MCQ 1011 Mark
Integration of $\frac{1}{1+(\log_\text{e}\text{x})^2}$ with respect to $\log_\text{e}\text{x}$ is:
- A
$\frac{\tan^{-1}(\log_\text{e}\text{x})}{\text{x}}+\text{C}$
- ✓
$\tan^{-1}(\log_\text{e}\text{x})+\text{C}$
- C
$\frac{\tan^{-1}\text{x}}{\text{x}}+\text{C}$
- D
AnswerCorrect option: B. $\tan^{-1}(\log_\text{e}\text{x})+\text{C}$
$\frac{1}{1+(\log_\text{e}\text{x})^2}\text{ d}(\log_\text{e}\text{x})$
Put $\log_\text{e}\text{x}=\text{t}$
$\int\frac{\text{dt}}{1+\text{t}^2}=\tan^{-1}\text{t}+\text{C}$
$\tan^{-1}(\log_\text{e}\text{x})+\text{C}$
View full question & answer→MCQ 1021 Mark
Choose the correct option from given four options:
$\int\frac{\text{x}+\sin\text{x}}{1+\cos\text{x}}\text{dx}$ is equal to:
- A
$\log|1+\cos\text{x}|+\text{C}$
- B
$\log|\text{x}+\sin\text{x}|+\text{C}$
- C
$\text{x}-\tan\frac{\text{x}}{2}+\text{C}$
- ✓
$\text{x}\cdot\tan\frac{\text{x}}{2}+\text{C}$
AnswerCorrect option: D. $\text{x}\cdot\tan\frac{\text{x}}{2}+\text{C}$
$\text{I}=\int\frac{\text{x}+\sin\text{x}}{1+\cos\text{x}}\text{dx}$
$=\int\frac{\text{x}}{1+\cos\text{x}}\text{dx}+\int\frac{\sin\text{x}}{1+\cos\text{x}}\text{dx}$
$=\int\frac{\text{x}}{2\cos^2\frac{\text{x}}{2}}\text{dx}+\int\frac{2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}\text{dx}$
$=\int\text{x}\sec^2\frac{\text{x}}{2}\text{dx}+\int\tan\frac{\text{x}}{2}\text{dx}$
$=\frac{1}{2}\Big[\text{x}\cdot2\tan\frac{\text{x}}{2}-\int2\tan\frac{\text{x}}{2}\text{dx}\Big]+\int\tan\frac{\text{x}}{2}\text{dx}=\text{x}\cdot\tan\frac{\text{x}}{2}+\text{C}$
View full question & answer→MCQ 1031 Mark
If $\int\frac{3\text{x}+4}{\text{x}^3-2\text{x}-4}\text{dx}=\log|\text{x}-2|+\text{k}\log\text{f(x)}+\text{c},$ then:
- A
- B
- C
$\text{k}=-\frac{1}{2}$
- ✓
View full question & answer→MCQ 1041 Mark
What is the value of $\int_{1}^{\text{e}}\frac{1+\log\text{x}}{\text{x}}\text{dx}?$
- ✓
$\frac{3}{2}$
- B
$\frac{1}{2}$
- C
$\text{e}$
- D
$\frac{1}{\text{e}}$
AnswerCorrect option: A. $\frac{3}{2}$
$\frac{3}{2}$
View full question & answer→MCQ 1051 Mark
If $\frac{\text{dy}}{\text{dt}}=\text{ky}$ and $\text{k}\neq0$ which of the following could be the equation of $y $:
- A
$y = kx - 7$
- ✓
$y = 95e^{kt}$
- C
$y = 5 + ln\ k$
- D
$y = (x - k)^2$
AnswerCorrect option: B. $y = 95e^{kt}$
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 $\int\frac{\text{dy}}{\text{y}}=\int\text{kdt}$
$\Rightarrow ln (y) = kt + c \ ($where $c$ is constant$)$
$\Rightarrow y = e^ce^{kt} = ae^{kt} (a = e^c)$
The possible solution is $y = 95e^{kt}$
View full question & answer→MCQ 1061 Mark
$\int_{0}^{\frac{\pi}{2}}\sqrt{1+\sin2\text{x}}\text{dx}$ is equal to:
- A
$2\sqrt{2}$
- B
$2(\sqrt{2+1})$
- ✓
$0$
- D
$2(\sqrt{2-1})$
View full question & answer→MCQ 1071 Mark
Evaluate: $\int\frac{1}{\sqrt{9+8\text{x}-\text{x}^2}}\text{dx}.$
- A
$-\sin^{-1}(\frac{\text{x}-4}{5})+\text{c}$
- B
$\sin^{-1}(\frac{\text{x}+4}{5})+\text{c}$
- ✓
$\sin^{-1}(\frac{\text{x}-4}{5})+\text{c}$
- D
$\text{None of there}$
AnswerCorrect option: C. $\sin^{-1}(\frac{\text{x}-4}{5})+\text{c}$
$\sin^{-1}(\frac{\text{x}-4}{5})+\text{c}$
View full question & answer→MCQ 1081 Mark
$\int(\text{x}-1)\text{e}^{-\text{x}}\text{ dx}$ is equal to:
- ✓
$-\text{x}\text{e}^{\text{x}}+\text{C}$
- B
$\text{x}\text{e}^{\text{x}}+\text{C}$
