text{x}}\text{ dx}$

$=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\text{cosec }2\text{x}\text{ dx}$

$=\frac{1}{2}\int\limits^\frac{\pi}{3}_\frac{\pi}{6}2\text{cosec }2\text{x}\text{ dx}$

$=\frac{-1}{2}\big[\log(\text{cosec}2\text{x}+\cot2\text{x})\big]^\frac{\pi}{3}_\frac{\pi}{6}$

$=\frac{-1}{2}\big[-2\log\sqrt{3}\big]$

$=\log\sqrt{3}$

2. $\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$

Solution:

We have,

$=\int\limits^\frac{\pi}{2}_0\frac{1}{2+\cos\text{x}}\text{ dx}$

$=\int\limits^\frac{\pi}{2}_0\frac{1}{2+\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}}\text{ dx}$

$=\int\limits^\frac{\pi}{2}_0\frac{1+\tan^2\frac{\text{x}}{2}}{2+2\tan^2\frac{\text{x}}{2}+1-\tan^2\frac{\text{x}}{2}}\text{ dx}$

$=\int\limits^\frac{\pi}{2}_0\frac{\sec^2\frac{\text{x}}{2}}{3+\tan^2\frac{\text{x}}{2}}\text{dx}$

Putting $\tan\frac{\text{x}}{2}=\text{t}$

$\Rightarrow \frac{1}{2}\sec^2\frac{\text{x}}{2}\text{dx}=\text{dt}$

$\Rightarrow \sec^2\frac{\text{x}}{2}\text{dx}=2\text{dt}$

when $\text{x}\rightarrow0;\text{ t}\rightarrow0$

and $\text{x}\rightarrow\frac{\pi}{2};\text{ t}\rightarrow1$

$\therefore\ \text{I}=\int\limits^1_0\frac{2}{3+\text{t}^2}\text{ dt}$

$=2\int\limits^1_0\frac{1}{(\sqrt{3})^2+\text{t}^2}\text{dt}$

$=\frac{2}{\sqrt{3}}\Big[\tan^{-1}\frac{\text{t}}{\sqrt{3}}\Big]^1_0$

$=\frac{2}{\sqrt{3}}\Big[\tan^{-1}\frac{1}{\sqrt{3}}-\tan^{-1}\frac{0}{\sqrt{3}}\Big]$

$=\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$

4. 1

Solution:

We have,

$\text{I}=\int\limits^{\frac{\pi}{2}}_0\text{x}\sin\text{x dx}$

$=\big[-\text{x}\cos\text{x}\big]^{\frac{\pi}{2}}_0-\int\limits^{\frac{\pi}{2}}_01(-\cos\text{x})\text{dx}$

$=\big[-\text{x}\cos\text{x}\big]^{\frac{\pi}{2}}_0+\int\limits^{\frac{\pi}{2}}_0\cos\text{x dx}$

$=-\big[\text{x}\cos\text{x}\big]^{\frac{\pi}{2}}_0+\big[\sin\text{x}\big]^{\frac{\pi}{2}}_0$

$=-\big[0-0\big]+\big[1-0\big]$

$=1$

1. $\frac{\pi}{4}$

Solution:

Let, $\text{I}=\int\limits^\frac{\pi}{2}_0\frac{1}{1+\tan\text{x}}\text{dx}\ ...(\text{i})$

$=\int\limits^\frac{\pi}{2}_0\frac{1}{1+\tan\big(\frac{\pi}{2}-\text{x}\big)}\text{dx}$

$=\int\limits^\frac{\pi}{2}_0\frac{1}{1+\cot\text{x}}\text{dx}\ ...(\text{ii})$

Adding (i) and (ii) we get

$2\text{I}=\int\limits^\frac{\pi}{2}_0\Big[\frac{1}{1+\tan\text{x}}+\frac{1}{1+\cot\text{x}}\Big]\text{dx}$

$=\int\limits^\frac{\pi}{2}_0\bigg[\frac{(1+\cot\text{x})+(1+\tan\text{x})}{(1+\tan\text{x})(1+\cot\text{x})}\bigg]\text{dx}$

$=\int\limits^\frac{\pi}{2}_0\Big[\frac{2+\tan\text{x}+\cot\text{x}}{1+\tan\text{x}+\cot\text{x}+\tan\text{x}\cot\text{x}}\Big]\text{dx}$

$=\int\limits^\frac{\pi}{2}_0\Big[\frac{2+\tan\text{x}+\cot\text{x}}{2+\tan\text{x}+\cot\text{x}}\Big]\text{dx}$

$=\int\limits^\frac{\pi}{2}_0\text{dx}$

$=\big[\text{x}\big]^\frac{\pi}{2}_0=\frac{\pi}{2}$

Hence, $\text{I}=\frac{\pi}{4}$

1. $\frac{\pi^2}{4}$

Solution:

We have,

$\text{I}= \int\limits^\pi_0\frac{\text{x}\tan\text{x}}{\sec\text{x}+\cos\text{x}}\text{dx}\ ...(\text{i})$

$=\int\limits^\pi_0\frac{(\pi-\text{x)}\tan(\pi-\text{x})}{\sec(\pi-\text{x})+\cos(\pi-\text{x})}\text{dx}$

$=\int\limits^\pi_0\frac{(\pi-\text{x})\tan\text{x}}{\sec\text{x}+\cos\text{x}}\ ...(\text{ii})$

Adding (i) and (ii), we get

$2\text{I}=\int\limits^\pi_0\Big[\frac{\text{x}\tan\text{x}}{\sec\text{x}+\cos\text{x}}+\frac{(\pi-\text{x})\tan\text{x}}{\sec\text{x}+\cos\text{x}}\Big]\text{dx}$

$\Rightarrow\text{I}=\frac{1}{2}\int\limits^\pi_0\frac{\pi\tan\text{x}}{\sec\text{x}+\cos\text{x}}\text{dx}$

