Maths M.C.Q (1 Marks)

3. $\pi$

$\text{Let}\ \text{I}=\int^{\frac{\pi}{2}}\limits_{\frac{-\pi}{2}}\text{(x}^{3}+\text{x}\cos\text{x}+\tan^{5}\text{x}+1)\text{dx}=\int^{\frac{\pi}{2}}\limits_{\frac{-\pi}{2}}\text{x}^{3}\text{dx}+\int^{\frac{\pi}{2}}\limits_{\frac{-\pi}{2}}\text{x}\cos\text{x}\ \text{dx}+\int^{\frac{\pi}{2}}\limits_{\frac{-\pi}{2}}\tan^{5}\text{x}\ \text{dx}+\int^{\frac{\pi}{2}}\limits_{\frac{-\pi}{2}}1\text{dx}$

$\Rightarrow\ \ \text{I}=0+0+0+\text{(x)}^{\frac{\pi}{2}}_{\frac{-\pi}{2}}=\bigg(\frac{-\pi}{2}\bigg)=\pi$

1.  $1$

Solution:

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

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

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

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

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

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

$=1$

4. $\pm\log(\sin\text{x}-\cos\text{x})+\text{C}$

Solution:

Let $\text{I}=\int\frac{\sin\text{x}+\cos\text{x}}{\sqrt{1-\sin2\text{x}}}\text{ dx}$

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

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

$=\int\frac{(\sin\text{x}+\cos\text{x})\text{dx}}{|\sin\text{x}-\cos\text{x}|}$

$=\pm\int\Big(\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}-\cos\text{x}}\Big)\text{dx}$

Let $\sin\text{x}-\cos\text{x}=\text{t}$

$(\cos\text{x}+\sin\text{x})\text{dx}=\text{dt}$

$\therefore\ \text{I}=\pm\int\frac{\text{dt}}{\text{t}}$

$=\ln|\text{t}|+\text{C}$

$=\pm\ln(\sin\text{x}-\cos\text{x})+\text{C}$ $(\because\text{t}=\sin\text{x}-\cos\text{x})$

1. $-\frac{1}{4}\tan(7-4\text{x})+\text{c}$

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

Solution:

$\int\frac{\sin^8\text{x}-\cos^8\text{x}}{1-2\sin^2\text{x}\cos^2\text{x}}\text{ dx}=\text{a}\sin2\text{x}+\text{C}\ ...(\text{i})$

Considering LHS of eq. (i)

$\Rightarrow\int\frac{(\sin^4\text{x}-\cos^4\text{x})(\sin^4\text{x}+\cos^4\text{x})}{(1-2\sin^2\text{x}\cos^2\text{x})}$

$\Rightarrow\frac{(\sin^2\text{x}-\cos^2\text{x})(\sin^2\text{x}+\cos^2\text{x})\cdot(\sin^4\text{x}+\cos^4\text{x})\text{ dx}}{\big\{(\sin^2\text{x}+\cos^2\text{x})^2-2\sin^2\text{x}\cos^2\text{x}\big\}}$

$\Rightarrow\int\frac{(\sin^2\text{x}-\cos^2\text{x})\cdot(\sin^4\text{x}+\cos^4\text{x})\text{ dx}}{(\sin^4\text{x}+\cos^4\text{x}+2\sin^2\text{x}\cos^2\text{x}-2\sin^2\text{x}\cos^2\text{x})}$

$\Rightarrow-\int\frac{(\cos^2\text{x}-\sin^2\text{x})\times(\sin^4\text{x}+\cos^4\text{x})\text{ dx}}{(\sin^4\text{x}+\cos^4\text{x})}$

$\Rightarrow-\int\cos(2\text{x})\text{ dx}\ ...(\text{ii})$ $(\because\cos^2\text{x}-\sin^2\text{x}=\cos2\text{x})$

Comparing the RHS of eq. (i) with eq. (ii) we get

$\text{a}=-\frac{1}{2}$

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

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

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

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

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

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

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

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

3. $\sin^{-1}(\frac{\text{x}-4}{5})+\text{c}$

2. $-\log|\text{e}^{-\text{x}}+\sqrt{\text{e}^{2\text{x}}-1}|+\text{c}$

4. $\frac{1}{2}(\text{x}-4)\sqrt{\text{x}^2-8\text{x}+7}+9\text{log}\Bigg|\text{x}-4+\sqrt{\text{x}^2-8\text{x}+7}\Bigg|+\text{C}$