- C
$-\text{x}\text{e}^{-\text{x}}+\text{C}$
- D
$\text{x}\text{e}^{-\text{x}}+\text{C}$
AnswerCorrect option: A. $-\text{x}\text{e}^{\text{x}}+\text{C}$
$\int(\text{x}-1)\text{e}^{-\text{x}}\text{ dx}$
$=(\text{x}-1)\int\text{e}^{-\text{x}}\text{ dx}-\int\Big(\Big[\frac{\text{d}(\text{x}-1)}{\text{dx}}\Big]\int\text{e}^{-\text{x}}\text{dx}\Big)\text{ dx}$
$=(\text{x}-1)\frac{\text{e}^{-\text{x}}}{-1}-\int\frac{\text{e}^{-\text{x}}}{-1}\text{ dx}$
$=-(\text{x}-1)\text{e}^{-\text{x}}+\frac{\text{e}^{-\text{x}}}{-1}+\text{C}$
$=-\text{x}\text{e}^{-\text{x}}+\text{e}^{-\text{x}}+\text{C}$
$=-\text{x}\text{e}^{-\text{x}}+\text{C}$
View full question & answer→MCQ 1091 Mark
What is the value of $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\text{dx}}{\sin2\text{x}}?$
- A
$\frac{1}{2}\log(-1)$
- B
$\log(-1)$
- ✓
$\log3$
- D
$\log\sqrt{3}$
AnswerCorrect option: C. $\log3$
$\log3$
View full question & answer→MCQ 1101 Mark
Solve: $\int\frac{\text{x}^2+1}{\text{x}^2+1}\text{dx}=$
- A
$1 + C$
- B
$x^2 + C$
- ✓
$x + C$
- D
$0$
AnswerCorrect option: C. $x + C$
View full question & answer→MCQ 1111 Mark
Evaluate: $\int(2\tan\text{x}-3\cot\text{x})^2\text{dx}.$
- A
$-4\tan\text{x}-\cot\text{x}-25\text{x}+\text{c}$
- B
$4\tan\text{x}-9\cot\text{x}-25\text{x}+\text{c}$
- C
$-4\tan\text{x}+9\cot\text{x}+25\text{x}+\text{c}$
- ✓
$4\tan\text{x}+9\cot\text{x}+25\text{x}+\text{c}$
AnswerCorrect option: D. $4\tan\text{x}+9\cot\text{x}+25\text{x}+\text{c}$
$4\tan\text{x}+9\cot\text{x}+25\text{x}+\text{c}$
View full question & answer→MCQ 1121 Mark
The value of $\int\frac{\sin\text{x}+\cos\text{x}}{\sqrt{1-\sin2\text{x}}}\text{ dx}$ is equal to:
- A
$\sqrt{\sin2\text{x}}+\text{C}$
- B
$\sqrt{\cos2\text{x}}+\text{C}$
- C
$\pm(\sin\text{x}-\cos\text{x})+\text{C}$
- ✓
$\pm\log(\sin\text{x}-\cos\text{x})+\text{C}$
AnswerCorrect option: D. $\pm\log(\sin\text{x}-\cos\text{x})+\text{C}$
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 $\sin\text{x}-\cos\text{x}=\text{t}$
$(\cos\text{x}+\sin\text{x})\text{dx}=\text{dt}$
$\therefore\ \text{I}=\pm\int\frac{\text{dt}}{\text{t}}$
$=\ln|\text{t}|+\text{C}$
$=\pm\ln(\sin\text{x}-\cos\text{x})+\text{C}$ $(\because\text{t}=\sin\text{x}-\cos\text{x})$
View full question & answer→MCQ 1131 Mark
$\int_{0}^{\frac{\pi^2}{4}}\frac{\sin\sqrt{\text{y}}}{\sqrt{\text{y}}}.$
- A
$1$
- ✓
$2$
- C
$\frac{\pi}{4}$
- D
$\frac{\pi^2}{8}$
View full question & answer→MCQ 1141 Mark
Evaluate: $\int\sec^2(7-4\text{x})\text{dx}.$
- ✓
$-\frac{1}{4}\tan(7-4\text{x})+\text{c}$
- B
$\frac{1}{4}\tan(7-4\text{x})+\text{c}$
- C
$\frac{1}{4}\tan(7+4\text{x})+\text{c}$
- D
$-\frac{1}{4}\tan(7\text{x}-4)+\text{c}$
AnswerCorrect option: A. $-\frac{1}{4}\tan(7-4\text{x})+\text{c}$
$-\frac{1}{4}\tan(7-4\text{x})+\text{c}$
View full question & answer→MCQ 1151 Mark
If $\int\frac{\sin^8\text{x}-\cos^8\text{x}}{1-2\sin^2\text{x}\cos^2\text{x}}\text{ dx}=\text{a}\sin2\text{x}+\text{C},$ then a =
- ✓
$-\frac{1}{2}$
- B
$\frac{1}{2}$
- C
$-1$
- D
$1$
AnswerCorrect option: A. $-\frac{1}{2}$
$\int\frac{\sin^8\text{x}-\cos^8\text{x}}{1-2\sin^2\text{x}\cos^2\text{x}}\text{ dx}=\text{a}\sin2\text{x}+\text{C}\ ...(\text{i})$
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
$\text{a}=-\frac{1}{2}$
View full question & answer→MCQ 1161 Mark
