MCQ 11 Mark
$\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sec^2\text{x dx}$ is equal to:
Answer$\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sec^2\text{x dx}$
$\Rightarrow\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sec^2\text{x dx}$ $=\Big[\tan\text{x}\Big]^{\frac{\pi}{4}}_{-\frac{\pi}{4}}$
$\Rightarrow\Big[\tan\big(\frac{\pi}{4}\big)-\tan\big(-\frac{\pi}{4}\big)\Big]$
$\Rightarrow[1-(-1)]$
$\Rightarrow2$
View full question & answer→MCQ 21 Mark
The value of $\int\frac{\cos2\text{x}}{{\cos}{\text{ x}}}\text{dx}$ is equal to:
- ✓
$2\sin\text{x}-\ell\text{ n }\mid\sec\text{x}+\tan\text{x}\mid+\text{ c}$
- B
$2\sin\text{x}-\ell\text{ n }\mid\sec\text{x}-\tan\text{x}\mid+\text{ c}$
- C
$2\sin\text{x}+\ell\text{ n }\mid\sec\text{x}+\tan\text{x}\mid+\text{ c}$
- D
$3\sin\text{x}-\ell\text{ n }\mid\sec\text{x}+\tan\text{x}\mid+\text{ c}$
AnswerCorrect option: A. $2\sin\text{x}-\ell\text{ n }\mid\sec\text{x}+\tan\text{x}\mid+\text{ c}$
View full question & answer→MCQ 31 Mark
Choose the correct option from given four options:
$\int\frac{\text{x}}{\text{x}+1}$ is equal to:
- A
$\text{x}+\frac{\text{x}^2}{2}+\frac{\text{x}^3}{3}-\log|1-\text{x}|+\text{C}$
- B
$\text{x}+\frac{\text{x}^2}{2}-\frac{\text{x}^3}{3}-\log|1-\text{x}|+\text{C}$
- C
$\text{x}-\frac{\text{x}^2}{2}-\frac{\text{x}^3}{3}-\log|1+\text{x}|+\text{C}$
- ✓
$\text{x}-\frac{\text{x}^2}{2}+\frac{\text{x}^3}{3}-\log|1+\text{x}|+\text{C}$
AnswerCorrect option: D. $\text{x}-\frac{\text{x}^2}{2}+\frac{\text{x}^3}{3}-\log|1+\text{x}|+\text{C}$
Let $\text{I}=\int\frac{\text{x}}{\text{x}+1}\text{dx}$
We know that, $\frac{\text{x}^3}{\text{x}+1}$ is an improper fraction.
To convert it into proper fraction, we have to divide numerator by denominator.
After performing long division, we get
$\frac{\text{x}^3}{\text{x}+1}=(\text{x}^2-\text{x}+1)-\frac{1}{(\text{x}+1)}$
$\therefore\ \text{I}=\int\Big((\text{x}^2-\text{x}+1)-\frac{1}{(\text{x}+1)}\Big)\text{dx}$
$=\frac{\text{x}^3}{3}-\frac{\text{x}^2}{2}+\text{x}-\log|\text{x}+1|+\text{C}$
View full question & answer→MCQ 41 Mark
$\int\limits^1_0\frac{\text{d}}{\text{dx}}\Big\{\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\Big\}\text{dx}$ is equal to:
- A
$0$
- B
${\pi}$
- ✓
$\frac{\pi}{2}$
- D
$\frac{\pi}{4}$
AnswerCorrect option: C. $\frac{\pi}{2}$
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, $\text{I}=\Big[\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\Big]^1_0$
$=\sin^{-1}(1)-\sin^{-1}(0)$
$=\frac{\pi}{2}$
View full question & answer→MCQ 51 Mark
$\int\limits^\sqrt{3}_1\frac{1}{1+\text{x}^2}\text{ dx}$ is equal to:
- ✓
$\frac{\pi}{12}$
- B
$\frac{\pi}{6}$
- C
$\frac{\pi}{4}$
- D
$\frac{\pi}{3}$
AnswerCorrect option: A. $\frac{\pi}{12}$
$\int\limits^\sqrt{3}_1\frac{1}{1+\text{x}^2}\text{dx}$
$=\big[\tan^{-1}\text{x}\big]^\sqrt{3}_1$
$=\frac{\pi}{3}-\frac{\pi}{4}$
$=\frac{\pi}{12}$
View full question & answer→MCQ 61 Mark
$\int\text{x}\sec\text{x}^2\text{ dx}$ is equal to:
- ✓
$\frac{1}{2}\log\big(\sec\text{x}^2+\tan\text{x}^2\big)+\text{C}$
- B
$\frac{\text{x}^2}{2}\log\big(\sec\text{x}^2+\tan\text{x}^2\big)+\text{C}$
- C
$2\log\big(\sec\text{x}^2+\tan\text{x}^2\big)+\text{C}$
- D
AnswerCorrect option: A. $\frac{1}{2}\log\big(\sec\text{x}^2+\tan\text{x}^2\big)+\text{C}$
$\text{I}=\int\text{x}\sec\text{x}^2\text{ dx}$
Put $\text{x}^2=\text{t}$
$=\text{x}=\sqrt{\text{t}}$
$2\text{xdx}=\text{dt}$
$\text{xdx}=\frac{\text{dt}}{2}$
$\text{I}=\int\sec\text{t}\frac{\text{dt}}{2}$
$\text{I}=\frac{1}{2}\log(\sec\text{t}+\tan\text{t})+\text{C}$
$\frac{1}{2}\log\big(\sec\text{x}^2+\tan\text{x}^2\big)+\text{C}$
View full question & answer→MCQ 71 Mark
$\int\frac{2\text{dx}}{\sqrt{1-4\text{x}^2}}=$
- A
$\tan^{-1}(2\text{x})+\text{c}$
- B
$\cot^{-1}(2\text{x})+\text{c}$
- C
$\cos^{-1}(2\text{x})+\text{c}$
- ✓
$\sin^{-1}(2\text{x})+\text{c}$