$=\Bigg(\frac{\text{x}-4}{2}\Bigg)\sqrt{(\text{x}-4)^2-3^2}-\frac{3^2}{2}\text{log}\Big|\text{x}-4+\sqrt{(\text{x}-4)^2-3^2}\Big|+\text{C}$​​​​​​​

$\Bigg[\therefore\ \int\sqrt{\text{x}^2-\text{a}^2}\text{dx}=\frac{\text{x}}{2}\sqrt{\text{x}^2-\text{a}^2}-\frac{\text{a}^2}{2}\text{log}\Big|\text{x}+\sqrt{\text{x}^2-\text{a}^2}\Big|\Bigg]$

$=\Bigg(\frac{\text{x}-4}{2}\Bigg)\sqrt{\text{x}^2-8\text{x}+7}-\frac{9}{2}\text{log}\Big|\text{x}-4+\sqrt{\text{x}^2-8​​\text{x}+7}\Big|+\text{C}$

2. $2\sqrt{\text{x}}+\text{k}$

2. $\frac{1}{4}\tan^{-1}\Big(\frac{\text{x}^2}{2}\Big)$

Solution:

$\text{I}=\int\frac{\text{x}}{4+\text{x}^4}\text{ dx}$

Put $\text{x}^2=\text{t}$

$2\text{xdx}=\text{dt}$

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

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

$\text{I}=\frac{1}{2}\tan^{-1}\Big(\frac{\text{t}}{2}\Big)+\text{C}$

$\text{I}=\frac{1}{2}\tan^{-1}\Big(\frac{\text{x}^2}{2}\Big)+\text{C}$

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,

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

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

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

$=\pi\log2$

4. $\text{x}-\frac{\text{x}^2}{2}+\frac{\text{x}^3}{3}-\log|1-\text{x}|+\text{C}$

Solution:

$\text{I}=\int\frac{\text{x}^3}{\text{x}+1}\text{ dx}$

$\text{I}=\int\frac{\text{x}^3+1-1}{\text{x}+1}\text{ dx}$

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

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

$\text{I}=\text{x}-\frac{\text{x}^2}{2}+\frac{\text{x}^3}{3}-\log|1-\text{x}|+\text{C}$

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

Solution:

We know that. $\frac{\text{d}}{\text{dx}}(\tan\text{x})=\sec^2\text{x}$

$\therefore\text{d }(\tan\text{x})=\sec^2\text{xdx}$

$\Rightarrow\text{I}=\int\sec^2\text{xdx}=\tan\text{x+c}$

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

1. $-\frac{1}{\log_\text{e}2}$

Solution:

$\text{k}=\int\frac{2^{\frac{1}{\text{x}}}}{\text{x}^2}\text{ dx}$

Put $\frac{1}{\text{x}}=\text{t}$

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

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

$\text{k}=\int2^{\text{t}}(-\text{dt})$

$\text{k}=\frac{-2^{\text{t}}}{\log_\text{e}2}+\text{C}$

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

$\text{k}=\frac{-1}{\log_\text{e}2}$

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

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

Solution:

Using substitution method.

u = 2x

du = 2dx

$\frac{\text{du}}{2}=\text{dx}$

Plug in.

$\text{dy}=\frac{1}{2}\cos(\text{u})\text{du}$

Integrate.

$\text{y}=\frac{\sin(2\text{x})}{2}+\text{c}$

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, $\text{I}=\frac{\pi}{4}$

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. $0$

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

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

$=0$

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

1. $-\frac{\cos^4\text{x}}{4}+\text{C}$

4. $-\frac{1}{(\text{n}+1)(1+\text{lnx})^{\text{n}-1}}+\text{c}$

Solution:

Given,

$\int\frac{1}{\text{x}(1+\text{ln}(\text{x}))^\text{n}}\text{dx}$

apply u = 1 + ln (x)

