Maths M.C.Q (1 Marks)

4. 2

Solution:

$\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sec^2\text{x dx}$

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

$\Rightarrow\Big[\tan\big(\frac{\pi}{4}\big)-\tan\big(-\frac{\pi}{4}\big)\Big]$

$\Rightarrow[1-(-1)]$

$\Rightarrow2$

1. $\log2$

1. $\frac{-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+\text{C}$

Solution:

$\text{I}=\int\frac{2}{(\text{e}^{\text{x}}+\text{e}^{-\text{x}})^2}\text{ dx}$

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

Put $\text{t}=\text{e}^{2\text{x}}+1$

$\text{dt}=2\text{e}^{2\text{x}}\text{ dx}$

$\text{I}=\int\frac{\text{dt}}{\text{t}^2}$

$\text{I}=\frac{-1}{\text{t}}+\text{C}$

$\text{I}=\frac{-1}{\text{e}^{2\text{x}}+1}+\text{C}$

$\text{I}=\frac{-\frac{1}{\text{e}^{\text{x}}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+\text{C}$

$\text{I}=\frac{-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+\text{C}$

3. $\tan(\text{xe}^{\text{x}})+\text{C}$

Solution:

$\text{I}=\int\frac{\text{e}^{\text{x}}(1+\text{x})}{\cos^{2}(\text{xe}^{\text{x}})}\text{ dx}$

Put $\text{xe}^{\text{x}}=\text{t}$

$\text{e}^{\text{x}}(1+\text{x})\text{dx}=\text{dt}$

$\text{I}=\int\frac{\text{dt}}{\cos^2\text{t}}$

$=\text{I}=\int\sec^2\text{t dt}$

$\text{t}=\tan\text{t}+\text{C}$

 $\text{I}=\tan(\text{xe}^{\text{x}})+\text{C}$

1. $2(\sin\text{x}+\text{x}\cos\theta)+\text{c}$

1. 6

$\text{Let}\ \text{I}=\int\limits_{\frac{1}{3}}^{1}\frac{(\text{x}-\text{x}^{3})^{\frac{1}{3}}}{\text{x}^{4}}\text{dx}=\int\limits_{3}^{1}\frac{\text{x}\bigg(\frac{1}{\text{x}^{2}}-1\bigg)^{\frac{1}{3}}}{\text{x}^{4}}\text{dx} $

$=\int\limits_{\frac{1}{3}}^{1}\frac{\bigg(\frac{1}{\text{x}^{2}}-1\bigg)^{\frac{1}{3}}}{\text{x}^{3}}\text{dx}=\int\limits_{\frac{1}{3}}^{1}\bigg(\frac{1}{\text{x}^{2}}-1\bigg)^{\frac{1}{3}}.\frac{1}{\text{x}^{3}}\text{dx}$

$\text{put}\frac{1}{\text{x}^{2}}=\text{t},\ \therefore-\frac{2}{\text{x}^{3}}\text{dx}=\text{dt}\Rightarrow\frac{1}{\text{x}^{3}}\text{dx}=-\frac{1}{2}\text{dt}$

$\text{when}\ \text{x}=\frac{1}{3},\text{t}=9$

$\text{when}\ \text{x}=1,\text{t}=1$

$\therefore\ \ \text{I}=-\frac{1}{2}\int^{1}_{9}\big(\text{t}-1\big)^{\frac{1}{3}}\text{dt}=-\frac{1}{2}\Bigg[\frac{\big(\text{t}-1\big)^{\frac{4}{3}}}{\frac{4}{3}}\Bigg]^{1}_{0}$

$=-\frac{3}{8}\bigg[\big(\text{t}-1\big)^{\frac{4}{3}}\bigg]^{1}_{9}=-\frac{3}{8}\bigg[\big(1-1\big)^{\frac{4}{3}}-\big(9-1\big)^{\frac{4}{3}}\bigg]$

$=-\frac{3}{8}\bigg[{0}-(8)^{\frac{4}{3}}\bigg]^{1}_{9}=\frac{3}{8}(2)^{3^{\frac{4}{3}}}=\frac{3}{8}\text{x}\ 2^{4}=\frac{3}{8}\text{x}16=6$