AnswerCorrect option: D. $\sin^{-1}(2\text{x})+\text{c}$
View full question & answer→MCQ 81 Mark
The value of $\int\frac{1}{\text{x}+\text{x}\log\text{x}}\text{ dx}$ is:
AnswerCorrect option: D. $\log(1+\log\text{x})$
$\text{I}=\int\frac{1}{\text{x}+\text{x}\log\text{x}}\text{ dx}$
$\text{I}=\int\frac{\text{dx}}{\text{x}+(1+\log\text{x})}$
Put $1+\log\text{x}=\text{t}$
$\frac{1}{\text{x}}\text{dx}=\text{dt}$
$\text{I}=\int\frac{1}{\text{t}}\text{ dt}$
$\text{I}=\log|\text{t}|+\text{C}$
$\text{I}=\log(1+\log\text{x})+\text{C}$
View full question & answer→MCQ 91 Mark
$\int\limits^{\frac{\pi}{2}}_0\sin2\text{x }\log\tan\text{x dx}$ is equal to:
- A
$\pi$
- B
$\frac{\pi}{2}$
- ✓
$0$
- D
$2\pi$
Answer$\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin2\text{x }\log\tan\text{x dx}\ ....(\text{i})$
$\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin(\pi-2\text{x})\log\tan\big(\frac{\pi}{2}-\text{x}\big)\text{dx}$
$\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin2\text{x}\log\cot\text{x dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin2\text{x}\big(\log\tan\text{x}+\log\cot\text{x}\big)\text{dx}$
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin2\text{x}\big(\log\tan\text{x}\cot\text{x}\big)\text{dx}$
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin\text{x}(\log1)\text{dx}$
$\text{I}=0$
View full question & answer→MCQ 101 Mark
The value of the integral $\int\limits^{\frac{\pi}{2}}_0\frac{\sqrt{\cos\text{x}}}{\sqrt{\cos\text{x}}+\sqrt{\sin\text{x}}}\text{ dx}$ is:
- A
$0$
- B
$\frac{\pi}{2}$
- ✓
$\frac{\pi}{4}$
- D
AnswerCorrect option: C. $\frac{\pi}{4}$
$\text{ I}=\int\limits^\frac{\pi}{2}_0\frac{\sqrt{\cos\text{x}}}{\sqrt\cos\text{x}+\sqrt{\sin\text{x}}}\text{dx}\ ...(\text{i})$
$=\int\limits^\frac{\pi}{2}_0\frac{\sqrt{\cos\big(\frac{\pi}{2}-\text{x}\big)}}{\sqrt{\cos\big(\frac{\pi}{2}-\text{x}\big)+\sqrt{\sin\big(\frac{\pi}{2}-\text{x}\big)}}}\text{dx}$
$=\int\limits^\frac{\pi}{2}_0\frac{\sqrt{\sin\text{x}}}{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}\text{dx}$
$=\int\limits^\frac{\pi}{2}_0\frac{\sqrt{\sin{\text{x}}}}{\sqrt{\cos\text{x}}+\sqrt{\sin\text{x}}}\text{dx}...(\text{ii})$
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, $\text{I}=\frac{\pi}{4}$
View full question & answer→MCQ 111 Mark
What is the value of $\int_{-1}^{1}\sin^3\text{x}\cos^2\text{xdx}?$
- ✓
$0$
- B
$1$
- C
$\frac{1}{2}$
- D
$2$
View full question & answer→MCQ 121 Mark
$\int\frac{\text{x}^9}{(4\text{x}^2+1)^6}\text{dx}$ is equal to:
- A
$\frac{1}{5\text{x}}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{c}$
- B
$\frac{1}{5}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{c}$
- C
$\frac{1}{10\text{x}}\Big(\frac{1}{\text{x}}+4\Big)^{-5}+\text{c}$
- ✓
$\frac{1}{10\text{x}}\Big(\frac{1}{\text{x}^2}+4\Big)^{-5}+\text{c}$
AnswerCorrect option: D. $\frac{1}{10\text{x}}\Big(\frac{1}{\text{x}^2}+4\Big)^{-5}+\text{c}$
View full question & answer→MCQ 131 Mark
Choose the correct answer in exercise$:\ \int\frac{\text{dx}}{\text{x}(\text{x}^2+1)}$ equals
- ✓
$\log|x|-\frac{1}{2}\log(\text{x}^2+1)+\text{C}$
- B
$\log|x|+\frac{1}{2}\log(\text{x}^2+1)+\text{C}$
- C
$-\log|\text{x}|+\frac{1}{2}\log(\text{x}^2+1)+\text{C}$
- D
$\log|\text{x}|+\frac{1}{2}\log(\text{x}^2+1)+\text{C}$
AnswerCorrect option: A. $\log|x|-\frac{1}{2}\log(\text{x}^2+1)+\text{C}$
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 $x^2 = t$
$\Rightarrow 2x dx = dt$
Putting this value in $\text{eq. (i),}$
$\text{I}=\int\frac{\text{dt}}{2\text{t}(\text{t}+1)}=\frac{1}{2}\int\frac{(\text{t}+1)-\text{t}}{\text{t}(\text{t}+1)}\text{dt}$
$\text{I}=\frac{1}{2}\int\Bigg(\frac{\text{t}+1}{\text{t}(\text{t}+1)}-\frac{\text{t}}{\text{t}(\text{t}+1)}\Bigg)\text{dt}=\frac{1}{2}\int\Bigg(\frac{1}{\text{t}}-\frac{1}{\text{t}+1}\Bigg)\text{dt}=\frac{1}{2}\Bigg[\int\frac{1}{\text{t}}\text{dt}-\int\frac{1}{\text{t}+1}\text{dt}\Bigg]$