$=\frac{\pi}{2}\int\limits^\pi_0\frac{\sin\text{x}}{1+\cos^2\text{x}}\text{dx}$

Putting $\cos\text{x}=\text{t}$

$\Rightarrow-\sin\text{x}\text{ dx}=\text{dt}$

$\Rightarrow \sin\text{x}\text{ dx}=-\text{dt}$

When $\text{x}\rightarrow0;\text{ t}\rightarrow1$

and $\text{x}\rightarrow\pi;\text{ t}\rightarrow-1$

$\Rightarrow\text{I}=\frac{\pi}{2}\int\limits^{-1}_1\frac{-\text{dt}}{1+\text{t}^2}$

$=\frac{\pi}{2}\int\limits^1_{-1}\frac{\text{dt}}{1+\text{t}^2}$

$=\frac{\pi}{2}\big[\tan^{-1}\text{t}\big]^1_{-1}$

$=\frac{\pi}{2}\big[\tan^{-1}(1)-\tan^{-1}(-1)\big]$

$=\frac{\pi}{2}\Big[\frac{\pi}{4}-\Big(-\frac{\pi}{4}\Big)\Big]$

$=\frac{\pi}{2}\times\frac{\pi}{2}=\frac{\pi^2}{4}$

Hence, $\text{I}=\frac{\pi^2}{4}$

2. 0

Solution:

Let, $\text{I}=\int\limits^1_0\tan^{-1}\frac{2\text{x}-1}{1+\text{x}-\text{x}^2}\text{ dx}\ ....(\text{i})$

$=\int\limits^1_0\tan^{-1}\frac{2(1-\text{x})-1}{1+(1-\text{x})-(1-\text{x})^2}\text{ dx}$

$=\int\limits^1_0\tan^{-1}\frac{1-2\text{x}}{2-\text{x}-1-\text{x}^2+2\text{x}}\text{ dx}$

$=\int\limits^1_0\tan^{-1}\frac{1-2\text{x}}{1+\text{x}-\text{x}^2}\text{ dx}$

$=-\int\limits^1_0\tan^{-1}\frac{2\text{x}-1}{1+\text{x}-\text{x}^2}\text{ dx}\ ....(\text{ii})$

Adding (i) and (ii)

$2\text{I}=\int\limits^1_0\tan^{-1}\frac{2\text{x}-1}{1+\text{x}-\text{x}^2}\text{ dx}-\int\limits^1_0\tan^{-1}\frac{2\text{x}-1}{1+\text{x}-\text{x}^2}\text{ dx}$

Hence, $\text{I}=0$

2. $\frac{1}{2}$

Solution:

$\int\limits^\alpha_0\frac{1}{1+4\text{x}^2}\text{ dx}=\frac{\pi}{8}$

$\Rightarrow\int\limits^\alpha_0\frac{1}{1+(2\text{x)}^2}\text{ dx}=\frac{\pi}{8}$

$\Rightarrow \frac{1}{2}\big[\tan^-2\text{x}\big]^\alpha_0=\frac{\pi}{8}$

$\Rightarrow \frac{1}{2}\tan^{-1}2\alpha=\frac{\pi}{8}$

$\Rightarrow 2\alpha=\tan\frac{\pi}{4}$

$\Rightarrow2\alpha=1$

$\therefore\alpha=\frac{1}{2}$

3. $0$

Solution:

$\int\limits^\pi_{-\pi}\sin^3\text{x}\cos^2\text{x}\text{ dx}$

$=\int\limits^\pi_{-\pi}\sin\text{x}(1-\cos^2\text{x})\cos^2\text{x}\text{ dx}$

Let $\cos\text{x}=\text{t},$ then $-\sin\text{x}\text{ dx}=\text{dt}$

When, $\text{x}=-\pi,\text{t}-1,\text{x}=\pi,\text{t}=-1$

Therefore the integral becomes

$\int\limits^{-1}_{-1}(1-\text{t}^2)\text{t}^2\text{ dt}$

$=0$

2. $\frac{\pi}{4}$

Solution:

We have, 

$\text{I}=\int\limits^\infty_0\frac{\text{x}}{(1+\text{x})(1+\text{x}^2)}\text{ dx}$

Putting $\text{x}=\tan\theta$

$\Rightarrow \text{dx}=\sec^2\theta\text{ d}\theta$

when $\text{x}\rightarrow0;\theta\rightarrow0$

and $\text{x}\rightarrow\infty;\theta\rightarrow\frac{\pi}{2}$

Now, integral becomes

$\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\tan\theta}{(1+\tan\theta)\sec^2\theta}\sec^2\theta\text{ d}\theta$

$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\tan\theta}{1+\tan\theta}\text{d}\theta$

$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\frac{\sin\theta}{\cos\theta}}{1+\frac{\sin\theta}{\cos\theta}}\text{d}\theta$

$\Rightarrow \text{I}=\int\limits^\frac{\pi}{2}_0\frac{\sin\theta}{\sin\theta+\cos\theta}\text{d}\theta\ ...(\text{i})$

$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\sin\big(\frac{\pi}{2}-\theta\big)}{\sin\big(\frac{\pi}{2}-\theta\big)+\cos\big(\frac{\pi}{2}-\theta\big)}\text{ d}\theta$ $\bigg[\therefore\ \int\limits^\text{a}_0\text{f}(\text{x)}\text{dx}=\int\limits^\text{a}_0\text{f}(\text{a}-\text{x})\text{dx}\bigg]$

$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\cos\theta}{\cos\theta+\sin\theta}\text{d}\theta$

$\Rightarrow \text{I}=\int\limits^\frac{\pi}{2}_0\frac{\cos\theta}{\sin\theta+\cos\theta}\text{d}\theta\ ....(\text{ii})$

Adding (i) and (ii), we get

$2\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\sin\theta+\cos\theta}{\sin\theta+\cos\theta}\text{d}\theta$

$\Rightarrow 2\text{I}=\int\limits^\frac{\pi}{2}_0\text{d}\theta$

$\Rightarrow 2\text{I}=\frac{\pi}{2}$

$\Rightarrow \text{I}=\frac{\pi}{4}$

3. 0

Solution

Let $\text{I}=\int\limits^\frac{\pi}{2}_0\log\Big(\frac{4+3\sin\text{x}}{4+3\cos\text{x}}\Big)\text{dx}\ ...(\text{i})$