$\text{Let}\ \text{I}=\int^{1}_{\frac{1}{3}}\frac{\big(\text{x}-\text{x}^{3}\big)^{\frac{1}{3}}}{\text{x}^{4}}\text{dx}=\int^{1}_{3}\frac{\text{x}\bigg(\frac{1}{\text{x}^{2}}-1\bigg)^{\frac{1}{3}}}{\text{x}^{4}}\text{dx}$

$=\int^{1}_{\frac{1}{3}}\frac{\bigg(\frac{1}{\text{x}^{2}}-1\bigg)^{\frac{1}{3}}}{\text{x}^{3}}\text{dx}=\int^{1}_{\frac{1}{3}}\bigg(\frac{1}{\text{x}^{2}}-1\bigg)^{\frac{1}{3}}.\frac{1}{\text{x}^{3}}\text{dx}$

$\text{put}\frac{1}{\text{x}^{2}}=\text{t},\ \therefore-\frac{2}{\text{x}^{3}}\text{dx}=\text{dt}\Rightarrow\frac{1}{\text{x}^{3}}\text{dx}=-\frac{1}{2}\text{dt}$

$\text{when}\ \text{x}=\frac{1}{3},\text{t}=9$

$\text{when}\ \text{x}=1,\text{t}=1$

$\therefore\ \ \text{I}=-\frac{1}{2}\int^{1}_{9}\big(\text{t}-1\big)^{\frac{1}{3}}\text{dt}=-\frac{1}{2}\bigg[\frac{\text{(t}-1)^{\frac{4}{3}}}{\frac{4}{3}}\bigg]^{1}_{0}$

$=-\frac{3}{8}\bigg[(\text{t}-1)^{\frac{4}{3}}\bigg]^{1}_{9}=-\frac{3}{8}\bigg[(1-1)^{\frac{4}{3}}-(9-1)^{\frac{4}{3}}\bigg]$

$=-\frac{3}{8}\bigg[0-(8)^{\frac{4}{3}}\bigg]=-\frac{3}{8}(2)^{3^{\frac{4}{3}}}=\frac{3}{8}\times2^{4}=\frac{3}{8}\times16=6$

1. $\frac{1}{3}(1+\text{log})^3+\text{c}$

1. 4

Solution:

We have,

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

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

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

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

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

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

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

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

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

$=4$

4. $\frac{\text{x}^3}{3}-\text{x}+2\tan^{-1}\text{x}+\text{c}$

4. $\sqrt{\frac{\text{x}}{3}}$

1. $\frac{-1}{2\text{x}^{2}}+\text{c}$

Solution:

we have $\frac{\text{dy}}{\text{dx}}=\text{x}^{-3}$

⇒ dy = x ^-3 dx

Integrating both sides 

$\therefore\text{y}=\int\text{x}^{-3}\text{dx}=\frac{\text{x}^{-3 + 1}}{-3+1}+\text{c}=-\frac{1}{2\text{x}^2}+\text{c}$

2. $\frac{1}{10100}$

1. $\tan\frac{\text{x}}{2}+\text{k}$

2. An even function

Solution:

The anti derivative of an odd function is even. Let f(x) be odd

eg = f(x) = x odd function

$\int\text{xdx}=\frac{\text{x}^2}{2}+\text{c}$

$\text{g}'(\text{x})=\frac{{\text{x}}^{2}}{\text{x}}+\text{c}$ is even.

2. $\text{x}$

2. $\text{x + c}$

Solution:

$\int\frac{\sin\text{x} + \cos\text{x}}{\sqrt{(1 + \sin2\text{x})}}\text{dx}$

$\int\frac{\sin\text{x} + \cos\text{x}}{\sqrt{(\sin\text{x}+\cos\text{x})^2}}\text{dx}$

$=\int1\text{dx}$

$=\text{x + c}$

3. $\text{a}=\frac{-1}{10},\text{b}=\frac{2}{5}$

Solution:

Given that, $\int\frac{\text{dx}}{(\text{x}+2)(\text{x}^2+1)}=\text{a}\log|1+\text{x}^2|+\text{b}\tan^{-1}\text{x}+\frac{1}{5}\log|\text{x}+2|+\text{C}$

