Maths M.C.Q (1 Marks)

1. $\frac{1}{2}\log\big(\sec\text{x}^2+\tan\text{x}^2\big)+\text{C}$

Solution:

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

Put $\text{x}^2=\text{t}$

$=\text{x}=\sqrt{\text{t}}$

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

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

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

$\text{I}=\frac{1}{2}\log(\sec\text{t}+\tan\text{t})+\text{C}$

$\frac{1}{2}\log\big(\sec\text{x}^2+\tan\text{x}^2\big)+\text{C}$

4. $\log(1+\log\text{x})$

Solution:

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

$\text{I}=\int\frac{\text{dx}}{\text{x}+(1+\log\text{x})}$

Put $1+\log\text{x}=\text{t}$

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

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

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

$\text{I}=\log(1+\log\text{x})+\text{C}$

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

Solution:

$\text{I}=\int\frac{\text{x}+3}{(\text{x}+4)^2}\text{e}^{\text{x}}\text{ dx}$

$\text{I}=\int\bigg(\frac{\text{x}+4-1}{(\text{x}+4)^2}\bigg)\text{e}^{\text{x}}\text{ dx}$

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

$\text{f(x)}=\frac{1}{\text{x}+4}$

$\text{f}'(\text{x})=-\frac{1}{(\text{x}+4)^2}$

$\text{I}=\frac{\text{e}^{\text{x}}}{\text{x}+4}+\text{C}$

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

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

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

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

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

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

Solution:

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

Consider,

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

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

Comaring coefficeints and solving it simultaneously we get

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

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

Integrating we get as,

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

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

3. $\text{x}-\tan\text{x}+\text{C}$

Solution:

$\text{I}=\int\frac{\cos2\text{x}-1}{\cos2\text{x}+1}\text{ dx}$

$\text{I}=-\int\frac{1-\cos2\text{x}}{1+\cos2\text{x}}\text{ dx}$

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

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

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

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

$\text{I}=\text{x}-\tan\text{x}+\text{C}$

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

Solution:

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

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

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

Here, $\text{f(x)}=-\cot\text{x}$

$\text{f}'(\text{x})=\text{cosec}^2\text{x}$

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

1. $\frac{\text{a}^{\text{x}+\frac{1}{\text{x}}}}{\log_\text{e}\text{a}}$

Solution:

$\text{f(x)}=\Big(1-\frac{1}{\text{x}^2}\Big)\text{a}^{\text{x}+\frac{1}{\text{x}}}$

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

Let $\Big(\text{x}+\frac{1}{\text{x}}\Big)=\text{t}$

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

$\therefore\ \int\text{f(x)}\text{dx}=\int\text{a}^{\text{t}}\text{ dt}$

$=\frac{\text{a}^{\text{t}}}{\log_\text{e}\text{a}}+\text{C}$

$=\frac{\text{a}^{\text{x}+\frac{1}{\text{x}}}}{\log_\text{e}\text{a}}+\text{C}$ $\Big(\because\text{t}=\text{x}+\frac{1}{\text{x}}\Big)$

4.  none of these.

Solution:

$\int|\text{x}|^3\text{ dx}$

$|\text{x}|=\begin{cases}\text{x},\text{ x}\geq0\\-\text{x},\text{ x}<0\end{cases}$

Case I:

When $\text{x}\geq0$

$\therefore\ \int|\text{x}|^3\text{ dx}$

$=\int\text{x}^3\text{ dx}$

$=\frac{\text{x}^4}{4}+\text{C}$

Case II:

$\text{x}<0$

$\int|\text{x}|^3\text{ dx}$

$=-\int\text{x}^3\text{ dx}$

$=\frac{-\text{x}^4}{4}+\text{C}$ 

4. $\frac{1}{10}\Big(\frac{1}{\text{x}^2}+4\Big)^{-5}+\text{C}$

Solution:

Let $\text{I}=\int\frac{\text{x}^9}{(4\text{x}^2+1)^6}\text{ dx}$

$=\int\frac{\text{x}^{9}}{\text{x}^{\frac{1}{2}}\big(4+\frac{1}{\text{x}^2}\big)^6}\text{ dx}$

$=\int\frac{\text{x}^{\frac{1}{3}}}{\big(4+\frac{1}{\text{x}^2}\big)^6}\text{ dx}$

Let $\Big(4+\frac{1}{\text{x}^2}\Big)=\text{t}$

On differentiating both sides, we get

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

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

$=-\frac{1}{2}\Big(-\frac{1}{5}\Big)\text{t}^{-5}+\text{C}$

$=\frac{1}{10}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{C}$

Therefore, $\int\frac{\text{x}^9}{(4\text{x}^2+1)^6}\text{ dx}=\frac{1}{10}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{C}$