$=\int\limits^\frac{\pi}{2}_0\log\Bigg[\frac{4+3\sin\big(\frac{\pi}{2}-\text{x}\big)}{4+3\cos\big(\frac{\pi}{2}-\text{x}\big)}\Bigg]\text{dx}$

$=\int\limits^\frac{\pi}{2}_0\log\Big(\frac{4+3\cos\text{x}}{4+3\sin\text{x}}\Big)\text{dx}\ ....(\text{ii})$

Adding (i) and (ii)

$2\text{I}=\int\limits^\frac{\pi}{2}_0\Big[\log\Big(\frac{4+3\sin\text{x}}{4+3\cos\text{x}}\Big)+\log\Big(\frac{4+3\cos\text{x}}{4+3\sin\text{x}}\Big)\Big]\text{dx}$

$=\int\limits^\frac{\pi}{2}_0\log1\text{ dx}=0$

Hence, $\text{I}=0$

4. $\frac{\text{a}+\text{b}}{2}\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx}$

Solution:

Let, $\text{I}=\int\limits^\text{b}_\text{a}\text{x}\text{ f}(\text{x})\text{dx}\ ....(\text{i})$

$=\int\limits^\text{b}_\text{a}(\text{a}+\text{b}-\text{x})\text{f}(\text{a}+\text{b}-\text{x})\text{dx}$

$=\int\limits^\text{b}_\text{a}(\text{a}+\text{b}-\text{x})\text{f}(\text{x})\text{dx}\ ....(\text{ii})$

Adding (i) and (ii)

$2\text{I}=\int\limits^\text{b}_\text{a}(\text{x}+\text{a}+\text{b}-\text{x})\text{f}(\text{x})\text{dx}$

$=(\text{a}+\text{b})\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx}$

Hence $\text{I}=\frac{\text{a}+\text{b}}{2}\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx}$

2. $\text{e}-1$

Solution:

Let, $\text{I}=\int\limits^\frac{\pi}{4}_0\cos\text{x}\text{ e}^{\sin\text{x}}\text{dx}$

Let $\sin\text{x}=\text{t},$ then $\cos\text{x}\text{ dx}=\text{dt}$

when $\text{x}=0,\text{t}=0$ and $\text{x}=\frac{\pi}{2},\text{t}=1$

Therefore the integrel becomes

$\text{I}=\int\limits^1_0\text{e}^\text{t}\text{dt}$

$=\big[\text{e}^\text{t}\big]^1_0$

$=\text{e}-1$

2. $\frac{\pi}{60}$

Solution:

$\int\limits^\infty_0\frac{1}{\big(\text{x}^2+4\big)\big(\text{x}^2+9\big)}\text{dx}$

$=\frac{1}{5}\int\limits^\infty_0\frac{1}{\big(\text{x}^2+4\big)}-\frac{1}{\big(\text{x}^2+9\big)}\text{dx}$

$=\frac{1}{5}\bigg[\frac{1}{2\tan^{-1}{}}\frac{\text{x}}{2}-\frac{1}{3}\tan^{-1}\frac{\text{x}}{3}\bigg]^\infty_0$

$=\frac{1}{5}\bigg[\frac{1}{2}\times\frac{\pi}{2}-\frac{1}{3}\times\frac{\pi}{2}\bigg]$

$=\frac{1}{5}\times\frac{\pi}{12}$

$=\frac{\pi}{60}$

3. $\pi$

Solution:

$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\big(\text{x}^3+\text{x}\cos\text{x}+\tan^5\text{x}+1\big)\text{dx}$

$=\Big[\frac{\text{x}^4}{4}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}+\Big[\text{x}\sin\text{x}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}-\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin\text{x}\text{dx}\\+\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\tan^3\text{x}(\sec^2\text{x}-1)\text{dx}+\Big[\text{x}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}$

$=\frac{\pi^4}{64}-\frac{\pi^4}{64}+\frac{\pi}{2}-\frac{\pi}{2}-\Big[-\cos\text{x}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}\\+\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\tan^3\text{x}\sec^2\text{x}\text{ dx}-\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\tan^3\text{x}\text{ dx}+\frac{\pi}{2}+\frac{\pi}{2}$

$=+0+\Big[\frac{\tan^4\text{x}}{4}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}-\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\tan\text{x}\sec^2\text{x}\text{ dx}-\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\tan\text{x}\text{ dx}$

$=\pi-\Big[\frac{\tan^2\text{x}}{2}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}-\Big[-\log\big(\cos\text{x}\big)\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}$

$=\pi$

1. 4

Solution:

We have,

$\text{I}=\int\limits^2_{-2}\big|1-\text{x}^2\big|\text{dx}$

$\big|1-\text{x}^2\big|=\begin{cases}-\big(1-\text{x}^{2}\big), & -2<\text{x}<-1\\\big(1-\text{x}^2\big), & -1<\text{x}<1 \\-\big(1-\text{x}^2\big),&1<\text{x}<2\end{cases}$

$\therefore\ \text{I}=\int\limits^{-1}_{-2}\big|1-\text{x}^2\big|\text{dx}+\int\limits^1_{-1}\big|1-\text{x}^2\big|\text{dx}+\int\limits^2_1\big|1-\text{x}^2\big|\text{dx}$

$=\int\limits^{-1}_{-2}-\big(1-\text{x}^2)\text{dx}+\int\limits^1_{-1}(1-\text{x}^2)\text{dx}+\int\limits^2_1-(1-\text{x}^2)\text{dx}$

$=-\int\limits^{-1}_{-2}(1-\text{x}^2)\text{dx}+\int\limits^1_{-1}(1-\text{x}^2)\text{dx}-\int\limits^2_1(1-\text{x}^2)\text{dx}$

$=\Big[\text{x}-\frac{\text{x}^3}{3}\Big]^{-1}_{-2}+\Big[\text{x}-\frac{\text{x}^3}{3}\Big]^1_{-1}-\Big[\text{x}-\frac{\text{x}^3}{3}\Big]^2_1$

$=\Big[-1+\frac{1}{3}+2-\frac{8}{3}\Big]+\Big[1-\frac{1}{3}+1-\frac{1}{3}\Big]-\Big[2-\frac{8}{3}-1+\frac{1}{3}\Big]$