$=\int\frac{1}{\text{u}^{\text{n}}}\text{du}$

$=\int\text{u}^{-\text{n}}\text{du}$

$=\frac{\text{u}^{-\text{n}+1}}{-\text{n}+1}$

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

$-\frac{1}{(\text{n}+1)(1+\text{lnx})^{\text{n}-1}}+\text{c}$

2. $-\text{e}^{\text{x}}\cot\frac{\text{x}}{2}+\text{C}$

Solution:

Let $\text{I}=\int\text{e}^{\text{x}}\Big(\frac{1-\sin\text{x}}{1-\cos\text{x}}\Big)\text{dx}$

$\int\text{e}^{\text{x}}\Big(\frac{1}{1-\cos\text{x}}-\frac{\sin\text{x}}{1-\cos\text{x}}\Big)\text{dx}$

$\int\text{e}^{\text{x}}\bigg(\frac{1}{2\sin^2\frac{\text{x}}{2}}-\frac{2\sin\frac{\pi}{2}\cos\frac{\pi}{2}}{2\sin^2\frac{\pi}{2}}\bigg)\text{dx}$

$\int\text{e}^\text{x}\Big(\frac{1}{2}\text{cosec}^2\frac{\text{x}}{2}-\cot\frac{\text{x}}{2}\Big)\text{dx}$

As, we know that $\int\text{e}^{\text{x}}\{\text{f(x)}+\text{f}'(\text{x})\}\text{dx}=\text{e}^{\text{x}}\text{f(x)}+\text{C}$

$\therefore\ \text{I}=-\text{e}^{\text{x}}\cot\Big(\frac{\text{x}}{2}\Big)+\text{C}$

1. $\text{g}'(0)=\cos (\log2)$

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

Solution:

Let $\text{I}=\int\frac{\cos2\text{x}-\cos2\theta}{\cos\text{x}-\cos\theta}\text{dx}$

$=\int\frac{\big(2\cos^2\text{x}-1-2\cos^2\theta+1)}{\cos\text{x}-\cos\theta}\text{dx}$

$=2\int\frac{\big(\cos\text{x}+\cos\theta)(\cos\text{x}-\cos\theta)}{\cos\text{x}-\cos\theta}\text{dx}$

$=2\int(\cos\text{x}+\cos\theta)\text{dx}$

$=2(\sin\text{x}+\text{x}\cos\theta)+\text{C}$

4. $\pi^\pi$

Solution:

$\text{g}'(\text{x})=\int\text{x}^{\text{x}}(1+\log{\text{e}^\text{x}})\text{dx}$

$=\int\text{d}(\text{x}^{\text{x}})$

$\text{g'}(\text{x})=\text{x}^{\text{x}}$

$\text{g'}({\pi})={\pi}^{\pi}$

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

1. $\frac{1}{\sqrt{6}}\tan^{-1}\Big(\frac{1}{\sqrt{6}}\tan\frac{\text{x}}{2}\Big)+\text{C}$

Solution:

Let $\text{I}=\int\frac{\text{dx}}{7+5\cos\text{x}}$

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

$\therefore\ \text{I}=\int\frac{\text{dx}}{7+5\times\bigg(\frac{1-\tan^2\frac{\pi}{2}}{1+\tan^2\frac{\pi}{2}}\bigg)}$

$=\int\frac{\big(1+\tan^2\frac{\pi}{2}\big)\text{dx}}{7\big(1+\tan^2\frac{\pi}{2}\big)+5-5\tan^2\frac{\pi}{2}}$

$=\int\frac{\sec^2\frac{\pi}{2}\text{dx}}{2\tan^2\frac{\pi}{2}+12}$

$=\frac{1}{2}\int\frac{\sec^2\frac{\pi}{2}\text{dx}}{\tan^2\frac{\pi}{2}+(\sqrt{6})^2}$

Let $\tan\frac{\text{x}}{2}=\text{t}$

$\frac{1}{2}\sec^2\Big(\frac{\text{x}}{2}\Big)\text{dx}=\text{dt}$

$\sec^2\Big(\frac{\text{x}}{2}\Big)\text{dx}=2\text{dt}$

$\therefore\ \text{I}=\frac{1}{2}\int\frac{2\text{ dt}}{\text{t}^2+(\sqrt{6})^2}$