$=\frac{1}{2}[\log|\text{t}|-\log|\text{t}+1|]+\text{c}$
$=\frac{1}{2}[\log|\text{x}^2|-\log(\text{x}^2+1)]+\text{c}$
$=\log|\text{x}|-\frac{1}{2}\log|\text{x}^2+1|+\text{c}$
View full question & answer→MCQ 141 Mark
$\int\frac{1}{1-\cos\text{x}-\sin\text{x}}\text{ dx}=$
- A
$\log\big|1+\cot\frac{\pi}{2}\big|+\text{C}$
- B
$\log\big|1-\tan\frac{\pi}{2}\big|+\text{C}$
- ✓
$\log\big|1-\cot\frac{\pi}{2}\big|+\text{C}$
- D
$\log\big|1+\tan\frac{\pi}{2}\big|+\text{C}$
AnswerCorrect option: C. $\log\big|1-\cot\frac{\pi}{2}\big|+\text{C}$
$\int\frac{\text{dx}}{1-\cos\text{x}-\sin\text{x}}$
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
$\text{I}=\int\frac{\frac{2\text{dt}}{1+\text{t}^2}}{1-\frac{1-\text{t}^2}{1+\text{t}^2}-\frac{2\text{t}}{1+\text{t}^2}}$
$\text{I}=\int\frac{\text{dt}}{\text{t}^2-1}$
$\text{I}=\int\frac{\text{dt}}{\text{t}^2-\text{t}+\frac{1}{4}-\frac{1}{4}}$
$\text{I}=\ln\Big|1-\cot\frac{\text{x}}{2}\Big|+\text{C}$
View full question & answer→MCQ 151 Mark
If $\int\frac{1}{(\text{x}+2)(\text{x}^2+1)}\text{ dx}=\text{a}\log|1+\text{x}^2|+\text{b}\tan^{-1}\text{x}+\frac{1}{5}\log|\text{x}+2|+\text{C},$ then
- A
$\text{a}=-\frac{1}{10},\text{ b}=\frac{2}{5}$
- B
$\text{a}=\frac{1}{10},\text{ b}=\frac{2}{5}$
- ✓
$\text{a}=-\frac{1}{10},\text{ b}=\frac{2}{5}$
- D
$\text{a}=\frac{1}{10},\text{ b}=\frac{2}{5}$
AnswerCorrect option: C. $\text{a}=-\frac{1}{10},\text{ b}=\frac{2}{5}$
$\text{I}=\int\frac{1}{(\text{x}+2)(\text{x}^2+1)}\text{ dx}$
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,
$\frac{1}{5}\log|\text{x}+2|-\frac{1}{10}\log|\text{x}^2+1|+\frac{2}{5}\tan^{-1}\text{x}+\text{C}$
$\text{a}=-\frac{1}{10},\text{ b}=\frac{2}{5}$
View full question & answer→MCQ 161 Mark
$\int\frac{1}{7}\sin\Big(\frac{\text{x}}{7}+10\Big)\text{dx}$ is equal to:
- A
$\frac{1}{7}\cos\Big(\frac{\text{x}}{7}+10\Big)\text{C}$
- B
$-\frac{1}{7}\cos\Big(\frac{\text{x}}{7}+10\Big)\text{C}$
- ✓
$\cos\Big(\frac{\text{x}}{7}+10\Big)\text{C}$
- D
$-7\cos\Big(\frac{\text{x}}{7}+10\Big)\text{C}$
AnswerCorrect option: C. $\cos\Big(\frac{\text{x}}{7}+10\Big)\text{C}$
Let $\text{I}=\int\frac{1}{7}\sin\Big(\frac{\text{x}}{7}+10\Big)\text{dx}$
$=\frac{1}{7}\int\sin\Big(\frac{\text{x}}{7}+10\Big)=\frac{1}{7}\frac{-\cos\Big(\frac{\text{x}}{7}+10\Big)}{\frac{1}{7}}$
$=-\cos\Big(\frac{\text{x}}{7}+10\Big)+\text{C}$
View full question & answer→MCQ 171 Mark
Choose the correct answer in Exercise: The value of $\int^{\frac{\pi}{2}}\limits_{0}\log\bigg(\frac{4+3\sin\text{x}}{4+3\cos\text{x}}\bigg)\text{dx}\ $is
Answer$\text{Let}\ \text{I}=\int^{\frac{\pi}{2}}\limits_{0}\log\bigg(\frac{4+3\sin\text{x}}{4+3\cos\text{x}}\bigg)\text{dx}\ \ \ ....(1)$
$\Rightarrow\ \ \ \text{I}=\int^{\frac{\pi}{2}}\limits_{0}\log\begin{bmatrix} \frac{4+3\sin\bigg(\frac{\pi}{2}-\text{x}\bigg)}{4+3\cos\bigg(\frac{\pi}{2}-\text{x}\bigg)}\text{dx}\\ \end{bmatrix}\text{dx}=\int^{\frac{\pi}{2}}\limits_{0}\log\bigg[\frac{4+3\cos\text{x}}{4+3\sin\text{x}}\bigg]\text{dx}\ \ \ ....(2)$
Adding eq. (i) and (ii),
$2\text{I}=\int^{\frac{\pi}{2}}\limits_{0}\bigg[\log\bigg(\frac{4+3\sin\text{x}}{4+3\cos\text{x}}\bigg)+\log\bigg(\frac{4+3\cos\text{x}}{4+3\sin\text{x}}\bigg)\bigg]\text{dx}=\int^{\frac{\pi}{2}}\limits_{0}\bigg[\log\bigg(\frac{4+3\sin\text{x}}{4+3\sin\text{x}}\bigg)\bigg(\frac{4+3\cos\text{x}}{4+3\sin\text{x}}\bigg)\bigg]\text{dx}$
$\Rightarrow\ \ \ 2\text{I}=\int^{\frac{\pi}{2}}\limits_{0}(\log1)\text{dx}=\int^{\frac{\pi}{2}}\limits_{0}0\ \text{dx}=0\ \ \ \ \ \Rightarrow\text{I}=0$
View full question & answer→MCQ 181 Mark
Choose the correct answer in Exercise:
If $\text{f (a}+\text{b)}-\text{x}=\text{f (x)},$ then $\int^{\text{b}}_{\text{a}}\text{x f (x)}\ \text{dx}$ is equal to
- A
$\frac{\text{a}+\text{b}}{2}\int^{\text{a}}_{\text{b}}\text{f (b}-\text{x)}\text{dx}$
- B