$\int\limits^\frac{\pi}{2}_0\frac{1}{1+\tan\text{x}}\text{dx}$ is equal to:
- ✓
$\frac{\pi}{4}$
- B
$\frac{\pi}{3}$
- C
$\frac{\pi}{2}$
- D
$\pi$
AnswerCorrect option: A. $\frac{\pi}{4}$
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, $\text{I}=\frac{\pi}{4}$
View full question & answer→MCQ 1171 Mark
$\int\limits^\pi_0\frac{1}{\text{a}+\text{b}\cos\text{x}}\text{ dx}=$
- ✓
$\frac{\pi}{\sqrt{\text{a}^2-\text{b}^2}}$
- B
$\frac{\pi}{\text{ab}}$
- C
$\frac{\pi}{\text{a}^2-\text{b}^2}$
- D
$({\text{a}+\text{b}})\pi$
AnswerCorrect option: A. $\frac{\pi}{\sqrt{\text{a}^2-\text{b}^2}}$
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 $\text{x}\rightarrow\pi;\text{ t}\rightarrow\infty$
$\therefore\ \text{I}=\int\limits^\pi_0\frac{2\text{dt}}{(\text{a}+\text{b})+(\text{a}-\text{b})\text{t}^2}$
$=\frac{2}{\text{a}-\text{b}}\int\limits^\pi_0\frac{1}{\big(\frac{\text{a}+\text{b}}{\text{a}-\text{b}}\big)+\text{t}^2}\text{dt}$
$=\frac{2}{(\text{a}-\text{b})}\int\limits^\infty_0\frac{1}{\Big(\sqrt{\frac{\text{a}+\text{b}}{\text{a}-\text{b}}}\Big)^2+\text{t}^2}\text{ dt}$
$=\frac{2}{(\text{a}-\text{b})}\times\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}$ $\Bigg[\tan^{-1}\frac{\text{t}}{\sqrt{\frac{\text{a}+\text{b}}{\text{a}-\text{b}}}}\Bigg]^\infty_0$
$=\frac{2}{\sqrt{\text{a}^2-\text{b}^2}}\Big[\frac{\pi}{2}\Big]$
$=\frac{\pi}{\sqrt{\text{a}^2-\text{b}^2}}$
View full question & answer→MCQ 1181 Mark
Evaluate: $\int\frac{1}{\sqrt{1-\text{e}^{\text{2x}}}}\text{dx}.$
- A
$\log|\text{e}^{-\text{x}}+\sqrt{\text{e}^{2\text{x}}-1}|+\text{c}$
- ✓
$-\log|\text{e}^{-\text{x}}+\sqrt{\text{e}^{2\text{x}}-1}|+\text{c}$
- C
$-\log|\text{e}^{-\text{x}}-\sqrt{\text{e}^{2\text{x}}-1}|+\text{c}$
- D
$\text{None of these}$
AnswerCorrect option: B. $-\log|\text{e}^{-\text{x}}+\sqrt{\text{e}^{2\text{x}}-1}|+\text{c}$
$-\log|\text{e}^{-\text{x}}+\sqrt{\text{e}^{2\text{x}}-1}|+\text{c}$
View full question & answer→MCQ 1191 Mark
$\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}\frac{\text{dx}}{1+\cos2\text{x}}\text{dx}$ is equal to:
View full question & answer→MCQ 1201 Mark
$\int\frac{\sin^6\text{x}}{\cos^8\text{x}}\text{ dx}=$
- A
$\tan7\text{x}+\text{C}$
- ✓
$\frac{\tan^7\text{x}}{7}+\text{C}$
- C
$\frac{\tan7\text{x}}{7}+\text{C}$
- D
$\sec^7\text{x}+\text{C}$
AnswerCorrect option: B. $\frac{\tan^7\text{x}}{7}+\text{C}$
$\text{I}=\int\frac{\sin^6\text{x dx}}{\cos^8\text{x}}$
$\text{I}=\int\tan^6\text{x}\sec^2\text{x dx}$
Put $\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\text{I}=\int\text{t}^6\text{dt}$
$\text{I}=\frac{\text{t}^7}{7}+\text{C}$
$\text{I}=\frac{\tan^7\text{x}}{7}+\text{C}$
View full question & answer→MCQ 1211 Mark
Evaluate: $\int\frac{3\text{x}^2+1}{(\text{x}^2-1)^3}\text{dx}$
- ✓
$\text{c}-\frac{\text{x}}{(\text{x}^2-1)^2}$
- B
$\text{c}-\frac{\text{x}}{(\text{x}^2+1)^2}$
- C
$\frac{\text{x}}{(\text{x}^2-1)^2}$
- D
$\frac{\text{x}}{(\text{x}^2+1)^2}$
AnswerCorrect option: A. $\text{c}-\frac{\text{x}}{(\text{x}^2-1)^2}$
$\text{c}-\frac{\text{x}}{(\text{x}^2-1)^2}$
View full question & answer→MCQ 1221 Mark
Evaluate: $\int\sqrt{1+\text{y}}^2.\text{2ydy:}$
- ✓
$\text{I}=\frac{2}{3}(1+\text{y}^2)^\frac{3}{2}+\text{c}$
- B
$\text{I}=\frac{2}{5}(1-\text{y}^2)\frac{3}{2}+\text{c}$
- C