$=\frac{1}{\sqrt{6}}\tan^{-1}\Big(\frac{\text{t}}{\sqrt{6}}\Big)+\text{C}$ $\Big(\because\int\frac{1}{\text{a}^2+\text{x}^2}=\frac{1}{\text{a}}\tan^{-1}\frac{\text{x}}{\text{a}}+\text{C}\Big)$

$=\frac{1}{\sqrt{6}}\tan^{-1}\bigg(\frac{\tan\frac{\pi}{2}}{\sqrt{16}}\bigg)+\text{C}$ $\Big(\because\text{t}=\tan\frac{\text{x}}{2}\Big)$

2. $\log|\sin\text{x}+\cos\text{x}|+\text{C}$ 

$\text{Let I}=\int\frac{\cos2\text{x}}{(\cos\text{x}+\sin\text{x)}^{2}}$

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

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

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

$\text{Let}\cos\text{x}+\sin\text{x}=\text{t}\Rightarrow(\cos\text{x}-\sin\text{x)}\text{dx}=\text{dt}$

$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}}$

$=\log|\text{t}|+\text{C}$

$=\log|\cos\text{x}+\sin\text{x}|+\text{C}$

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. $0$

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,

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

2.  $\text{x}\sin\text{x}$

$\text{f}\text{(x)}=\int\limits_{0}^{\text{x}}\text{t}\sin\text{t}\ \text{dt}$

$\therefore\text{f}\text{'(x)}=\text{x}\sin\text{x}\ $ $\big[$$\therefore$ of first fundamental theorem$\big]$

$\text{f}\text{(x)}=\int\limits_{0}^{\text{x}}\text{t}\sin\text{t}\ \text{dt}$

$\therefore\text{f}\text{'(x)}=\text{x}\sin\text{x}\ \ $ $\big[$$\therefore$ of first fundamental theorem$\big]$

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

1. t³ - t² + C

Solution:

$\int(3\text{t}^2-2\text{t})\text{dt}=\text{t}^3-\text{t}^2+\text{c}$

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

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

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

$=2$

1. $\text{x}^{\sin\text{x}}+\text{C}$

Solution:

$\int\text{x}^{\sin\text{x}}\Big(\frac{\sin\text{x}}{\text{x}}+\cos\text{x}\cdot\log\text{x}\Big)\text{dx}$

Put $\text{x}^{\sin\text{x}}=\text{t}$

Taking $\log$ on both sides,

$\log\text{t}=\sin\text{x}\log\text{x}$

$\frac{1}{\text{t}}=\frac{\sin\text{x}}{\text{x}}+\cos\text{x}\log\text{x}$

$1=\int\text{t}\cdot\frac{\text{dt}}{\text{t}}$

$1=\text{t}+\text{C}$

$1=\text{x}^{\sin\text{x}}+\text{C}$

1. $1-\log2$

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

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

Also, let $\text{e}^{\text{x}}=\text{t}\Rightarrow\text{e}^{\text{x}}\ \text{dx}=\text{dt}$

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

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

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

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 $\text{I}=\frac{\pi}{4}$

4. $2\sin\sqrt{\text{x}}+\text{C}$

Solution:

$\text{I}=\int\frac{\cos\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$

Put $\sqrt{\text{x}}=\text{t}$

$\frac{1}{2\sqrt{\text{x}}}\text{ dx}=\text{dt}$

$\frac{1}{\sqrt{\text{x}}}\text{ dx}=2\text{dt}$

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

$\text{I}=2\sin\text{t}+\text{C}$

$\text{I}=2\sin\sqrt{\text{x}}+\text{C}$

1. x³ - x

Solution:

$\int(3\text{x}^2-1)\text{dx:}$

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

= x³ - x

2. $\log_{10}\text{e}.\text{x}\log_\text{e}(\frac{\text{x}}{\text{e}})+\text{c}$

4. $\log_\text{e}(10^\text{x}+\text{x}^{10}+\text{c}$

1. $\text{e}^{\text{x}}\text{f(x)}+\text{C}$

Solution:

$\int\text{e}^{\text{x}}\{\text{f(x)}+\text{f}'(\text{x})\}\text{dx}=\text{e}^{\text{x}}\text{f(x)}+\text{C}$