$\frac{\text{a}+\text{b}}{2}\int^{\text{b}}_{\text{a}}\text{f (b}+\text{x)}\text{dx}$
- C
$\frac{\text{b}-\text{a}}{2}\int^{\text{b}}_{\text{a}}\text{f (x)}\text{dx}$
- ✓
$\frac{\text{a}+\text{b}}{2}\int^{\text{b}}_{\text{a}}\text{f (x)}\text{dx}$
AnswerCorrect option: D. $\frac{\text{a}+\text{b}}{2}\int^{\text{b}}_{\text{a}}\text{f (x)}\text{dx}$
$\text{Let I}=\int^{\text{b}}\limits_{a}\text{x f (x)}\text{dx}$
$\text{I}=\int^{\text{b}}\limits_{\text{a}}\text{(a}+\text{b}-\text{x)}\text{f (a}+\text{b}-\text{x)}\text{dx}\ \ \ \ \ \Bigg[\int^{\text{b}}\limits_{\text{a}}\text{f (x)dx}=\int^{\text{b}}\limits_{\text{a}}\text{f (a}+\text{b}-\text{x)}\text{dx}\Bigg]$
$\Rightarrow\text{I}=\int^{\text{b}}\limits_{\text{a}}\text{(a}+\text{b}-\text{x)}\ \text{f (x)}\text{dx}$
$\Rightarrow\text{I}=\text{(a}+\text{b)}\int^{\text{b}}\limits_{\text{a}}\ \text{f (x)}\text{dx}\ \ \ -\text{I}$ [Using(1)]
$\Rightarrow\text{I}+\text{I}=\text{(a}+\text{b)}\int^{\text{b}}\limits_{\text{a}}\ \text{f (x)}\text{dx}$
$\Rightarrow2\text{I}=\text{(a}+\text{b)}\int^{\text{b}}\limits_{\text{a}}\ \text{f (x)}\text{dx}$
$\Rightarrow\text{I}=\bigg(\frac{\text{a}+\text{b}}{2}\bigg)\int^{\text{b}}\limits_{\text{a}}\ \text{f (x)}\text{dx}$
View full question & answer→MCQ 191 Mark
$\int\frac{\cos2\text{x}-1}{\cos2\text{x}+1}\text{ dx}=$
- A
$\tan\text{x}-\text{x}+\text{C}$
- B
$\text{x}+\tan\text{x}+\text{C}$
- ✓
$\text{x}-\tan\text{x}+\text{C}$
- D
$-\text{x}-\cot\text{x}+\text{C}$
AnswerCorrect option: C. $\text{x}-\tan\text{x}+\text{C}$
$\text{I}=\int\frac{\cos2\text{x}-1}{\cos2\text{x}+1}\text{ dx}$
$\text{I}=-\int\frac{1-\cos2\text{x}}{1+\cos2\text{x}}\text{ dx}$
$\text{I}=-\int\frac{2\sin^2\text{x}}{2\cos^2\frac{\text{x}}{2}}\text{ dx}$
$\text{I}=-\int\tan^2\text{x dx}$
$\text{I}=-\int(\sec^2\text{x}-1)\text{dx}$
$\text{I}=-(\tan\text{x}-\text{x})+\text{C}$
$\text{I}=\text{x}-\tan\text{x}+\text{C}$
View full question & answer→MCQ 201 Mark
Choose the correct option from given four options:
$\int\tan^{-1}\sqrt{\text{x}}\text{ dx}$ is equal to:
- ✓
$(\text{x}+1)\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\text{C}$
- B
$\text{x}\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\text{C}$
- C
$\sqrt{\text{x}}-\text{x}\tan^{-1}\sqrt{\text{x}}+\text{C}$
- D
$\sqrt{\text{x}}-(\text{x}+1)\tan^{-1}\sqrt{\text{x}}+\text{C}$
AnswerCorrect option: A. $(\text{x}+1)\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\text{C}$
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;
$=\text{x}\tan^{-1}\sqrt{\text{x}}-\int\frac{\text{t}^2}{1+\text{t}^2}\text{dt}$
$=\text{x}\tan^{-1}\sqrt{\text{x}}-\int\Big(1-\frac{1}{1+\text{t}^2}\Big)\text{dt}$
$=\text{x}\tan^{-1}\sqrt{\text{x}}-\text{t}+\tan^{-1}\text{t}+\text{C}$
$=\text{x}\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\tan^{-1}\text{t}+\text{C}$
$=\text{x}\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\tan^{-1}\sqrt{\text{x}}+\text{C}$
$=(\text{x}+1)\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\text{C}$
View full question & answer→MCQ 211 Mark
The value of $\int\frac{\text{d}(\text{x}^2+1)}{\sqrt{\text{x}^2+2}}$ is:
- ✓
$2\sqrt{\text{x}^2+2}+\text{c}$
- B
$\sqrt{\text{x}^2+2}+\text{c}$
- C
$\text{x}\sqrt{\text{x}^2+2}+\text{c}$
- D
${4}\sqrt{\text{x}^2+2}+\text{c}$
AnswerCorrect option: A. $2\sqrt{\text{x}^2+2}+\text{c}$
View full question & answer→MCQ 221 Mark
The value of $\int\limits^\frac{\pi}{4}_0\cos\text{x}\text{ e}^{\sin\text{x}}\text{ dx}$ is:
AnswerLet, $\text{I}=\int\limits^\frac{\pi}{4}_0\cos\text{x}\text{ e}^{\sin\text{x}}\text{dx}$
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
$\text{I}=\int\limits^1_0\text{e}^\text{t}\text{dt}$
$=\big[\text{e}^\text{t}\big]^1_0$
$=\text{e}-1$
View full question & answer→MCQ 231 Mark
$\int\big(1+5\text{x}+10\text{x}^2+10\text{x}^3+5\text{x}^4+\text{x}^5\big)\text{dx}=\frac{(1+\text{x})\text{p}}{6}+\text{c}$ then p is:
AnswerGiven, $\int\big(1+5\text{x}+10\text{x}^2+10\text{x}^3+5\text{x}^4+\text{x}^5\big)\text{dx}$
$\int(\text{x+1})^5\text{dx}$ [by binomial theorem]
$=(\text{x+1})^5\text{dx}=\frac{(\text{x+1})6}{6}+\text{c}$
So, the value of p = 6
View full question & answer→MCQ 241 Mark
$\int\text{e}^{\text{x}}(1-\cot\text{x}+\cot^2\text{x})\text{dx}=$
- A
$\text{e}^{\text{x}}\cot\text{x}+\text{C}$
- ✓
$-\text{e}^{\text{x}}\cot\text{x}+\text{C}$
- C
$\text{e}^{\text{x}}\text{cosec x}+\text{C}$
- D
$-\text{e}^{\text{x}}\text{cosec x}+\text{C}$
AnswerCorrect option: B. $-\text{e}^{\text{x}}\cot\text{x}+\text{C}$
$\text{I}=\int\text{e}^{\text{x}}(1-\cot\text{x}+\cot^2\text{x})\text{dx}$
$\text{I}=\int\text{e}^\text{x}(1+\cot^2\text{x}-\cot\text{x})\text{dx}$
$\text{I}=\int\text{e}^{\text{x}}(\text{cosec}^2\text{x}-\cot\text{x})\text{dx}$
Here, $\text{f(x)}=-\cot\text{x}$
$\text{f}'(\text{x})=\text{cosec}^2\text{x}$
$\text{I}=-\text{e}^{\text{x}}\cot\text{x}+\text{C}$
View full question & answer→MCQ 251 Mark
If $\frac{\text{dy}}{\text{dx}}=3$ then y is equal to:
Answer$\frac{\text{dy}}{\text{dx}}=3$
$\text{dy}=3\text{dx}$
$\int\text{dy}=\int3\text{dx}=3\text{x}+\text{c}$
$\therefore\text{y}=3\text{x}+\text{c}$
View full question & answer→MCQ 261 Mark
$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\sin^9\text{xdx}=$
View full question & answer→MCQ 271 Mark
Evaluate: $\int\frac{1}{1+3\sin^2\text{x}+8\cos^2\text{x}}\text{dx}.$
- A
$\frac{1}{6}\tan^{-1}(2\tan\text{x})+\text{c}$
- B
$\tan^{-1}(2\tan\text{x})+\text{c}$
- ✓
$\frac{1}{6}\tan^{-1}(\frac{2\tan\text{x}}{3})+\text{c}$
- D
AnswerCorrect option: C. $\frac{1}{6}\tan^{-1}(\frac{2\tan\text{x}}{3})+\text{c}$
View full question & answer→MCQ 281 Mark
$\int\frac{\text{xdx}}{(\text{x-1)}(\text{x-2})}$ equals:
- A
$\log\begin{vmatrix}\frac{(\text{x}-1)^2}{\text{x}-2}\end{vmatrix}+\text{c}$
- ✓
$\log\begin{vmatrix}\frac{(\text{x}-2)^2}{\text{x}-2}\end{vmatrix}+\text{c}$
- C
$\log\begin{vmatrix}\Big(\frac{\text{x}-1}{\text{x}-2}\Big)^2\end{vmatrix}+\text{c}$
- D
$\log|(\text{x}-1)(\text{x}-2)+\text{c}$
AnswerCorrect option: B. $\log\begin{vmatrix}\frac{(\text{x}-2)^2}{\text{x}-2}\end{vmatrix}+\text{c}$
View full question & answer→MCQ 291 Mark
$\int\limits^{\pi}_0\frac{1}{1+\sin\text{x}}\text{ dx}$ equals:
- A
$0$
- B
$\frac{1}{2}$
- ✓
$2$
- D
$\frac{3}{2}$
Answer$\int\limits^{\pi}_0\frac{1}{1+\sin\text{x}}\text{ dx}$
$=\int\limits^\pi_0\frac{1}{1+\sin\text{x}}\times\frac{1-\sin\text{x}}{1-\sin\text{x}}\text{dx}$
$= \int\limits^\pi_0\frac{1-\sin\text{x}}{1-\sin^2\text{x}}\text{dx}$
$= \int\limits^\pi_0\frac{1-\sin\text{x}}{\cos^2\text{x}}\text{dx}$
$=\int\limits^\pi_0(\sec^2\text{x}-\sec\text{x}\tan\text{x})\text{dx}$
$=\big[\tan\text{x}-\sec\text{x}\big]^\pi_0$
$=0+1-0+1$
$=2$
View full question & answer→MCQ 301 Mark
Choose the correct answer in Exercise: The value of $\int^{\frac{\pi}{2}}\limits_{\frac{-\pi}{2}}\text{(x}^{3}+\text{x}\cos\text{x}+\tan^{5}\text{x}+1)\text{dx}\ $is
Answer$\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}$
$\Rightarrow\ \ \text{I}=0+0+0+\text{(x)}^{\frac{\pi}{2}}_{\frac{-\pi}{2}}=\bigg(\frac{-\pi}{2}\bigg)=\pi$
View full question & answer→MCQ 311 Mark
The primitive of the function $\text{f(x)}=\Big(1-\frac{1}{\text{x}^2}\Big)\text{a}^{\text{x}+\frac{1}{\text{x}}},\text{ a}>0$ is:
- ✓
$\frac{\text{a}^{\text{x}+\frac{1}{\text{x}}}}{\log_\text{e}\text{a}}$
- B
$\log_\text{e}\text{a}\cdot\text{a}^{\text{x}+\frac{1}{\text{x}}}$
- C
$\frac{\text{a}^{\text{x}+\frac{1}{\text{x}}}}{\text{x}}{\log_\text{e}\text{a}}$
- D
$\text{x}\frac{\text{a}^{\text{x}+\frac{1}{\text{x}}}}{\log_\text{e}\text{a}}$
AnswerCorrect option: A. $\frac{\text{a}^{\text{x}+\frac{1}{\text{x}}}}{\log_\text{e}\text{a}}$
$\text{f(x)}=\Big(1-\frac{1}{\text{x}^2}\Big)\text{a}^{\text{x}+\frac{1}{\text{x}}}$
