MCQ 1511 Mark
$\int\text{e}^{\text{x}}\Big(\frac{1-\sin\text{x}}{1-\cos\text{x}}\Big)\text{dx}=$
- A
$-\text{e}^{\text{x}}\tan\frac{\text{x}}{2}+\text{C}$
- ✓
$-\text{e}^{\text{x}}\cot\frac{\text{x}}{2}+\text{C}$
- C
$-\frac{1}{2}\text{e}^{\text{x}}\tan\frac{\text{x}}{2}+\text{C}$
- D
$-\frac{1}{2}\text{e}^{\text{x}}\cot\frac{\text{x}}{2}+\text{C}$
AnswerCorrect option: B. $-\text{e}^{\text{x}}\cot\frac{\text{x}}{2}+\text{C}$
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 $\int\text{e}^{\text{x}}\{\text{f(x)}+\text{f}'(\text{x})\}\text{dx}=\text{e}^{\text{x}}\text{f(x)}+\text{C}$
$\therefore\ \text{I}=-\text{e}^{\text{x}}\cot\Big(\frac{\text{x}}{2}\Big)+\text{C}$
View full question & answer→MCQ 1521 Mark
$\int\frac{\text{a}}{(1+\text{x}^2)\tan^{-1}\text{x}}\text{dx}=$
- ✓
$\text{a}\log|\tan^{-1}\text{x}|+\text{c}$
- B
$\frac{1}{2}(\tan^{-1}\text{x})^2+\text{c}$
- C
$\text{a}\log(1+\text{x}^2)+\text{c}$
- D
AnswerCorrect option: A. $\text{a}\log|\tan^{-1}\text{x}|+\text{c}$
View full question & answer→MCQ 1531 Mark
For $\text{x, }\in\text{ R},\text{f}(\text{x})=\mid\log2-\sin\text{x}\mid$ and $g(x) = f(f(x))$ then:
- ✓
$\text{g}\ '(0)=\cos (\log2)$
- B
$\text{g}\ '(0)=-\cos (\log2)$
- C
$\text{g }$ is diffrerentible at $x = 0$ and $\text{g}\ '(0)=-\sin(\log2)$
- D
$\text{g }$ is diffrerentible at $x = 0$
AnswerCorrect option: A. $\text{g}\ '(0)=\cos (\log2)$
View full question & answer→MCQ 1541 Mark
Choose the correct option from given four options:
$\int\frac{\cos2\text{x}-\cos2\theta}{\cos\text{x}-\cos\theta}\text{dx}$ is equal to:
- ✓
$2(\sin+\text{x}\cos\theta)+\text{C}$
- B
$2(\sin-\text{x}\cos\theta)+\text{C}$
- C
$2(\sin+2\text{x}\cos\theta)+\text{C}$
- D
$2(\sin-2\text{x}\cos\theta)+\text{C}$
AnswerCorrect option: A. $2(\sin+\text{x}\cos\theta)+\text{C}$
Let $\text{I}=\int\frac{\cos2\text{x}-\cos2\theta}{\cos\text{x}-\cos\theta}\text{dx}$
$=\int\frac{\big(2\cos^2\text{x}-1-2\cos^2\theta+1)}{\cos\text{x}-\cos\theta}\text{dx}$
$=2\int\frac{\big(\cos\text{x}+\cos\theta)(\cos\text{x}-\cos\theta)}{\cos\text{x}-\cos\theta}\text{dx}$
$=2\int(\cos\text{x}+\cos\theta)\text{dx}$
$=2(\sin\text{x}+\text{x}\cos\theta)+\text{C}$
View full question & answer→MCQ 1551 Mark
Calculate: $\int(\text{x}^3-\frac{1}{\text{x}}+{3\text{x}})\text{dx:}$
- A
$\frac{\text{x}^{4}}{3}-\log\text{x}+\frac{\text{5x}^{2}}{2}+\text{c}$
- B
$\frac{\text{x}^{4}}{4}-\log\text{x}+\frac{\text{2x}^{2}}{3}+\text{c}$
- C
$\frac{\text{x}^{4}}{4}-\log\text{x}+\frac{\text{3x}^{2}}{4}+\text{c}$
- ✓
$\frac{\text{x}^{4}}{4}-\log\text{x}+\frac{\text{3x}^{2}}{2}+\text{c}$
AnswerCorrect option: D. $\frac{\text{x}^{4}}{4}-\log\text{x}+\frac{\text{3x}^{2}}{2}+\text{c}$
$\int(\text{x}^3-\frac{1}{\text{x}}+{3\text{x}})\text{dx}=\frac{x^4}{4}-\text{ln}\mid{\text{x}}\mid{+}\frac{3\text{x}^2}{2}+\text{c}$