Hence, the correct option is (d)

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 $\text{xe}^{\text{x}}=\text{t}$

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

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

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

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

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

4. $\frac{1}{2}\big[\text{x}+\log(\sin\text{x}+\cos\text{x})\big]+\text{C}$

Solution:

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

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

Numerator can be written as,

$\cos\text{x}=\text{A}(\sin\text{x}+\cos\text{x})+\text{B}\frac{\text{d}(\sin\text{x}+\cos\text{x})}{\text{dx}}$

$\cos\text{x}=(\text{A}-\text{B})\sin\text{x}+(\text{A}+\text{B})\cos\text{x}$

$\text{A}-\text{B}=0$ and $\text{A}+\text{B}=1$

$\text{A}=\frac{1}{2}=\text{B}$

$\text{I}=\int\frac{\big[\frac{1}{2}(\sin\text{x}+\cos\text{x})+\frac{1}{2}(\cos\text{x}-\sin\text{x})\big]\text{dx}}{\sin\text{x}+\cos\text{x}}$

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

$\text{I}=\frac{1}{2}\big[1+\int(\sin\text{x}+\cos\text{x})\big]+\text{C}$

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

Solution:

$\text{I}=\int\sqrt{\frac{\text{x}}{1-\text{x}}}\text{ dx}$

$\text{I}=\int\sqrt{\frac{\text{x}}{1-\text{x}}\cdot\frac{\text{x}}{\text{x}}}\text{ dx}$

$\text{I}=\int\frac{\text{x dx}}{\sqrt{\text{x}-\text{x}^2}}$

Consider,

$\text{x}=\text{A}\frac{\text{d}(\text{x}-\text{x}^2)}{\text{dx}}+\text{B}$

$\text{x}=\text{A}(1-2\text{x})+\text{B}$

$\text{x}=-2\text{Ax}+\text{A}+\text{B}$

$-2\text{A}=1$

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

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

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

$\text{I}=\frac{-1}{2}\times2\sqrt{\text{x}-\text{x}^2}+\frac{1}{2}\int\frac{1}{\sqrt{\text{x}-\text{x}^2}}\text{ dx}$

Second term after completing square method you will get as

$\text{I}=-\sqrt{\text{x}-\text{x}^2}+\sin^{-1}\sqrt{\text{x}}+\text{C}$

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 $\log_\text{e}\text{x}=\text{t}$

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

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

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

Solution:

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

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

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

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

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

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

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

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. $\text{A}=\frac{2}{3},\text{ B}=\frac{5}{3}$

Solution:

$\int\frac{1}{5+4\sin\text{x}}\text{ dx}=\text{A}\tan^{-1}\Big(\text{B}\tan\frac{\pi}{2}+\frac{4}{3}\Big)+\text{C}\ ...(\text{i})$

Considering the LHS of eq. (1)

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

$\Rightarrow\int\frac{1}{5+\frac{8\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}}\text{dx}$

$\Rightarrow\frac{\Big(1+\tan^2\frac{\text{x}}{2}\Big)}{5\Big(1+\tan^2\frac{\text{x}}{2}\Big)+8\tan\frac{\text{x}}{2}}\text{dx}$

$\Rightarrow\frac{\sec^2\frac{\text{x}}{2}}{5\tan^2\frac{\text{x}}{2}+8\tan\frac{\text{x}}{2}+5}\text{dx}\dots(2)$

Let $\tan\frac{\text{x}}{2}=\text{t}$

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

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

$\Rightarrow\frac{2}{5}\int\frac{\text{dt}}{\text{t}^2+\frac{8}{5}\text{t}+1}$

$\Rightarrow\frac{2}{5}\int\frac{\text{dt}}{\text{t}^2+\frac{8}{5}\text{t}+\big(\frac{4}{5}\big)^2-\big(\frac{4}{5}\big)^2+1}$

$\Rightarrow\frac{2}{5}\int\frac{\text{dt}}{\big(\text{t}+\frac{4}{5}\big)^2+1-\frac{16}{25}}$

$\Rightarrow\frac{2}{5}\int\frac{\text{dt}}{\big(\text{t}+\frac{4}{5}\big)^2+\big(\frac{3}{5}\big)^2}$

$\Rightarrow\frac{2}{5}\times\frac{2}{3}\tan^{-1}\bigg(\frac{\text{t}+\frac{4}{5}}{\frac{3}{5}}\bigg)+\text{C}$