$\therefore\ \int\text{f(x)}\text{dx}=\int\Big(1-\frac{1}{\text{x}^2}\Big)\text{a}^{\text{x}+\frac{1}{\text{x}}}$
Let $\Big(\text{x}+\frac{1}{\text{x}}\Big)=\text{t}$
$\Big(1-\frac{1}{\text{x}^2}\Big)\text{dx}=\text{dt}$
$\therefore\ \int\text{f(x)}\text{dx}=\int\text{a}^{\text{t}}\text{ dt}$
$=\frac{\text{a}^{\text{t}}}{\log_\text{e}\text{a}}+\text{C}$
$=\frac{\text{a}^{\text{x}+\frac{1}{\text{x}}}}{\log_\text{e}\text{a}}+\text{C}$ $\Big(\because\text{t}=\text{x}+\frac{1}{\text{x}}\Big)$
View full question & answer→MCQ 321 Mark
The value of $\int\limits^\frac{\pi}{2}_0\log\Big(\frac{4+3\sin\text{x}}{4+3\cos\text{x}}\Big)\text{dx}$ is:
AnswerLet $\text{I}=\int\limits^\frac{\pi}{2}_0\log\Big(\frac{4+3\sin\text{x}}{4+3\cos\text{x}}\Big)\text{dx}\ ...(\text{i})$
$=\int\limits^\frac{\pi}{2}_0\log\Bigg[\frac{4+3\sin\big(\frac{\pi}{2}-\text{x}\big)}{4+3\cos\big(\frac{\pi}{2}-\text{x}\big)}\Bigg]\text{dx}$
$=\int\limits^\frac{\pi}{2}_0\log\Big(\frac{4+3\cos\text{x}}{4+3\sin\text{x}}\Big)\text{dx}\ ....(\text{ii})$
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, $\text{I}=0$
View full question & answer→MCQ 331 Mark
$\int\limits^\text{e}_1\log\text{x}\text{ dx}=$
Answer$\int\limits^\text{e}_1\log\text{x}\text{ dx}$
$=\int\limits^\text{e}_1\log\text{x}\text{ x}^0\text{dx}$
$=\big[\text{x}\log\text{x}\big]^\text{e}_1-\int\limits^\text{e}_1\frac{1}{\text{x}}\text{dx}$
$=\big[\text{x}\log\text{x}\big]^\text{e}_1-\big[\text{x}\big]^\text{e}_1$
$=(\text{e}-0)-(\text{e}-1)$
$= \text{e}-\text{e}+1$
$=1$
View full question & answer→MCQ 341 Mark
$\int|\text{x}|^3\text{ dx}$ is equal to:
- A
$\frac{-\text{x}^4}{4}+\text{C}$
- B
$\frac{|\text{x}|^4}{4}+\text{C}$
- C
$\frac{\text{x}^4}{4}+\text{C}$
- ✓
Answer$\int|\text{x}|^3\text{ dx}$
$|\text{x}|=\begin{cases}\text{x},\text{ x}\geq0\\-\text{x},\text{ x}<0\end{cases}$
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:
$\text{x}<0$
$\int|\text{x}|^3\text{ dx}$
$=-\int\text{x}^3\text{ dx}$
$=\frac{-\text{x}^4}{4}+\text{C}$
View full question & answer→MCQ 351 Mark
Choose the correct option from given four options:
$\int\frac{\text{x}^9}{(4\text{x}^2+1)^6}\text{dx}$ is equal to:
- A
$\frac{1}{5\text{x}}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{C}$
- B
$\frac{1}{5}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{C}$
- C
$\frac{1}{10\text{x}}(1+4)^{-5}+\text{C}$
- ✓
$\frac{1}{10}\Big(\frac{1}{\text{x}^2}+4\Big)^{-5}+\text{C}$
AnswerCorrect option: D. $\frac{1}{10}\Big(\frac{1}{\text{x}^2}+4\Big)^{-5}+\text{C}$
Let $\text{I}=\int\frac{\text{x}^9}{(4\text{x}^2+1)^6}\text{dx}=\int\frac{\text{x}^9}{\text{x}^{12}\Big(4+\frac{1}{\text{x}^2}\Big)^6}\text{dx}=\int\frac{\text{dx}}{\text{x}^3\Big(4+\frac{1}{\text{x}^2}\Big)^6}$
Put $4+\frac{1}{\text{x}^2}=\text{t}\Rightarrow\frac{-2}{\text{x}^3}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\frac{1}{2}\int\frac{\text{dt}}{\text{t}^6}=-\frac{1}{2}\Big[\frac{\text{t}^{-6+1}}{-6+1}\Big]+\text{C}$
$=\frac{1}{10}\Big[\frac{1}{\text{t}^5}\Big]+\text{C}=\frac{1}{10}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{C}$
View full question & answer→MCQ 361 Mark
The value of $\int\limits^1_0\tan^{-1}\Big(\frac{2\text{x}-1}{1+\text{x}-\text{x}^2}\Big)\text{ dx},$ is:
AnswerLet, $\text{I}=\int\limits^1_0\tan^{-1}\frac{2\text{x}-1}{1+\text{x}-\text{x}^2}\text{ dx}\ ....(\text{i})$
$=\int\limits^1_0\tan^{-1}\frac{2(1-\text{x})-1}{1+(1-\text{x})-(1-\text{x})^2}\text{ dx}$
$=\int\limits^1_0\tan^{-1}\frac{1-2\text{x}}{2-\text{x}-1-\text{x}^2+2\text{x}}\text{ dx}$
$=\int\limits^1_0\tan^{-1}\frac{1-2\text{x}}{1+\text{x}-\text{x}^2}\text{ dx}$
$=-\int\limits^1_0\tan^{-1}\frac{2\text{x}-1}{1+\text{x}-\text{x}^2}\text{ dx}\ ....(\text{ii})$
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, $\text{I}=0$
View full question & answer→MCQ 371 Mark
$\int\frac{2\text{x}\log(1+\text{x}^2)}{1+\text{x}^2}\text{dx:}$