$=-\Big[1-\frac{7}{3}\Big]+\Big[2-\frac{2}{3}\Big]-\Big[1-\frac{7}{3}\Big]$

$=-1+\frac{7}{3}+2-\frac{2}{3}-1+\frac{7}{3}$

$=4$

1. $\frac{\pi}{4}$

Solution:

$\int\limits^{\pi}_0\frac{1}{5+3\cos\text{x}}\text{ dx}$

$=\int\limits^{\pi}_0\frac{1}{5+3\frac{1-\tan^{2}\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}}}\text{ dx}$

$=\int\limits^{\pi}_0\frac{1+\tan^{2}\frac{\text{x}}{2}}{5+5\tan^{2}\frac{\text{x}}{2}+3-3\tan^{2}\frac{\text{x}}{2}}$

$=\int\limits^{\pi}_0\frac{\sec^2\frac{\text{x}}{2}}{8+2\tan^{2}\frac{\text{x}}{2}}\text{ dx}$

Let $\tan\frac{\text{x}}{2}=\text{t},$ then $\sec^2\frac{\text{x}}{2}\text{ dx}=2\text{dt}$

When $\text{x}=0,\text{ t}=0,\text{x}=\pi,\text{ t}=\infty$

Therefore the integral becomes

$\frac{1}{2}\int\limits^{\infty}_0\frac{\text{dt}}{4+\text{t}^2}$

$=\frac{1}{2}\Big[\tan^{-1}\frac{\text{t}}{2}\Big]^{\infty}_0$

$=\frac{1}{2}\Big(\frac{\pi}{2}-0\Big)$

$=\frac{\pi}{4}$

2. $0$

Solution:

$\int\limits^1_0(\text{a}-\text{x})^2\text{ f}(\text{x})\text{dx}$

$=\text{a}^2\int\limits^1_0\text{f}(\text{x})\text{dx}+\int\limits^1_0\text{x}^2\text{f}(\text{x})\text{dx}-2\text{a}\int\limits^1_0\text{x}\text{f}(\text{x})\text{dx}$

$=\text{a}^2\times1+\text{a}^2-2\text{aa}$ (As per given values)

$=2\text{a}^2-2\text{a}^2$

$=0$

2. $10\Big(\frac{\pi}{2}\Big)^9$

Solution:

We have,

$\text{I}_{10}=\int\limits^\frac{\pi}{2}_0\text{x}^{10}\sin\text{x}\text{ dx}$

$=\big[\text{x}^{10}(-\cos\text{x})\big]^\frac{\pi}{2}_0-\int\limits^\frac{\pi}{2}_0\big[10\text{x}^9\int\sin\text{x}\text{ dx}\big]\text{dx}$

$=\big[-\text{x}^{10}\cos\text{x}\big]^\frac{\pi}{2}_0-10\int\limits^\frac{\pi}{2}_0\text{x}^9(-\cos\text{x})\text{dx}$

$=-\big[\text{x}^{10}\cos\text{x}\big]^\frac{\pi}{2}_0+10\int\limits^\frac{\pi}{2}_0\text{x}^9\cos\text{x}\text{ dx}$

$=-\big[\text{x}^{10}\cos\text{x}\big]^\frac{\pi}{2}_0+10\big[\text{x}^9\sin\text{x}\big]^\frac{\pi}{2}_0-10\int\limits^\frac{\pi}{2}_09\text{x}^8\sin\text{x}\text{ dx}$

$=-\Big[\Big(\frac{\pi}{2}\Big)^{10}\times0-0^{10}\cos0\Big]+10\Big[\Big(\frac{\pi}{2}\Big)^9\times1-0^9\times0\Big]\\-90\int\limits^\frac{\pi}{2}_0\text{x}^8\sin\text{x dx}$

$=10\Big[\Big(\frac{\pi}{2}\Big)^9\times1\Big]-90\text{I}_8$

$=10\Big(\frac{\pi}{2}\Big)^9-90\text{I}_8$

$\therefore\ \text{I}_{10}+90\text{I}_8$

$=10\Big(\frac{\pi}{2}\Big)^9$

3. $8$

Solution:

$\int\limits^{2\pi}_0\sqrt{1+\sin\frac{\text{x}}{2}}\text{dx}$

$=\int\limits^{2\pi}_0\sqrt{\sin^2\frac{\text{x}}{4}+\cos^2\frac{\text{x}}{4}+2\sin\frac{\text{x}}{4}\cos\frac{\text{x}}{4}}\text{dx}$

$=\int\limits^{2\pi}_0\Big(\sin\frac{\text{x}}{4}+\cos\frac{\text{x}}{4}\Big)\text{dx}$

$=\Bigg[\frac{-\cos^\frac{\text{x}}{4}}{\frac{1}{4}}+\frac{\sin\frac{\text{x}}{4}}{\frac{1}{4}}\Bigg]^{2\pi}_0$

$=4\Big[\frac{\text{x}}{4}-\cos\frac{\text{x}}{4}\Big]^{2\pi}_0$

$=4\Big[\sin\frac{2\pi}{4}-\cos\frac{2\pi}{4}-\sin0+\cos0\Big]$

$=4\Big[\sin\frac{\pi}{2}-\cos\frac{\pi}{2}-0+1\Big]$

$=4\big[1-0-0+1\big]$

$=4\times2$

$=8$

3. 0

Solution:

$\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin2\text{x }\log\tan\text{x dx}\ ....(\text{i})$

$\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin(\pi-2\text{x})\log\tan\big(\frac{\pi}{2}-\text{x}\big)\text{dx}$

$\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin2\text{x}\log\cot\text{x dx}\ ...(\text{ii})$

Adding (i) and (ii) we get

$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin2\text{x}\big(\log\tan\text{x}+\log\cot\text{x}\big)\text{dx}$

$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin2\text{x}\big(\log\tan\text{x}\cot\text{x}\big)\text{dx}$

$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin\text{x}(\log1)\text{dx}$

$\text{I}=0$

2. $2$

Solution:

$\int\limits^1_{-1}|1-\text{x}|\text{dx}$

$=\int\limits^0_{-1}(1-\text{x})\text{dx}+\int\limits^1_0(1-\text{x})\text{dx}$

$=\Big[\text{x}-\frac{\text{x}^2}{2}\Big]^0_{-1}+\Big[\text{x}-\frac{\text{x}^2}{2}\Big]^1_0$

$=0+1+\frac{1}{2}+1-\frac{1}{2}-0$

$=2$

4. $-\frac{16}{3}$

solution:

$\text{I}=\int\limits^1_0\frac{\text{x}}{(1-\text{x})^{\frac{5}{4}}}\text{ dx}$

Put, 1 - x = t ⇒ x =1 -t

⇒ dx = -dt

$\text{I}=\int\limits^0_1\frac{(1-\text{t})(-\text{dt})}{\text{t}^{\frac{5}{4}}}$

$\text{I}=\int\limits^1_0\Big(\text{t}^\frac{5}{4}-\text{t}^\frac{-1}{4}\Big)\text{dt}$

$\text{I}=\Bigg[\frac{\text{t}^{-\frac{5}{4}}}{\frac{-1}{4}}-\frac{\text{t}^\frac{3}{4}}{\frac{3}{4}}\Bigg]^1_0$

$\text{I}=-4-\frac{4}{3}$

$\text{I}=\frac{-16}{3}$

1. $\pi\ln 2$

Solution:

$\int\limits^{\infty}_0\log\Big(\text{x}+\frac{1}{\text{x}}\Big)\frac{1}{1+\text{x}^2}\text{ dx}$

Substitute $\text{x}=\tan\theta$

$\text{dx}=\sec^2\theta\text{d}\theta$

When,

$\text{x}=0\Rightarrow\theta=0$

$\text{x}=\infty\Rightarrow\theta=\frac{\pi}{2}$

$\int\limits^{\frac{\pi}{2}}_0\Big(\tan\theta+\frac{1}{\tan\theta}\Big)\frac{1}{1+\tan^{2}}\times\sec^2\theta\text{d}\theta$

$\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{\tan^2\theta+1}{\tan\theta}\Big)\frac{1}{1+\tan^{2}\theta}\times\sec^2\theta\text{d}\theta$

$\Rightarrow\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{\sec^2\theta}{\tan\theta}\Big)\frac{1}{\sec^2\theta}\times\sec^2\theta\text{d}\theta$ $\big[\because1+\tan^2\theta=\sec^2\theta\big]$

$\Rightarrow\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{\sec^2\theta}{\tan\theta}\Big)\text{d}\theta$

$\Rightarrow\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{1}{\sin\theta\cdot\cos\theta}\Big)\text{d}\theta$

$\Rightarrow\int\limits^{\frac{\pi}{2}}_0\log\big(\sin\theta\cos\theta\big)\text{d}\theta$

$\Rightarrow-\int\limits^{\frac{\pi}{2}}_0\big[\log\sin\theta+\log\cos\theta\big]\text{d}\theta$

$\Rightarrow-\int\limits^{\frac{\pi}{2}}_0\log\sin\theta\text{d}\theta-\int\limits^{\frac{\pi}{2}}_0\log\cos\theta\text{d}\theta$

Let us conside,

$\int\limits^{\frac{\pi}{2}}_0\log\sin\theta\text{d}\theta=\text{I}\ ....(\text{i})$

$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\log\Big(\sin\Big(\frac{\pi}{2}-\theta\Big)\Big)\text{d}\theta$

$\Rightarrow\int\limits^{\frac{\pi}{2}}_0\log\cos\theta\text{d}\theta\ ...(\text{ii})$

Adding (i) and (ii)

$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\log\sin\theta\text{ d}\theta+\int\limits^{\frac{\pi}{2}}_0\log\cos\theta\text{ d}\theta$

$=\int\limits^{\frac{\pi}{2}}_0\log(\sin\theta\cdot\cos\theta)\text{d}\theta$

$=\int\limits^{\frac{\pi}{2}}_0\log\big(\sin2\theta\big)\text{d}\theta-\int\limits^{\frac{\pi}{2}}_0\log2\text{ d}\theta$

Let us consider $2\theta=\text{t}$

$2\text{d}\theta=\text{dt}$

$2\text{I}=\frac{1}{2}\int\limits^{\pi}_{0}\log(\sin\text{t})\text{dt}-\frac{\pi}{2}\log2$

$2\text{I}=\frac{2}{2}\int\limits^{\pi}_{0}\log(\sin\text{t})\text{dt}-\frac{\pi}{2}\log2$ $\big[\because\sin\theta$ is positive in both 1^st and 2^nd quadrants$\big]$

$2\text{I}=\text{I}-\frac{\pi}{2}\log2$

$2\text{I}-\text{I}=-\frac{\pi}{2}\log2$

$\text{I}=-\frac{\pi}{2}\log2,$ where $\text{I}=\int\limits^{\frac{\pi}{2}}_0\log\sin\theta\text{ d}\theta$

Now,

$-\int\limits^{\frac{\pi}{2}}_0\log(\sin\theta)\text{ d}\theta-\int\limits^{\frac{\pi}{2}}_0\log\cos\theta\text{ d}\theta$

$-2\int\limits^{\frac{\pi}{2}}_0\log\sin\theta\text{ d}\theta=-2\times\text{I}$

$=-2\times-\frac{\pi}{2}\log2$ $\Big[\text{Where I}=-\frac{\pi}{2}\log2\Big]$

$=\pi\log2$

1. $\sqrt{1-\pi^2}-1$

Solution:

We have,

$\text{I}=\int\limits^\pi_0\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{dx}$

$=\int\limits^\pi_0\Big[\sqrt{\frac{1-\text{x}}{1+\text{x}}}\times\frac{\sqrt{1-\text{x}}}{\sqrt{1-\text{x}}}\Big]\text{dx}$

$=\int\limits^\pi_0\frac{1-\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}$

$=\int\limits^\pi_0\frac{1}{\sqrt{1-\text{x}^2}}\text{dx}-\int\limits^\pi_0\frac{\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}$