$\Rightarrow\frac{2}{3}\tan^{-1}\Big(\frac{5\text{t}+4}{3}\Big)+\text{C}$

$\Rightarrow\frac{2}{3}\tan^{-1}\Big(\frac{5}{3}\tan\frac{\text{x}}{2}+\frac{4}{3}\Big)+\text{C}\ ...\text{(ii)}$ $\Big(\because\text{t}=\tan\frac{\text{x}}{2}\Big)$

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

$\therefore\ \text{A}=\frac{2}{3},\text{ B}=\frac{5}{3}$

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 $\tan\text{x}=\text{t}$

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

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

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

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

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

Solution:

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

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

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

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

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

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

Solution:

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

Put $\frac{1}{\text{x}}=\text{t}$

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

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

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

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

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

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

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

Solution:

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

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

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

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

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

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

Solution:

$\text{I}=\int\frac{\sin^2\text{x}}{\cos^4\text{x}}\text{ dx}$

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

Put $\tan\text{x}=\text{t}$

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

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

$\text{I}=\frac{\text{t}^3}{3}+\text{C}$

$\text{I}=\frac{\tan^3\text{x}}{3}+\text{C}$

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

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 $\tan\frac{\text{x}}{2}=\text{t}$

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

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

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

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

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

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,

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

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

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

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

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

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

Solution:

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

4. $\text{a}=\frac{1}{3},\text{ b}=-1$

Solution:

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

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

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

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

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

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

$\text{I}=\frac{1}{2}\Big(\frac{2}{3}\text{t}^{\frac{3}{2}}-2\sqrt{\text{t}}\Big)+\text{C}$

$\text{I}=\frac{1}{3}(1+\text{x}^{2})^{\frac{3}{2}}-\sqrt{1+\text{x}^2}+\text{C}$

$\text{a}=\frac{1}{3},\text{ b}=-1$

3. $-\frac{1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\cos\text{x}}{\sqrt{3}}\Big)+\text{C}$

Solution:

$\text{I}=\int\frac{\sin\text{x}}{3+4\cos^2\text{x}}\text{ dx}$

Put $\cos\text{x}=\text{t}$

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

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

$\text{I}=\int\frac{-\text{dt}}{3+4\text{t}^2}$

$\text{I}=\frac{-1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\text{t}}{\sqrt{3}}\Big)+\text{C}$

$\text{I}=\frac{-1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\cos\text{x}}{\sqrt{3}}\Big)+\text{C}$

3. $\frac{1}{16}$

Solution:

$\int\frac{\cos8\text{x}+1}{\tan2\text{x}-\cot2\text{x}}\text{ dx}$

$=\int\frac{2\cos^24\text{x}}{\frac{\sin2\text{x}}{\cos2\text{x}}-\frac{\cos2\text{x}}{\sin2\text{x}}}\text{ dx}$

$=\int\frac{2\cos^24\text{x}}{\sin^22\text{x}-\cos^22\text{x}}\times\sin2\text{x}\cos2\text{x dx}$

$=\int-\frac{\cos^24\text{x}\sin4\text{x}}{\cos4\text{x}}\text{ dx}$

$=\frac{-1}{2}\int\sin8\text{x dx}$

$=\frac{\cos8\text{x}}{16}+\text{C}$

$\text{a}=\frac{1}{16}$

3. $\frac{1}{2}\log\tan\Big(\frac{\pi}{2}+\frac{\pi}{3}\Big)+\text{C}$

Solution:

$\text{I}=\int\frac{1}{\cos\text{x}+\sqrt{3}\sin\text{x}}\text{ dx}$

$\text{I}=\frac{1}{2}\int\frac{2}{\frac{\cos\text{x}}{2}+\frac{\sqrt{3}}{2}\sin\text{x}}\text{ dx}$

$\text{I}=\frac{1}{2}\int\frac{1}{\cos\big(\text{x}-\frac{\pi}{6}\big)}\text{ dx}$

$\text{I}=\frac{1}{2}\int\sec\Big(\text{x}-\frac{\pi}{6}\Big)\text{dx}$

$\text{I}=\frac{1}{2}\ln\Big|\tan\Big(\frac{\text{x}}{2}+\frac{\pi}{3}\Big)\Big|+\text{C}$

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

Solution:

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

$\text{I}=\int\text{x}\sin\text{x dx}$

$\text{I}=\text{x}\int\sin\text{x dx}-\int\Big(\frac{\text{dx}}{\text{dx}}\int\sin\text{x dx}\Big)\text{dx}$

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

$\text{I}=\text{x}\cos\text{x}+\sin\text{x}+\text{C}$

$\text{a}=\sin\text{x}+\text{C}$