$\text{I}=\frac{2}{3}(1-\text{y}^2)^\frac{3}{2}+\text{c}$
- D
AnswerCorrect option: A. $\text{I}=\frac{2}{3}(1+\text{y}^2)^\frac{3}{2}+\text{c}$
$\text{I}=\frac{2}{3}(1+\text{y}^2)^\frac{3}{2}+\text{c}$
View full question & answer→MCQ 1231 Mark
Integrate the following function with respect to x $\int\big(5\text{x}^2+3\text{x}-2\big)\text{dx:}$
- A
$\frac{5\text{x}^3}{3}-\frac{3\text{x}^3}{2}-\text{2x}+\text{c}$
- ✓
$\frac{5\text{x}^3}{3}+\frac{3\text{x}^2}{2}-\text{2x}+\text{c}$
- C
$\frac{5\text{x}^3}{3}-\frac{3\text{x}^2}{2}+\text{2x}-\text{c}$
- D
$\frac{5\text{x}^3}{3}+\frac{3\text{x}^3}{2}+\text{2x}+\text{c}$
AnswerCorrect option: B. $\frac{5\text{x}^3}{3}+\frac{3\text{x}^2}{2}-\text{2x}+\text{c}$
$\int\big(5\text{x}^2+3\text{x}-2\big)\text{dx}=5\int{\text{x}^2}\text{dx}+3\int\text{xdx}-2\int\text{dx}$
$=\frac{5\text{x}^3}{3}+\frac{3\text{x}^2}{2}-\text{2x}+\text{c}$
View full question & answer→MCQ 1241 Mark
$\int\frac{\text{x}^3}{\text{x}+1}\text{ dx}$ 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. $\text{x}-\frac{\text{x}^2}{2}+\frac{\text{x}^3}{3}-\log|1-\text{x}|+\text{C}$
$\text{I}=\int\frac{\text{x}^3}{\text{x}+1}\text{ dx}$
$\text{I}=\int\frac{\text{x}^3+1-1}{\text{x}+1}\text{ dx}$
$\text{I}=\int\frac{(\text{x}+1)(\text{x}^2-\text{x}+1)}{\text{x}+1}\text{ dx}$
$\text{I}=\int\Big(\text{x}^2-\text{x}+1-\frac{1}{\text{x}+1}\Big)\text{dx}$
$\text{I}=\text{x}-\frac{\text{x}^2}{2}+\frac{\text{x}^3}{3}-\log|1-\text{x}|+\text{C}$
View full question & answer→MCQ 1251 Mark
$\int\cos(\log_\text{e}.\text{x})\text{dx}$ is equal to:
- A
$\frac{1}{2}\text{x}[\cos(\log_\text{e}\text{x})+\sin(\log_\text{e}\text{x})]$
- ✓
$\text{x}[\cos(\log_\text{e}\text{x})+\sin(\log_\text{e}\text{x})]$
- C
$\frac{1}{2}\text{x}[\cos(\log_\text{e}\text{x})-\sin(\log_\text{e}\text{x})]$
- D
$\text{x}[\cos(\log_\text{e}\text{x})-\sin(\log_\text{e}\text{x})]$
AnswerCorrect option: B. $\text{x}[\cos(\log_\text{e}\text{x})+\sin(\log_\text{e}\text{x})]$
$\text{x}[\cos(\log_\text{e}\text{x})+\sin(\log_\text{e}\text{x})]$
View full question & answer→MCQ 1261 Mark
Choose the correct answer in Exercise. $\int\sqrt{\text{x}^2-8\text{x}+7}\text{dx}$ is equal to
- A
$\frac{1}{2}(\text{x}-4)\sqrt{\text{x}^2-8\text{x}+7}+9\text{log}\Bigg|\text{x}-4+\sqrt{\text{x}^2-8\text{x}+7}\Bigg|+\text{C}$
- B
$\frac{1}{2}(\text{x}+4)\sqrt{\text{x}^2-8\text{x}+7}+9\text{log}\Bigg|\text{x}+4+\sqrt{\text{x}^2-8\text{x}+7}\Bigg|+\text{C}$
- C
$\frac{1}{2}(\text{x}-4)\sqrt{\text{x}^2-8\text{x}+7}-3\sqrt2\text{log}\Bigg|\text{x}-4+\sqrt{\text{x}^2-8\text{x}+7}\Bigg|+\text{C}$
- ✓
$\frac{1}{2}(\text{x}-4)\sqrt{\text{x}^2-8\text{x}+7}-\frac{9}{2}\text{log}\Bigg|\text{x}-4+\sqrt{\text{x}^2-8\text{x}+7}\Bigg|+\text{C}$
AnswerCorrect option: D. $\frac{1}{2}(\text{x}-4)\sqrt{\text{x}^2-8\text{x}+7}-\frac{9}{2}\text{log}\Bigg|\text{x}-4+\sqrt{\text{x}^2-8\text{x}+7}\Bigg|+\text{C}$
$\frac{1}{2}(\text{x}-4)\sqrt{\text{x}^2-8\text{x}+7}+9\text{log}\Bigg|\text{x}-4+\sqrt{\text{x}^2-8\text{x}+7}\Bigg|+\text{C}$
$=\Bigg(\frac{\text{x}-4}{2}\Bigg)\sqrt{(\text{x}-4)^2-3^2}-\frac{3^2}{2}\text{log}\Big|\text{x}-4+\sqrt{(\text{x}-4)^2-3^2}\Big|+\text{C}$
$\Bigg[\therefore\ \int\sqrt{\text{x}^2-\text{a}^2}\text{dx}=\frac{\text{x}}{2}\sqrt{\text{x}^2-\text{a}^2}-\frac{\text{a}^2}{2}\text{log}\Big|\text{x}+\sqrt{\text{x}^2-\text{a}^2}\Big|\Bigg]$