- A
$\log(1+\text{x}^2)+\text{c}$
- ✓
$\frac{[\log(1+\text{x}^2)^2]}{2}+\text{c}$
- C
$2\log(1+\text{x}^2)+\text{c}$
- D
AnswerCorrect option: B. $\frac{[\log(1+\text{x}^2)^2]}{2}+\text{c}$
$\int\frac{2\text{x}\log(1+\text{x}^2)}{1+\text{x}^2}\text{dx}$
Let $\log(1+\text{x}^2) =\text{z}$ and $\frac{\text{2x}}{1+\text{x}^2}\text{dx}=\text{dz}$ Using these in the above integration we get,
$=\int\text{z, }{\text{dz}}$
$\frac{\text{z}^2}{2}+\text{c}$ [Where c is integrating constant]
$=\frac{[\log(1+\text{x}^2)^2]}{2}+\text{c}$
View full question & answer→MCQ 381 Mark
$\int\limits^3_0\frac{3\text{x}+1}{\text{x}^2+9}\text{ dx}=$
- ✓
$\frac{\pi}{12}+\log\big(2\sqrt{2}\big)$
- B
$\frac{\pi}{2}+\log\big(2\sqrt{2}\big)$
- C
$\frac{\pi}{6}+\log\big(2\sqrt{2}\big)$
- D
$\frac{\pi}{3}+\log\big(2\sqrt{2}\big)$
AnswerCorrect option: A. $\frac{\pi}{12}+\log\big(2\sqrt{2}\big)$
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 $I_1$
$\Rightarrow 2\text{x}\text{ dx}=\text{dt}$
$\Rightarrow \text{x}\text{ dx}=\frac{\text{dt}}{2}$
when $\text{x}\rightarrow0;\text{t}\rightarrow9$
and $\text{x}\rightarrow3;\text{t}\rightarrow18$
$\therefore\text{I}=\int\limits^{18}_9\frac{3\text{ dt}}{2\text{ t}}+\int\limits^3_0\frac{1}{\text{x}^2+9}\text{ dx}$
$=\frac{3}{2}\int\limits^{18}_9\frac{\text{dt}}{\text{t}}+\int\limits^3_0\frac{1}{\text{x}^2+3^2}\text{ dx}$
$=\frac{3}{2}\big[\log(\text{t})\big]^{18}_9+\frac{1}{3}\Big[\tan^{-1}\Big(\frac{\pi}{3}\Big)\Big]^3_0$
$=\frac{3}{2}\big[\log18-\log9\big]+\frac{1}{3}\Big(\frac{\pi}{4}-0\Big)$
$=\frac{3}{2}\Big[\log\frac{18}{9}\Big]+\frac{\pi}{12}$
$=\frac{3}{2}\big[\log2\big]+\frac{\pi}{12}$
$=\log(\sqrt{8})+\frac{\pi}{12}$
$=\log(2\sqrt{2})+\frac{\pi}{12}$
$=\frac{\pi}{12}+\log(2\sqrt{2})$
View full question & answer→MCQ 391 Mark
What is the value of $\int_{0}^{\frac{\pi}{2}}\frac{\sin\text{x}-\cos\text{x}}{1+\sin\text{x}\cos\text{x}}\text{dx}?$
- A
$1$
- B
$\frac{\pi}{2}$
- ✓
$0$
- D
$-\frac{\pi}{2}$
View full question & answer→MCQ 401 Mark
Choose the correct answer in Exercises:
$\int\frac{\text{dx}}{\sin^2\text{x}\cos^2\text{x}}\text{ equals}$
- A
$\tan\text{x}+\cot\text{x}+\text{C}$
- ✓
$\tan\text{x}-\cot\text{x}+\text{C}$
- C
$\tan\text{x}\cot\text{x}+\text{C}$
- D
$\tan\text{x}-\cot\text{2x}+\text{C}$
AnswerCorrect option: B. $\tan\text{x}-\cot\text{x}+\text{C}$
$\int\frac{\text{dx}}{\sin^2\text{x}\cos^2\text{x}}= \int\frac{\sin^2\text{x}+\cos^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{ dx}$
$=\int\frac{\sin^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}+ \frac{\cos^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{ dx}$
$=\int\frac{1}{\cos^2\text{x}}+\frac{1}{\sin^2\text{x}}\text{ dx}=\int\sec^2\text{x dx}+\int\text{cosec}^2\text{x}\text{ dx}$
$=\int\sec^2\text{x}\text{ dx} +\text{ dx}\int\ \text{cosec}^2\text{x}\text{ dx}$
$=\tan\text{x}-\cot\text{x}+\text{c} $
Therefore, option (B) is correct.
View full question & answer→MCQ 411 Mark
Choose the correct answer in Exercises:
$\int\frac{10\text{x}^9+10^{\text{x}}\log_\text{e}10}{\text{x}^{10}+10^{\text{x}}}\text{ equals}$
- A
$10^\text{x}-\text{x}^{10}+\text{C}$
- B
$10^\text{x}+\text{x}^{10}+\text{C}$
- C
$(10^\text{x}-\text{x}^{10})^{-1}+\text{C}$
- ✓
$\log(10^\text{x}+\text{x}^{10})+\text{C}$
AnswerCorrect option: D. $\log(10^\text{x}+\text{x}^{10})+\text{C}$
$\text{Let I}=\int\frac{10\text{x}^9+10^{\text{x}}\log_\text{e}10}{\text{x}^{10}+10^{\text{x}}}\text{ dx} \ \ \ \ ...\text{(i)} $
Putting ${\text{x}^{10}+10^{\text{x}}}=\text{t}\ \ \ \ \Rightarrow\ \ \ \ (10\text{x}^9+10^{\text{x}}\log_\text{e}10)\text{ dx = dt} $
$\therefore\ \ \ \ \ $From eq. (i), $\text{I}=\int\frac{\text{dt}}{\text{t}}=\log\begin{vmatrix}\text {t}\end{vmatrix}+\text{c}=\log\begin{vmatrix}\text{x}^{10}+10^\text{x}\end{vmatrix}+\text{c}$
Therefore, option (D) is correct.