Putting $1 -\text{x}^2=\text{t}$

$\Rightarrow - 2\text{x}\text{ dx}=\text{dt}$

$\Rightarrow \text{x}\text{ dx}=\frac{-\text{dt}}{2}$

when $\text{x}\rightarrow0;\text{t}\rightarrow1$

and $\text{x}\rightarrow\pi;\text{ t}\rightarrow-\pi^2$

$\therefore\ \text{I}=\int\limits^\pi_0\frac{1}{\sqrt{1-\text{x}^2}}-\int\limits^{(1-\pi^2)}_1\frac{-\text{dt}}{2\sqrt{\text{t}}}$

$=\big[\sin^{-1}\text{x}\big]^{\pi}_0+\frac{2}{2}\big[\sqrt{\text{t}}\big]^{1-\pi^2}_1$

$=\big[0-0\big]+\Big[\sqrt{1-\pi^2}-\sqrt{1}\Big]$

$=\sqrt{1-\pi^2}-1$

3. $2$

Solution:

$\int\limits^{\pi}_0\frac{1}{1+\sin\text{x}}\text{ dx}$

$=\int\limits^\pi_0\frac{1}{1+\sin\text{x}}\times\frac{1-\sin\text{x}}{1-\sin\text{x}}\text{dx}$

$= \int\limits^\pi_0\frac{1-\sin\text{x}}{1-\sin^2\text{x}}\text{dx}$

$= \int\limits^\pi_0\frac{1-\sin\text{x}}{\cos^2\text{x}}\text{dx}$

$=\int\limits^\pi_0(\sec^2\text{x}-\sec\text{x}\tan\text{x})\text{dx}$

$=\big[\tan\text{x}-\sec\text{x}\big]^\pi_0$

$=0+1-0+1$

$=2$

2. $\log2$

Solution:

We have,

$\text{I}=\int\limits^\infty_0\frac{1}{1+\text{e}^\text{x}}\text{ dx}$

Putting $\text{e}^\text{x}=\text{t}$

$\Rightarrow \text{e}^\text{x}\text{ dx}=\text{dt}$

$\Rightarrow \text{dx} = \frac{\text{dt}}{\text{t}}$

When $\text{x}\rightarrow0;\text{ t}\rightarrow1$

and $\text{x}\rightarrow\infty;\text{ t}\rightarrow\infty$

$\therefore\text{I}=\int\limits^\infty_1\frac{1}{\text{t}(1+\text{t})}\text{dt}$

$=\int\limits^\infty_1\frac{1}{\text{t}+\text{t}^2}\text{dt}$

$=\int\limits^\infty_1\frac{1}{\big(\text{t}+\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}\text{dt}$

$=\frac{1}{2\times\frac{1}{2}}\Bigg[\log\Bigg|\frac{\text{t}+\frac{1}{2}-\frac{1}{2}}{\text{t}+\frac{1}{2}+\frac{1}{2}}\Bigg|\Bigg]^\infty_1$

$=\Big[\log\Big|\frac{\text{t}}{\text{t+1}}\Big|\Big]^\infty_1$

$=\Bigg[\log\Bigg|\frac{\frac{\text{t}}{\text{t}}}{\frac{\text{t}}{\text{t}}+\frac{1}{\text{t}}}\Bigg|\Bigg]^\infty_1$

$=\Bigg[\log\Bigg|\frac{1}{1+\frac{1}{\text{t}}}\Bigg|\Bigg]^\infty_1$

$=\log\frac{1}{1+0}-\log\frac{1}{1+1}$

$=\log(1)-\log(\frac{1}{2})$

$=0-(-\log2)$

$=\log2$

1. $2$

Solution:

$\int\limits^\frac{\pi^2}{4}_0\frac{\sin\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{dx}$

Let $\sqrt{\text{x}}=\text{t},$ then $\frac{1}{2\sqrt{\text{x}}}\text{dx}=\text{dt}$

when $\text{x}=0,\text{t}=0,\text{x}=\frac{\pi^2}{4},\text{t}=\frac{\pi}{2}$

Therefore the integral becomes

$\int\limits^\frac{\pi}{2}_02\sin\text{t}\text{ dt}$

$=-2\big[\cos\text{t}\big]^\frac{\pi}{2}_0$

$=2$

2. $2$

Solution:

$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin|\text{x}|\text{dx}$

$=-\int\limits^0_{-\frac{\pi}{2}}\sin\text{x}\text{ dx}+\int\limits^\frac{\pi}{2}_0\sin\text{x}\text{ dx}$

$=-\big[-\cos\text{x}\big]^0_{-\frac{\pi}{2}}+\big[-\cos\text{x}\big]^\frac{\pi}{2}_0$

$=1-0-0+1$

$=2$

1.  $\frac{\pi}{\sqrt{\text{a}^2-\text{b}^2}}$

Solution:

We have,

$\text{I}=\int\limits^\pi_0\frac{1}{\text{a}+\text{b}\cos\text{x}}\text{ dx}$

$=\int\limits^\pi_0\frac{1}{\text{a}+\text{b}\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}}\text{ dx}$

$=\int\limits^\pi_0\frac{1+\tan^2\frac{\text{x}}{2}}{\text{a}\Big(1+\tan^2\frac{\text{x}}{2}\Big)+\Big(1-\tan^2\frac{\text{x}}{2}\Big)}\text{ dx}$

$=\int\limits^\pi_0\frac{1+\tan^2\frac{\text{x}}{2}}{(\text{a}+\text{b})+(\text{a}-\text{b})\tan^2\frac{\text{x}}{2}}\text{ dx}$

$=\int\limits^\pi_0\frac{\sec^2\frac{\text{x}}{2}}{(\text{a}+\text{b})+(\text{a}-\text{b})\tan^2\frac{\text{x}}{2}}\text{ dx}$

putting $\tan\frac{\text{x}}{2}=\text{t}$

$\Rightarrow \frac{1}{2}\sec^2\frac{\text{x}}{2}\text{dx}=\text{dt}$

$\Rightarrow\sec^2\frac{\text{x}}{2}\text{dx}=2\text{ dt}$

when $\text{x}\rightarrow0;\text{ t}\rightarrow0$

and $\text{x}\rightarrow\pi;\text{ t}\rightarrow\infty$

$\therefore\ \text{I}=\int\limits^\pi_0\frac{2\text{dt}}{(\text{a}+\text{b})+(\text{a}-\text{b})\text{t}^2}$