$=\Bigg(\frac{\text{x}-4}{2}\Bigg)\sqrt{\text{x}^2-8\text{x}+7}-\frac{9}{2}\text{log}\Big|\text{x}-4+\sqrt{\text{x}^2-8\text{x}+7}\Big|+\text{C}$
View full question & answer→MCQ 1271 Mark
$\int\frac{\text{dx}}{\sqrt{\text{x}}}=$
- A
$\sqrt{\text{x}}+\text{k}$
- ✓
$2\sqrt{\text{x}}+\text{k}$
- C
$\text{x}+\text{k}$
- D
$\frac{2}{3}\times\frac{3}{2}+\text{k}$
AnswerCorrect option: B. $2\sqrt{\text{x}}+\text{k}$
$2\sqrt{\text{x}}+\text{k}$
View full question & answer→MCQ 1281 Mark
$\int\frac{\text{x}}{4+\text{x}^4}\text{ dx}$ is equal to:
- A
$\frac{1}{4}\tan^{-1}\text{x}^2+\text{C}$
- ✓
$\frac{1}{4}\tan^{-1}\Big(\frac{\text{x}^2}{2}\Big)$
- C
$\frac{1}{2}\tan^{-1}\Big(\frac{\text{x}^2}{2}\Big)$
- D
AnswerCorrect option: B. $\frac{1}{4}\tan^{-1}\Big(\frac{\text{x}^2}{2}\Big)$
$\text{I}=\int\frac{\text{x}}{4+\text{x}^4}\text{ dx}$
Put $\text{x}^2=\text{t}$
$2\text{xdx}=\text{dt}$
$\text{x dx}=\frac{\text{dt}}{2}$
$\text{I}=\int\frac{\frac{\text{dt}}{2}}{4+\text{t}^2}$
$\text{I}=\frac{1}{2}\tan^{-1}\Big(\frac{\text{t}}{2}\Big)+\text{C}$
$\text{I}=\frac{1}{2}\tan^{-1}\Big(\frac{\text{x}^2}{2}\Big)+\text{C}$
View full question & answer→MCQ 1291 Mark
$\int\limits^{\infty}_0\log\Big(\text{x}+\frac{1}{\text{x}}\Big)\frac{1}{1+\text{x}^2}\text{ dx}=$
- ✓
$\pi\ln 2$
- B
$-\pi\ln2$
- C
$0$
- D
$-\frac{\pi}{2}\ln2$
AnswerCorrect option: A. $\pi\ln 2$
$\int\limits^{\infty}_0\log\Big(\text{x}+\frac{1}{\text{x}}\Big)\frac{1}{1+\text{x}^2}\text{ dx}$
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,
$-\int\limits^{\frac{\pi}{2}}_0\log(\sin\theta)\text{ d}\theta-\int\limits^{\frac{\pi}{2}}_0\log\cos\theta\text{ d}\theta$
$-2\int\limits^{\frac{\pi}{2}}_0\log\sin\theta\text{ d}\theta=-2\times\text{I}$
$=-2\times-\frac{\pi}{2}\log2$
$\Big[$Where $I=-\frac{\pi}{2}\log2\Big]$
$=\pi\log2$
View full question & answer→MCQ 1301 Mark
Evaluate the following integral: $\int\sec^2\text{xdx:}$
- A
$2\tan\text{x+c}$
- B
$\tan2\text{x}+\text{c}$
- ✓
$\tan\text{x}+\text{c}$
- D
AnswerCorrect option: C. $\tan\text{x}+\text{c}$
We know that. $\frac{\text{d}}{\text{dx}}(\tan\text{x})=\sec^2\text{x}$
$\therefore\text{d }(\tan\text{x})=\sec^2\text{xdx}$
$\Rightarrow\text{I}=\int\sec^2\text{xdx}=\tan\text{x+c}$
View full question & answer→MCQ 1311 Mark
If $\int\frac{\text{dx}}{(\text{x}+2)(\text{x}^2+1)}$ $=\text{a}\log|1+\text{x}^2|+\text{b}\tan^{-1}\text{x}+\frac{1}{5}\log|\text{x}+2|+\text{c},$ 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. $\text{a}=\frac{-1}{10},\text{b}=\frac{2}{5}$
$\text{a}=\frac{-1}{10},\text{b}=\frac{2}{5}$
View full question & answer→MCQ 1321 Mark
If $\int\frac{2^{\frac{1}{\text{x}}}}{\text{x}^2}\text{dx}=\text{k }2^{\frac{1}{\text{x}}}+\text{C},$ then k is equal to:
AnswerCorrect option: A. $-\frac{1}{\log_\text{e}2}$
$\text{k}=\int\frac{2^{\frac{1}{\text{x}}}}{\text{x}^2}\text{ dx}$
Put $\frac{1}{\text{x}}=\text{t}$
$\frac{-1}{\text{x}^2}\text{ dx}=\text{dt}$
$\frac{1}{\text{x}^2}\text{ dx}=-\text{dt}$
$\text{k}=\int2^{\text{t}}(-\text{dt})$
$\text{k}=\frac{-2^{\text{t}}}{\log_\text{e}2}+\text{C}$
$\text{k}=\frac{-2^{\frac{1}{\text{x}}}}{\log_\text{e}2}+\text{C}$
$\text{k}=\frac{-1}{\log_\text{e}2}$
View full question & answer→MCQ 1331 Mark