View full question & answer→MCQ 421 Mark
$\int\frac{\text{x+sin x}}{1+\text{cos x}} \ \text{dx}$ is equal to:
- A
$\log|1+\cos\text{x}|+\text{c}$
- B
$\log|\text{x}+\sin\text{x}|+\text{c}$
- C
$\text{x}-\tan+\text{c}$
- ✓
$\text{x}.\text{tan}\frac{\text{x}}{2}+\text{c}$
AnswerCorrect option: D. $\text{x}.\text{tan}\frac{\text{x}}{2}+\text{c}$
View full question & answer→MCQ 431 Mark
$\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\frac{1}{\sin2\text{x}}\text{ dx}$ is equal to:
- A
$\log_\text{e}{3}$
- ✓
$\log_\text{e}\sqrt{3}$
- C
$\frac{1}{2}\log(-1)$
- D
$\log(-1)$
AnswerCorrect option: B. $\log_\text{e}\sqrt{3}$
$\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\frac{1}{\sin2\text{x}}\text{ dx}$
$=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\text{cosec }2\text{x}\text{ dx}$
$=\frac{1}{2}\int\limits^\frac{\pi}{3}_\frac{\pi}{6}2\text{cosec }2\text{x}\text{ dx}$
$=\frac{-1}{2}\big[\log(\text{cosec}\ 2\text{x}+\cot2\text{x})\big]^\frac{\pi}{3}_\frac{\pi}{6}$
$=\frac{-1}{2}\big[-2\log\sqrt{3}\big]$
$=\log\sqrt{3}$
View full question & answer→MCQ 441 Mark
If $\int\text{f}(\text{x})\text{dx}=-2\cos\sqrt{\text{x}}+\text{c}$ then $f(x)$ is equal to:
AnswerCorrect option: B. $\frac{\sin\sqrt{\text{x}}}{\sqrt{\text{x}}}$
View full question & answer→MCQ 451 Mark
$\int\frac{\cos2\text{x dx}}{(\sin\text{x}+\cos\text{x})^2}=$
- A
$-\frac{1}{\sin\text{x}+\cos\text{x}}+\text{c}$
- ✓
$\log|\sin\text{x}+\cos\text{x }|+\text{c}$
- C
$\log|\sin\text{x}-\cos\text{x }|+\text{c}$
- D
$\frac{1}{(\sin\text{x}+\cos\text{x})^2}$
AnswerCorrect option: B. $\log|\sin\text{x}+\cos\text{x }|+\text{c}$
View full question & answer→MCQ 461 Mark
Choose the correct option from given four options$:\ \int\limits^{\frac{\pi}{4}}_{\frac{-\pi}{4}}\frac{\text{dx}}{1+\cos2\text{x}}$ is equal to:
AnswerWe have, $\int\limits^{\frac{\pi}{4}}_{\frac{-\pi}{4}}\frac{\text{dx}}{1+\cos2\text{x}}=\int\limits^{\frac{\pi}{4}}_{\frac{-\pi}{4}}\frac{\text{dx}}{2\cos^2\text{x}}$ $[\because1+\cos2\text{A}=2\cos^2\text{A}]$
$=\frac{1}{2}\int\limits^{\frac{\pi}{4}}_{\frac{-\pi}{4}}\sec^2\text{x dx}$
$=\int\limits^{\frac{\pi}{4}}_0\sec^2\text{x dx}$ $\int\limits^\text{a}_{-\text{a}}\text{f(x)dx}=\begin{cases}\int\limits^\text{a}_0\text{f(x)},&\text{if f(x) is even}\\0,&\text{if f(x) is odd}\end{cases}$
$=\Big[\tan\text{x}\Big]^{\frac{\pi}{4}}_0=1$
$=\tan\frac{\pi}{4}-\tan0$
$=1$
View full question & answer→MCQ 471 Mark
$\int\frac{\text{dx}}{\sin(\text{x-a})\sin(\text{x-b})}$ is equal to:
- A
$\sin(\text{b-a})\log|\frac{\sin(\text{x-b})}{\sin(\text{x-a})}|+\text{c}$
- B
$\text{cosec}(\text{b-a})\log|\frac{\sin(\text{x-b})}{\sin(\text{x-b})}|+\text{c}$
- ✓
$\text{cosec}(\text{b-a})\log|\frac{\sin(\text{x-b})}{\sin(\text{x-a})}|+\text{c}$
- D
$\sin(\text{b-a})\log|\frac{\sin(\text{x-a})}{\sin(\text{x-b})}|+\text{c}$
AnswerCorrect option: C. $\text{cosec}(\text{b-a})\log|\frac{\sin(\text{x-b})}{\sin(\text{x-a})}|+\text{c}$
View full question & answer→MCQ 481 Mark
Solve $\int\frac{1}{\sqrt{9-25\text{x}^2}}\text{dx}$
- ✓
$\frac{1}{5}\sin^{-1}\big(\frac{5\text{x}}{3}\big)+\text{c}$
- B
$\sin^{-1}\big(\frac{5\text{x}}{3}\big)+\text{c}$
- C
$\frac{1}{5}\sin^{-1}\big(\frac{3\text{x}}{3}\big)+\text{c}$
- D
$\sin^{-1}\big(\frac{3\text{x}}{3}\big)+\text{c}$
AnswerCorrect option: A. $\frac{1}{5}\sin^{-1}\big(\frac{5\text{x}}{3}\big)+\text{c}$
We have,
$\text{I}=\int\frac{\text{dx}}{\sqrt{9-25\text{x}^2}}$
$\text{I}=\int\frac{\text{dx}}{5\sqrt\frac{{9}}{{25}}\text{-x}^2}$
$\text{I}=\frac{1}{5}\int\frac{\text{dx}}{\sqrt{\frac{{3}}{{5}}^2\text{-x}^2}}$
We know that
$\int\frac{\text{dx}}{\text{a}^2-\text{x}^2}=\sin^{-1}\big(\frac{\text{x}}{\text{a}}\big)+\text{C}$
$\therefore\text{I} = \frac{1}{5}\sin^{-1}\Big(\frac{\text{x}}{\frac{{3}}{5}}\Big)+\text{C}$
$\therefore\text{I} = \frac{1}{5}\sin^{-1}\Big(\frac{5\text{x}}{3}\Big)+\text{C}$
View full question & answer→MCQ 491 Mark
$\int_{0}^{\infty}\frac{1}{1+\text{e}^\text{x}}\text{dx}=$
- ✓
$\log2$
- B
$-\log2$
- C
$\log2-1$
- D
$\log4-1$
AnswerCorrect option: A. $\log2$
View full question & answer→MCQ 501 Mark
If $\text{f}'(\text{x})=\text{x}+\frac{1}{\text{x}},$ then value of f(x) is:
- A
$\text{x}^2+\log\text{x + c}$
- ✓
$\frac{\text{x}^2}{2}+\log\text{x + c}$
- C
$\frac{\text{x}}{2}+\log\text{x + c}$
- D
AnswerCorrect option: B. $\frac{\text{x}^2}{2}+\log\text{x + c}$
Given,
$\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}$
On integrating both sides, we get
$\text{f}(\text{x})=\frac{\text{x}^{2}}{{2}}+\log\text{x + c}$
View full question & answer→