$=\frac{2}{\text{a}-\text{b}}\int\limits^\pi_0\frac{1}{\big(\frac{\text{a}+\text{b}}{\text{a}-\text{b}}\big)+\text{t}^2}\text{dt}$

$=\frac{2}{(\text{a}-\text{b})}\int\limits^\infty_0\frac{1}{\Big(\sqrt{\frac{\text{a}+\text{b}}{\text{a}-\text{b}}}\Big)^2+\text{t}^2}\text{ dt}$

$=\frac{2}{(\text{a}-\text{b})}\times\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}$ $\Bigg[\tan^{-1}\frac{\text{t}}{\sqrt{\frac{\text{a}+\text{b}}{\text{a}-\text{b}}}}\Bigg]^\infty_0$

$=\frac{2}{\sqrt{\text{a}^2-\text{b}^2}}\Big[\frac{\pi}{2}\Big]$

$=\frac{\pi}{\sqrt{\text{a}^2-\text{b}^2}}$

4. $\frac{\pi}{4}$

Solution:

We have,

$\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\cot^3\text{x}}\text{ dx}\ ...(\text{i})$

$=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\cot^3\big(\frac{\pi}{2}-{\text{x}}\big)}\text{ dx}$

$\therefore\ \text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\tan^3\text{x}}\text{ dx}\ ...(\text{ii})$

Adding (i) and (ii) we get

$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\Big[\frac{1}{1+\cot^3\text{x}}+\frac{1}{\tan^3\text{x}}\Big]\text{ dx}$

$=\int\limits^{\frac{\pi}{2}}_0\bigg[\frac{2+\tan^3\text{x}+1+\cot^3\text{x}}{(1+\cot^3\text{x})(1+\tan^3\text{x})}\bigg]\text{dx}$

$=\int\limits^{\frac{\pi}{2}}_0\bigg[\frac{2\tan^3\text{x}+\cot^3\text{x}}{1+\tan^3\text{x}+\cot^3\text{x}+\cot^3\text{x}\tan^3\text{x}}\bigg]\text{dx}$

$=\int\limits^{\frac{\pi}{2}}_0\Big[\frac{2\tan^3\text{x}+\cot^3\text{x}}{1+\tan^3\text{x}+\cot^3\text{x}+1}\Big]\text{dx}$

$=\int\limits^{\frac{\pi}{2}}_0\Big[\frac{2+\tan^3\text{x}+\cot^3\text{x}}{2+\tan^3\text{x}+\cot^3\text{x}}\Big]\text{dx}$

$=\int\limits^{\frac{\pi}{2}}_0\Big[1\Big]^{\frac{\pi}{2}}_0$

$=\Big[\text{x}\Big]^{\frac{\pi}{2}}_0$

$=\frac{\pi}{2}$

Hence, $\text{I}=\frac{\pi}{4}$

4. $\log\Big(\frac{4}{3}\Big)$

Solution:

Let, $\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\cos\text{x}}{(2+\sin\text{x})(1+\sin\text{x})}\text{dx}$

Let $\sin\text{x}$ then $\cos\text{x}\text{ dx}=\text{dt}$

When $\text{x}=0,\text{t}=0,\text{x}=\frac{\pi}{2},\text{t}=1$

Therefore the integral becomes

$\text{I}=\int\limits^1_0\frac{\text{dt}}{(2+\text{t})(1+\text{t})}$

$=\int\limits^1_0\Big[\frac{-1}{2+\text{t}}+\frac{1}{1+\text{t}}\Big]\text{dt}$

$=\big[-\log(2+\text{t})+\log(1+\text{t})\big]^1_0$

$= \big[\log(1+\text{t})-\log(2+\text{t})\big]^1_0$

$=\log2-\log3-\log1+\log2$

$=\log\frac{4}{3}$

3. $\int\limits^{\text{a}}_0\text{f}(\text{x})\text{dx}+\int\limits^{\text{a}}_0\text{f}(2\text{a}-\text{x})\text{dx}$

Solution:

$\int\limits^\text{a}_0\text{f}(\text{x})\text{dx}+\int\limits^\text{a}_0\text{f}(2\text{a}-\text{x})\text{dx}$

According to the additivity property of integrals,

$\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx}=\int\limits^\text{c}_\text{a}\text{f}(\text{x})+\int\limits^\text{b}_\text{c}\text{f}(\text{x})\text{dx},$ where a < c < b

Using this property,

$\int\limits^{2\text{a}}_0\text{f}(\text{x})\text{dx}=\int\limits^\text{a}_0\text{f}(\text{x})\text{dx}+\int\limits^{2\text{a}}_0\text{f}(\text{x})\text{dx}\ ....(\text{i})$

Now, consider the integral, $\int\limits^{2\text{a}}_0\text{f}(\text{x})\text{dx}$

Let x = 2a - t Then dx = d (2a - t), dx = - dt

Also, x = a, t = a and x = 2a, t = 0

Therefore, $\int\limits^{2\text{a}}_\text{a}\text{f}(\text{x})\text{dx}=-\int\limits^0_\text{a}\text{f}(2\text{a}-\text{t})\text{dt}$

$\Rightarrow \int\limits^{2\text{a}}_\text{a}\text{f}(\text{x})\text{dx}=\int\limits^\text{a}_0\text{f}(2\text{a}-\text{t})\text{dt}$

$\Rightarrow \int\limits^{2\text{a}}_\text{a}\text{f}(\text{x})\text{dx}=\int\limits^\text{a}_0\text{f}(2\text{a}-\text{x})\text{dx}$

Substituting this in equation (i) we get,

$\int\limits^{2\text{a}}_0\text{f}(\text{x})\text{dx}=\int\limits^\text{a}_0\text{f}(\text{x})\text{dx}+\int\limits^\text{a}_0\text{f}(2\text{a}-\text{x})\text{dx}$

4. $\frac{\pi}{4}$

Solution:

We have,

$\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{dx}\ ...(\text{i})$

$\Rightarrow \text{I}=\int\limits^\frac{\pi}{2}_0\frac{\sin\big(\frac{\pi}{2}-\text{x}\big)}{\sin\big(\frac{\pi}{2}-\text{x}+\cos\big(\frac{\pi}{2}-\text{x}\big)}\text{dx}$

$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\cos\text{x}}{\cos\text{x}+\sin\text{x}}\text{dx}$

$\therefore\ \text{I}=\int\limits^\frac{\pi}{2}_0\frac{\cos\text{x}}{\sin\text{x}+\cos\text{x}}\text{dx}\ ...(\text{ii})$

Adding (i) and (ii), we get

$2\text{I}=\int\limits^\frac{\pi}{2}_0\Big[\frac{\sin\text{x}}{\sin{\text{x}}+\cos\text{x}}+\frac{\cos\text{x}}{\cos\text{x}+\sin\text{x}}\Big]\text{dx}$

$=\int\limits^\frac{\pi}{2}_0\Big[\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}+\cos\text{x}}\Big]\text{dx}$

$=\int\limits^\frac{\pi}{2}_0\text{dx}$

$=\big[\text{x}\big]^\frac{\pi}{2}_0$

$=\frac{\pi}{2}$

Hence $\text{I}=\frac{\pi}{4}$

3. $\frac{\pi}{12}$

Solution:

Let, $\text{I}=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\frac{1}{1+\sqrt{\cot\text{x}}}\text{dx}\ ...{\text{(i)}}$

$=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\frac{1}{\sqrt{\cot\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}}\text{dx}$ $\bigg[\text{using}\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx}=\int\limits^\text{b}_\text{a}\text{f}\big(\text{a}+\text{b}-\text{x}\big)\text{dx}\bigg]$

$=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\frac{1}{1+\sqrt{\tan\text{x}}}\text{dx}\ ...(\text{ii})$

Adding (i) and (ii) we get

$2\text{I}=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\bigg[\frac{1}{1+\sqrt{\cot\text{x}}}+\frac{1}{1+\sqrt{\tan\text{x}}}\bigg]\text{dx}$

$=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\frac{2+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}{\big(1+\sqrt{\cot\text{x}}\big)+\big(1+\sqrt{\tan\text{x}}\big)}\text{ dx}$

$=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\Bigg[\frac{2+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}{2+\sqrt{\cot\text{x}+\sqrt{\tan\text{x}}}}\Bigg]\text{dx}$

$=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\text{dx}$

$=\big[\text{x}\big]^\frac{\pi}{3}_\frac{\pi}{6}$

$=\frac{\pi}{3}-\frac{\pi}{6}$

$=\frac{\pi}{6}$

Hence, $\text{I}=\frac{\pi}{12}$

3.  $\ln\Big(\frac{3}{2}\Big)$

Solution:

$\lim\limits_{\text{n}\rightarrow\infty}\Big\{\frac{1}{2\text{n}+1}+\frac{1}{2\text{n}+2}+\ .....\ +\frac{1}{2\text{n}+\text{n}}\Big\}$

$=\lim\limits_{\text{n}\rightarrow\infty}\sum\limits^\text{n}_{\text{r}=1}\frac{1}{2\text{n}+\text{r}}$

$=\lim\limits_{\text{n}\rightarrow\infty}\frac{1}{\text{n}}\sum\limits^\text{n}_{\text{r}=1}\frac{1}{2+\frac{\text{r}}{\text{n}}}$

let $\frac{\text{r}}{\text{n}}=\text{x}$

$=\int\limits^\infty_0\frac{1}{2+\text{x}}\text{dx}$

$=\Big[\log\big(2+\text{x}\big)\Big]^\infty_0$

$=\log3-\log2$

$=\log\frac{3}{2}$

$=\ln\Big(\frac{3}{2}\Big)$

3. $\big(\ln\text{x}\big)^{-1}\text{x}(\text{x}-1)$

Solution:

$\text{f}'(\text{x})=\frac{1}{\log_\text{e}\text{x}^3}(3\text{x}^2)-\frac{1}{\log_\text{e}\text{x}^2}(2\text{x})$

$=\frac{3\text{x}^2}{3\ln\text{ x}}-\frac{2\text{x}}{2\ln\text{ x}}$

$=\frac{\text{x}^2}{\ln\text{ x}^{-1}}-\frac{\text{x}}{\ln\text{ x}}$

$=\frac{1}{\ln\text{ x}}\text{x}(\text{x}-1)$

$=\big(\ln\text{ x}\big)^{-1}\text{x}(\text{x}-1)$

3. $\frac{\pi}{4}$

Solution:

$\text{ I}=\int\limits^\frac{\pi}{2}_0\frac{\sqrt{\cos\text{x}}}{\sqrt\cos\text{x}+\sqrt{\sin\text{x}}}\text{dx}\ ...(\text{i})$

$=\int\limits^\frac{\pi}{2}_0\frac{\sqrt{\cos\big(\frac{\pi}{2}-\text{x}\big)}}{\sqrt{\cos\big(\frac{\pi}{2}-\text{x}\big)+\sqrt{\sin\big(\frac{\pi}{2}-\text{x}\big)}}}\text{dx}$

$=\int\limits^\frac{\pi}{2}_0\frac{\sqrt{\sin\text{x}}}{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}\text{dx}$

$=\int\limits^\frac{\pi}{2}_0\frac{\sqrt{\sin{\text{x}}}}{\sqrt{\cos\text{x}}+\sqrt{\sin\text{x}}}\text{dx}...(\text{ii})$

Adding (i) and (ii)

$2\text{I}=\int\limits^\frac{\pi}{2}_0\bigg[\frac{\sqrt{\cos\text{x}}}{\sqrt{\cos\text{x}}+\sqrt{\sin\text{x}}}+\frac{\sqrt{\sin\text{x}}}{\sqrt{\cos\text{x}}+\sqrt{\sin\text{x}}}\bigg]\text{dx}$

$=\int\limits^\frac{\pi}{2}_0\text{dx}$

$=[\text{x}]^\frac{\pi}{2}_0=\frac{\pi}{2}$

Hence, $\text{I}=\frac{\pi}{4}$