$\int\frac{\text{x}^3}{\text{x}+1}$ 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. $\text{x}+\frac{\text{x}^2}{2}+\frac{\text{x}^3}{3}-\log|1+\text{x}|+\text{c}$
$\text{x}+\frac{\text{x}^2}{2}+\frac{\text{x}^3}{3}-\log|1+\text{x}|+\text{c}$
View full question & answer→MCQ 1341 Mark
$\int\frac{\text{dx}}{1-\cos\text{x}-\sin\text{x}}$ is equal to:
- A
$\log|1+\cot\frac{\text{x}}{2}|+\text{c}$
- B
$\log|1-\tan\frac{\text{x}}{2}|+\text{c}$
- ✓
$\log|1-\cot\frac{\text{x}}{2}|+\text{c}$
- D
$\log|1+\tan\frac{\text{x}}{2}|+\text{c}$
AnswerCorrect option: C. $\log|1-\cot\frac{\text{x}}{2}|+\text{c}$
$\log|1-\cot\frac{\text{x}}{2}|+\text{c}$
View full question & answer→MCQ 1351 Mark
Choose the correct answer in Exercise:
$\int\frac{\text{dx}}{\text{x}^2+2\text{x}+2}\text{equals}$
- A
$\text{x}\tan^{-1}(\text{x}+1)+\text{C}$
- ✓
$\tan^{-1}(\text{x}+1)+\text{C}$
- C
$(\text{x}+1)\tan^{-1}\text{x}+\text{C}$
- D
$\tan^{-1}\text{x}+\text{C}$
AnswerCorrect option: B. $\tan^{-1}(\text{x}+1)+\text{C}$
$\int\frac{\text{dx}}{\text{x}^2+2\text{x}+2}$
$=\int\frac{1}{\text{x}^2+2\text{x}+1+1}\text{dx}$
$=\int\frac{1}{(\text{x+1})^2+(1)^2}\text{ dx}$
$=\tan^{-1}(\text{x}+1)+\text{C}$
Therefore, option (B) is correct.
View full question & answer→MCQ 1361 Mark
$\int\frac{\cos2\text{x}-\cos2\theta}{\cos\text{x}-\cos\theta}\text{ dx}$ is equal to:
- ✓
$2(\sin\text{x}+\text{x}\cos\theta)+\text{C}$
- B
$2(\sin\text{x}-\text{x}\cos\theta)+\text{C}$
- C
$2(\sin\text{x}+2\text{x}\cos\theta)+\text{C}$
- D
$2(\sin\text{x}-2\text{x}\cos\theta)+\text{C}$
AnswerCorrect option: A. $2(\sin\text{x}+\text{x}\cos\theta)+\text{C}$
$\text{I}=\int\frac{\cos2\text{x}-\cos2\theta}{\cos\text{x}-\cos\theta}\text{ dx}$
$\text{I}=\int\frac{2\cos^2\text{x}-1-(2\cos^2\theta-1)}{\cos\text{x}-\cos\theta}\text{ dx}$
$\text{I}=\int\frac{2(\cos^2\text{x}-\cos^2\theta)}{\cos\text{x}-\cos\theta}\text{ dx}$
$\text{I}=2\int(\cos\text{x}+\cos\theta)\text{dx}$
$\text{I}=2(\sin\text{x}+\text{x}\cos\theta)+\text{C}$
View full question & answer→MCQ 1371 Mark
Choose the correct option from given four options:
$\int\text{e}^\text{x}\Big(\frac{1-\text{x}}{1+\text{x}^2}\Big)^2\text{dx}$ is equal to:
- ✓
$\frac{\text{e}^\text{x}}{1+\text{x}^2}+\text{C}$
- B
$\frac{-\text{e}^\text{x}}{1+\text{x}^2}+\text{C}$
- C
$\frac{\text{e}^\text{x}}{(1+\text{e}^2)^2}+\text{C}$
- D
$\frac{-\text{e}^\text{x}}{(1+\text{x}^2)^2}+\text{C}$
AnswerCorrect option: A. $\frac{\text{e}^\text{x}}{1+\text{x}^2}+\text{C}$
$\int\text{e}^\text{x}\Big(\frac{1-\text{x}}{1+\text{x}^2}\Big)^2\text{dx}$
$=\int\text{e}^\text{x}\frac{1+\text{x}^2-2\text{x}}{(1+\text{x}^2)^2}\text{dx}$
$=\int\text{e}^\text{x}\Big[\frac{1}{(1+\text{x}^2)}-\frac{2\text{x}}{(1+\text{x}^2)^2}\Big]\text{dx}$
$=\int\text{e}^\text{x}[\text{f(x)}+\text{f}'(\text{x})]\text{dx},$ where $\text{f(x)}=\frac{1}{1+\text{x}^2}$
$=\text{e}^\text{x}\text{f(x)}+\text{C}=\frac{\text{e}^\text{x}}{1+\text{x}^2}+\text{C}$
View full question & answer→MCQ 1381 Mark
If $\frac{\text{dy}}{\text{dx}}=\cos(2\text{x})$ then y =
- ✓
$\frac{\sin(2\text{x)}}{\text{2}}+\text{c}$
- B
$2\sin(2\text{x})+\text{c}$
- C
$\frac{\sin(\text{x})}{2}+\text{c}$
- D
AnswerCorrect option: A. $\frac{\sin(2\text{x)}}{\text{2}}+\text{c}$
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.