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

$\frac{1}{(\text{x}+2)(\text{x}^2+1)}=\frac{\text{A}}{\text{x}+2}+\frac{\text{Bx}+\text{C}}{\text{x}^2+1}$

$\Rightarrow1=\text{A}(\text{x}^2+1)+(\text{Bx}+\text{C})(\text{x}+2)$

$\Rightarrow1=\text{Ax}^2+\text{A}+\text{Bx}^2+2\text{Bx}+\text{Cx}+2\text{C}$

$\Rightarrow1=(\text{A}+\text{B})\text{x}^2+(2\text{B}+\text{C})\text{x}+\text{A}+2\text{C}$

$\Rightarrow\text{A}+\text{B}=0,\text{A}+2\text{C}=1,2\text{B}+\text{C}=0$

We have, $\text{A}=\frac{1}{5},\text{B}=-\frac{1}{5}$ and $\text{C}=\frac{2}{5}$

$\therefore\ \int\frac{\text{dx}}{(\text{x}+2)(\text{x}^2+1)}=\frac{1}{5}\int\frac{1}{\text{x}+2}\text{dx}+\int\frac{-\frac{1}{5}\text{x}+\frac{2}{5}}{\text{x}^2+1}\text{dx}$

$=\frac{1}{5}\int\frac{1}{\text{x}+2}\text{dx}-\frac{1}{5}\int\frac{\text{x}}{1+\text{x}^2}\text{dx}+\frac{1}{5}\int\frac{2}{1+\text{x}^2}\text{dx}$

$=\frac{1}{5}\log|\text{x}+2|-\frac{1}{10}\log|1+\text{x}^2|+\frac{2}{5}\tan^{-1}\text{x}+\text{C}$

$\therefore\ \text{b}=\frac{2}{5}$ and $\text{a}=\frac{-1}{10}$

3. $\theta-\tan\theta+\text{c}$

3. $\frac{1}{3}\tan^{-1}(3\text{x})+\text{C}$

2. $\text{e}^\text{x}\sec\text{x}+\text{C}$

$\int\text{e}^\text{x}\sec\text{x}(1+\tan\text{x})\text{dx}$

Let $\text{I}=\int\text{e}^\text{x}\sec\text{x}(1+\tan\text{x})\text{dx}=\int\text{e}^\text{x}(\sec\text{x}+\sec\text{x}\tan\text{x})\text{dx}​​$

Also, let $\sec\text{x}=\text{f}(\text{x})\Rightarrow \ \sec\text{x}\tan\text{x}=\text{f}'(\text{x})$

It is known that, $\int\text{e}^\text{x}\{\text{f}(\text{x})+\text{f}'(\text{x})\}\text{dx}=\text{e}^\text{x}\text{f}(\text{x})+\text{C}$

$\therefore\ \text{I}=\text{e}^\text{x}\sec\text{x}+\text{C}$​​​​​​​

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

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

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

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

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

1. $\text{ln }\text{x}+\text{c}$

Solution:

$\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}$

$\text{dy}=\frac{\text{dx}}{\text{x}}$

$\int\text{dy}=\int\frac{\text{dx}}{\text{x}}=\log_{\text{e}}\text{x+c}$

$\therefore\text{y}=\text{ln }\text{x + c}$

2. y = 95e^kt

Solution:

Given, $\frac{\text{dy}}{\text{dt}}=\text{ky}$ and $\text{k}\neq0$

We have $\frac{\text{dy}}{\text{y}}=\text{kdt}$

Apply integral on both sides, so we get $\int\frac{\text{dy}}{\text{y}}=\int\text{kdt}$

⇒ ln (y) = kt + c (where c is constant)

⇒ y = e^ce^kt = ae^kt (a = e^c)

The possible solution is y = 95e^kt

3. $0$

1. $-\text{x}\text{e}^{\text{x}}+\text{C}$

Solution:

$\int(\text{x}-1)\text{e}^{-\text{x}}\text{ dx}$

$=(\text{x}-1)\int\text{e}^{-\text{x}}\text{ dx}-\int\Big(\Big[\frac{\text{d}(\text{x}-1)}{\text{dx}}\Big]\int\text{e}^{-\text{x}}\text{dx}\Big)\text{ dx}$

$=(\text{x}-1)\frac{\text{e}^{-\text{x}}}{-1}-\int\frac{\text{e}^{-\text{x}}}{-1}\text{ dx}$

$=-(\text{x}-1)\text{e}^{-\text{x}}+\frac{\text{e}^{-\text{x}}}{-1}+\text{C}$

$=-\text{x}\text{e}^{-\text{x}}+\text{e}^{-\text{x}}+\text{C}$

$=-\text{x}\text{e}^{-\text{x}}+\text{C}$

2. $\frac{\tan^7\text{x}}{7}+\text{C}$

Solution:

$\text{I}=\int\frac{\sin^6\text{x dx}}{\cos^8\text{x}}$

$\text{I}=\int\tan^6\text{x}\sec^2\text{x dx}$

Put $\tan\text{x}=\text{t}$

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

$\text{I}=\int\text{t}^6\text{dt}$

$\text{I}=\frac{\text{t}^7}{7}+\text{C}$

$\text{I}=\frac{\tan^7\text{x}}{7}+\text{C}$

1. $\text{c}-\frac{\text{x}}{(\text{x}^2-1)^2}$

2. $\text{x}[\cos(\log_\text{e}\text{x})+\sin(\log_\text{e}\text{x})]$

3. $\log|1-\cot\frac{\text{x}}{2}|+\text{c}$

4. $\frac{\text{x}^{4}}{4}-\log\text{x}+\frac{\text{3x}^{2}}{2}+\text{c}$

Solution:

$\int(\text{x}^3-\frac{1}{\text{x}}+{3\text{x}})\text{dx}=\frac{x^4}{4}-\text{ln}\mid{\text{x}}\mid{+}\frac{3\text{x}^2}{2}+\text{c}$

3. $2$

Solution:

$\int\limits_{0}^{\frac{\pi}{2}}\sqrt{1+\sin2\text{x}\text{dx}}=$

$\int\limits_{0}^{\frac{\pi}{2}}\sqrt{\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}\cos\text{xdx}}$

$\int\limits_{0}^{\frac{\pi}{2}}\sqrt{(\sin\text{x}+\cos\text{x})^2}\text{dx}$

$\int\limits_{0}^{\frac{\pi}{2}}(\sin\text{x}+\cos\text{x})\text{dx}$

$=[-\cos\text{x}+\sin\text{x}]\frac{\frac{\pi}{2}}{{0}}$

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

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

$=1+1=2$

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

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

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

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

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

3. $\tan\text{x}+\text{cot}\text{x}+\text{c}$

2. $2$

Solution:

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

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

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

$=1-0-0+1$

$=2$

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 $\frac{\text{r}}{\text{n}}=\text{x}$

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

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

$=\log3-\log2$

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

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

2. $2$

Solution:

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

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

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

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

$=2$

2. $\tan^{-1}(\log_\text{e}\text{x})+\text{C}$

Solution:

$\frac{1}{1+(\log_\text{e}\text{x})^2}\text{ d}(\log_\text{e}\text{x})$

Put $\log_\text{e}\text{x}=\text{t}$

$\int\frac{\text{dt}}{1+\text{t}^2}=\tan^{-1}\text{t}+\text{C}$

$\tan^{-1}(\log_\text{e}\text{x})+\text{C}$

4. $\text{x}\cdot\tan\frac{\text{x}}{2}+\text{C}$

Solution:

$\text{I}=\int\frac{\text{x}+\sin\text{x}}{1+\cos\text{x}}\text{dx}$

$=\int\frac{​​\text{x}}{1+\cos\text{x}}\text{dx}+\int\frac{\sin\text{x}}{1+\cos\text{x}}\text{dx}$

$=\int\frac{\text{x}}{2\cos^2\frac{\text{x}}{2}}\text{dx}+\int\frac{2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}\text{dx}$

$=\int\text{x}\sec^2\frac{\text{x}}{2}\text{dx}+\int\tan\frac{\text{x}}{2}\text{dx}$