$\text{y}=\frac{\sin(2\text{x})}{2}+\text{c}$
View full question & answer→MCQ 1391 Mark
$\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\cot^3\text{x}}\text{ dx}$ is equal to:
- A
$0$
- B
$1$
- C
$\frac{\pi}{2}$
- ✓
$\frac{\pi}{4}$
AnswerCorrect option: D. $\frac{\pi}{4}$
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, $\text{I}=\frac{\pi}{4}$
View full question & answer→MCQ 1401 Mark
If $\text{I}_{10}=\int\limits^\frac{\pi}{2}_0\text{x}^{10}\sin\text{x}\text{ dx},$ then the value of $l_{10} + 90l_8$ is:
- A
$9\Big(\frac{\pi}{2}\Big)^9$
- ✓
$10\Big(\frac{\pi}{2}\Big)^9$
- C
$\Big(\frac{\pi}{2}\Big)^9$
- D
$9\Big(\frac{\pi}{2}\Big)^8$
AnswerCorrect option: B. $10\Big(\frac{\pi}{2}\Big)^9$
We have,
$\text{I}_{10}=\int\limits^\frac{\pi}{2}_0\text{x}^{10}\sin\text{x}\text{ dx}$
$=\big[\text{x}^{10}(-\cos\text{x})\big]^\frac{\pi}{2}_0-\int\limits^\frac{\pi}{2}_0\big[10\text{x}^9\int\sin\text{x}\text{ dx}\big]\text{dx}$
$=\big[-\text{x}^{10}\cos\text{x}\big]^\frac{\pi}{2}_0-10\int\limits^\frac{\pi}{2}_0\text{x}^9(-\cos\text{x})\text{dx}$
$=-\big[\text{x}^{10}\cos\text{x}\big]^\frac{\pi}{2}_0+10\int\limits^\frac{\pi}{2}_0\text{x}^9\cos\text{x}\text{ dx}$
$=-\big[\text{x}^{10}\cos\text{x}\big]^\frac{\pi}{2}_0+10\big[\text{x}^9\sin\text{x}\big]^\frac{\pi}{2}_0-10\int\limits^\frac{\pi}{2}_09\text{x}^8\sin\text{x}\text{ dx}$
$=-\Big[\Big(\frac{\pi}{2}\Big)^{10}\times0-0^{10}\cos0\Big]+10\Big[\Big(\frac{\pi}{2}\Big)^9\times1-0^9\times0\Big]$
$-90\int\limits^\frac{\pi}{2}_0\text{x}^8\sin\text{x dx}$
$=10\Big[\Big(\frac{\pi}{2}\Big)^9\times1\Big]-90\text{I}_8$
$=10\Big(\frac{\pi}{2}\Big)^9-90\text{I}_8$
$\therefore\ \text{I}_{10}+90\text{I}_8$
$=10\Big(\frac{\pi}{2}\Big)^9$
View full question & answer→MCQ 1411 Mark
The value of $\int\limits^\pi_{-\pi}\sin^3\text{x}\cos^2\text{x}\text{ dx}$ is:
- A
$\frac{\pi^4}{2}$
- B
$\frac{\pi^4}{4}$
- ✓
$0$
- D
Answer$\int\limits^\pi_{-\pi}\sin^3\text{x}\cos^2\text{x}\text{ dx}$
$=\int\limits^\pi_{-\pi}\sin\text{x}(1-\cos^2\text{x})\cos^2\text{x}\text{ dx}$
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
$\int\limits^{-1}_{-1}(1-\text{t}^2)\text{t}^2\text{ dt}$
$=0$
View full question & answer→MCQ 1421 Mark
$\int\text{e}^\text{x}(\frac{1-\text{x}}{1+\text{x}^2})^2\text{dx}$ is equal to:
- ✓
$\frac{\text{e}^\text{x}}{1+\text{x}^2}+\text{c}$
- B
$-\frac{-\text{e}^\text{x}}{1+\text{x}^2}+\text{c}$
- C
$\frac{\text{e}^\text{x}}{(1+\text{x}^2)^2}+\text{c}$
- D
$\frac{-\text{e}^\text{x}}{(1+\text{x}^2)^2}+\text{c}$
AnswerCorrect option: A. $\frac{\text{e}^\text{x}}{1+\text{x}^2}+\text{c}$
$\frac{\text{e}^\text{x}}{1+\text{x}^2}+\text{c}$
View full question & answer→MCQ 1431 Mark
Evaluate $\int\cos^3\text{xe}^{\log{\sin}\text{x}}\text{dx}$
- ✓
$-\frac{\cos^4\text{x}}{4}+\text{C}$
- B
$-\frac{\sin\text{x}}{\text{x}^2}+\text{C}$
- C
$-\frac{\cos^3\text{x}}{3}+\text{C}$
- D
AnswerCorrect option: A. $-\frac{\cos^4\text{x}}{4}+\text{C}$
$-\frac{\cos^4\text{x}}{4}+\text{C}$
View full question & answer→MCQ 1441 Mark
Choose the correct option from given four options:
If $\int\frac{\text{dx}}{(\text{x}+2)(\text{x}^2+1)}=\text{a}\log|1+\text{x}^2|+\text{b}\tan^{-1}\text{x}+\frac{1}{5}\log|\text{x}+2|+\text{C},$ 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. $\text{a}=\frac{-1}{10},\text{b}=\frac{2}{5}$
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 $\text{a}=\frac{-1}{10}$
View full question & answer→MCQ 1451 Mark
If $(\text{a}+\text{b}-\text{x})=\text{f}(\text{x}),$ then $\int\limits^\text{b}_\text{a}\text{x f}(\text{x})\text{dx}$ is equal to:
- A
$\frac{\text{a}+\text{b}}{2}\int\limits^\text{b}_\text{a}\text{f}(\text{b}-\text{x})\text{dx}$
- B
$\frac{\text{a}+\text{b}}{2}\int\limits^\text{b}_\text{a}\text{f}(\text{b}+\text{x})\text{dx}$
- C
$\frac{\text{b}-\text{a}}{2}\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx}$
- ✓
$\frac{\text{a}+\text{b}}{2}\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx}$
AnswerCorrect option: D. $\frac{\text{a}+\text{b}}{2}\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx}$
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 $\text{I}=\frac{\text{a}+\text{b}}{2}\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx}$
View full question & answer→MCQ 1461 Mark
$\int\frac{\text{dx}}{1+\text{cos x}}=$
- ✓
$\tan\frac{\text{x}}{2}+\text{k}$
- B
$\frac{1}{2}\tan\frac{\text{x}}{2}+\text{k}$
- C
$2\tan\frac{\text{x}}{2}+\text{k}$
- D
$\tan^2\frac{\text{x}}{2}+\text{k}$
AnswerCorrect option: A. $\tan\frac{\text{x}}{2}+\text{k}$
$\tan\frac{\text{x}}{2}+\text{k}$
View full question & answer→MCQ 1471 Mark
$\int\frac{\sin^2\text{x}}{\cos^4\text{x}}\text{ dx}=$
- A
$\frac{1}{3}\tan^2\text{x}+\text{C}$
- B
$\frac{1}{2}\tan^2\text{x}+\text{C}$
- ✓
$\frac{1}{3}\tan^3\text{x}+\text{C}$
- D
AnswerCorrect option: C. $\frac{1}{3}\tan^3\text{x}+\text{C}$
$\text{I}=\int\frac{\sin^2\text{x}}{\cos^4\text{x}}\text{ dx}$
$\text{I}=\int\tan^{2}\text{x}\sec^2\text{x dx}$
Put $\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\text{I}=\int\text{t}^2\text{dt}$
$\text{I}=\frac{\text{t}^3}{3}+\text{C}$
$\text{I}=\frac{\tan^3\text{x}}{3}+\text{C}$
View full question & answer→MCQ 1481 Mark
$\int_{0}^{1}\frac{(\tan^{-1}\text{x})^2}{1+\text{x}^2}\text{dx}=$
- A
$1$
- B
$\frac{\pi^2}{64}$
- ✓
$\frac{\pi^2}{192}$
- D
$\text{None of these}$
AnswerCorrect option: C. $\frac{\pi^2}{192}$
$\frac{\pi^2}{192}$
View full question & answer→MCQ 1491 Mark
Choose the correct answer in Exercises:
$\int\frac{\text{e}^{\text{x}}(1+\text{x})}{\cos^2(\text{e}^\text{x}\text{x})}\text{dx}$ equals
- A
$-\cot(\text{e}^{\text{x}}\text{x})+\text{C}$
- ✓
$\tan(\text{xe}^\text{x})+\text{C}$
- C
$\tan(\text{e}^\text{x})+\text{C}$
- D
$\cot(\text{e}^\text{x})+\text{C}$
AnswerCorrect option: B. $\tan(\text{xe}^\text{x})+\text{C}$
$\int\frac{\text{e}^{\text{x}}(1+\text{x})}{\cos^2(\text{e}^\text{x}\text{x})}\text{dx}$
$\text{Let }\text{e}^{\text{x}}\text{x}=\text{t}$
$\Rightarrow(\text{e}^\text{x}\cdot\text{x}+\text{e}^{\text{x}}\cdot1)\text{dx}=\text{dt}$
$\text{e}^{\text{x}}(\text{x}+1)\text{dx}=\text{dt}$
$\therefore\int\frac{\text{e}^{\text{x}}(1+\text{x})}{\cos^2(\text{e}^\text{x}\text{x})}\text{dx}=\int\frac{\text{dt}}{\cos^2\text{t}}$
$=\int\sec^2\text{t}\text{ dt}$
$=\tan\text{t}+\text{C}$
$=\tan(\text{e}^{\text{x}}\cdot\text{x})+\text{C}$
Hence, the correct answer is B.
View full question & answer→MCQ 1501 Mark
What is $\int\frac{\text{dx}}{\text{x}(1+\text{lnx})^\text{n}}$ equal to $(\text{n}\neq1)$
- A
$\frac{{1}}{{(\text{n}-1)}(1+\text{lnx})^{\text{n}-1}}+\text{c}$
- B
$\frac{1-\text{n}}{(1+\text{lnx})^{1-\text{n}}}+\text{c}$
- C
$\frac{{\text{n}+1}}{{(1+\text{lnx})}^{\text{n}+1}}+\text{c}$
- ✓
$-\frac{1}{(\text{n}+1)(1+\text{lnx})^{\text{n}-1}}+\text{c}$
AnswerCorrect option: D. $-\frac{1}{(\text{n}+1)(1+\text{lnx})^{\text{n}-1}}+\text{c}$
Given,
$\int\frac{1}{\text{x}(1+\text{ln}(\text{x}))^\text{n}}\text{dx}$
apply u = 1 + ln (x)
$=\int\frac{1}{\text{u}^{\text{n}}}\text{du}$
$=\int\text{u}^{-\text{n}}\text{du}$
$=\frac{\text{u}^{-\text{n}+1}}{-\text{n}+1}$
$=\frac{(1+\text{ln}(\text{x}))^{-{\text{n}+1}}}{{-\text{n}+1}}$
$-\frac{1}{(\text{n}+1)(1+\text{lnx})^{\text{n}-1}}+\text{c}$
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