$=\frac{1}{2}\Big[\text{x}\cdot2\tan\frac{\text{x}}{2}-\int2\tan\frac{\text{x}}{2}\text{dx}\Big]+\int\tan\frac{\text{x}}{2}\text{dx}=\text{x}\cdot\tan\frac{\text{x}}{2}+\text{C}$

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

3. $\log3$

3. x + C

Solution:

Now, $\int\frac{\text{x}^2+1}{\text{x}^2+1}\text{dx}$

$=\int\text{dx}$

= x + C [Where C is integrating constant]

3. $8$

Solution:

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

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

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

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

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

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

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

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

$=4\times2$

$=8$

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

Solution:

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

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

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

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

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, $\text{I}=\frac{\pi}{4}$

1. $\text{log}|\text{x}|-\frac{1}{2}\text{log}(\text{x}^2+1)+\text{C}$

Let $\text{I}=\int\frac{1}{\text{x}(\text{x}^2+1)}\text{dx}=\int\frac{2\text{x}}{2\text{x}^2(\text{x}^2+1)}\text{dx}=\int\frac{2\text{x}}{2\text{x}^2(\text{x}^2+1)}\text{dx}\dots(\text{i})$

Putting x² = t

⇒ 2x dx = dt

Putting this value in eq. (i),

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

$\text{I}=\frac{1}{2}\int\Bigg(\frac{\text{t}+1}{\text{t}(\text{t}+1)}-\frac{\text{t}}{\text{t}(\text{t}+1)}\Bigg)\text{dt}=\frac{1}{2}\int\Bigg(\frac{1}{\text{t}}-\frac{1}{\text{t}+1}\Bigg)\text{dt}=\frac{1}{2}\Bigg[\int\frac{1}{\text{t}}\text{dt}-\int\frac{1}{\text{t}+1}\text{dt}\Bigg]$

$=\frac{1}{2}[\text{log}|\text{t}|-\text{log}|\text{t}+1|]+\text{c}$

$=\frac{1}{2}[\text{log}|\text{x}^2|-\text{log}(\text{x}^2+1)]+\text{c}$

$=\text{log}|\text{x}|-\frac{1}{2}\text{log}|\text{x}^2+1|+\text{c}$

3. $\log\big|1-\cot\frac{\pi}{2}\big|+\text{C}$

Solution:

$\int\frac{\text{dx}}{1-\cos\text{x}-\sin\text{x}}$

Put $\text{t}=\tan\frac{\text{x}}{2}$

$\text{dx}=\frac{2\text{dt}}{1+\text{t}^2}$

$\cos\text{x}=\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}=\frac{1-\text{t}^2}{1+\text{t}^2}$

$\sin\text{x}=\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}=\frac{2\text{t}}{1+\text{t}^2}$

Put in the question

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

$\text{I}=\int\frac{\text{dt}}{\text{t}^2-1}$

$\text{I}=\int\frac{\text{dt}}{\text{t}^2-\text{t}+\frac{1}{4}-\frac{1}{4}}$

$\text{I}=\ln\Big|1-\cot\frac{\text{x}}{2}\Big|+\text{C}$

1.  $(\text{x}+1)\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\text{C}$

Solution:

Let $\text{I}=\int1\cdot\tan^{-1}\sqrt{\text{x}}\text{ dx}$

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

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

Put $\text{x}=\text{t}^2\Rightarrow\text{dx}=2\text{t dt}$

$\therefore$ we get;

$=\text{x}\tan^{-1}\sqrt{\text{x}}-\int\frac{\text{t}^2}{1+\text{t}^2}\text{dt}$

$=\text{x}\tan^{-1}\sqrt{\text{x}}-\int\Big(1-\frac{1}{1+\text{t}^2}\Big)\text{dt}$

$=\text{x}\tan^{-1}\sqrt{\text{x}}-\text{t}+\tan^{-1}\text{t}+\text{C}$

$=\text{x}\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\tan^{-1}\text{t}+\text{C}$

$=\text{x}\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\tan^{-1}\sqrt{\text{x}}+\text{C}$

$=(\text{x}+1)\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\text{C}$