View full question & answer→MCQ 1561 Mark
Solve: $\int\limits_{0}^{\frac{\pi}{2}}\sqrt{1+\sin2\text{x}\text{dx}}=$
- A
$\frac{1}{2}$
- B
$1$
- ✓
$2$
- D
$\frac{3}{2}$
Answer$\int\limits_{0}^{\frac{\pi}{2}}\sqrt{1+\sin2\text{x}\text{dx}}=$
$\int\limits_{0}^{\frac{\pi}{2}}\sqrt{\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}\cos\text{xdx}}$
$\int\limits_{0}^{\frac{\pi}{2}}\sqrt{(\sin\text{x}+\cos\text{x})^2}\text{dx}$
$\int\limits_{0}^{\frac{\pi}{2}}(\sin\text{x}+\cos\text{x})\text{dx}$
$=[-\cos\text{x}+\sin\text{x}]\frac{\frac{\pi}{2}}{{0}}$
$=\Big[-\Big(\cos\frac{\pi}{2}-\cos0\Big)+\Big(\sin\frac{\pi}{2}-\sin0\Big)\Big]$
$=\big[-\big(0-1\big)+\big(1-0\big)\big]$
$=1+1=2$
View full question & answer→MCQ 1571 Mark
The value of $\int\limits^{\pi}_0\frac{\text{x}\tan\text{x}}{\sec\text{x}+\cos\text{x}}\text{ dx}$ is:
- A
$\frac{\pi^2}{4}$
- B
$\frac{\pi^2}{2}$
- ✓
$\frac{3\pi^2}{2}$
- D
$\frac{\pi^2}{2}$
AnswerCorrect option: C. $\frac{3\pi^2}{2}$
$\int\limits^{2\pi}_0\sqrt{1+\sin\frac{\text{x}}{2}}\text{dx}$
$=\int\limits^{2\pi}_0\sqrt{\sin^2\frac{\text{x}}{4}+\cos^2\frac{\text{x}}{4}+2\sin\frac{\text{x}}{4}\cos\frac{\text{x}}{4}}\text{dx}$
$=\int\limits^{2\pi}_0\Big(\sin\frac{\text{x}}{4}+\cos\frac{\text{x}}{4}\Big)\text{dx}$
$=\Bigg[\frac{-\cos^\frac{\text{x}}{4}}{\frac{1}{4}}+\frac{\sin\frac{\text{x}}{4}}{\frac{1}{4}}\Bigg]^{2\pi}_0$
$=4\Big[\frac{\text{x}}{4}-\cos\frac{\text{x}}{4}\Big]^{2\pi}_0$
$=4\Big[\sin\frac{2\pi}{4}-\cos\frac{2\pi}{4}-\sin0+\cos0\Big]$
$=4\Big[\sin\frac{\pi}{2}-\cos\frac{\pi}{2}-0+1\Big]$
$=4\big[1-0-0+1\big]$
$=4\times2$
$=8$
View full question & answer→MCQ 1581 Mark
$\int_{0}^{1}\frac{\text{x}}{1+\text{x}}\text{dx}=$
- ✓
$1-\log2$
- B
$2$
- C
$1+\log 2$
- D
$\log2$
AnswerCorrect option: A. $1-\log2$
View full question & answer→MCQ 1591 Mark
$\int\frac{\cot\text{x}}{3\sqrt{\sin\text{x}}}\text{dx}=$
- ✓
$\frac{-3}{3\sqrt{\sin\text{x}}}+\text{c}$
- B
$\frac{-2}{\sqrt{\sin^3\text{x}}}+\text{c}$
- C
$\frac{3}{\sqrt{\sin^{\frac{1}{3}}\text{x}}}+\text{c}$
- D
AnswerCorrect option: A. $\frac{-3}{3\sqrt{\sin\text{x}}}+\text{c}$
View full question & answer→MCQ 1601 Mark
$\int\limits^\pi_0\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{ dx}=$
- ✓
$\sqrt{1-\pi^2}-1$
- B
$\frac{\pi}{2}-1$
- C
$\frac{\pi}{2}+1$
- D
${\pi}+{1}$
AnswerCorrect option: A. $\sqrt{1-\pi^2}-1$
We have,
$\text{I}=\int\limits^\pi_0\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{dx}$
$=\int\limits^\pi_0\Big[\sqrt{\frac{1-\text{x}}{1+\text{x}}}\times\frac{\sqrt{1-\text{x}}}{\sqrt{1-\text{x}}}\Big]\text{dx}$
$=\int\limits^\pi_0\frac{1-\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}$
$=\int\limits^\pi_0\frac{1}{\sqrt{1-\text{x}^2}}\text{dx}-\int\limits^\pi_0\frac{\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}$
Putting $1 -\text{x}^2=\text{t}$
$\Rightarrow - 2\text{x}\text{ dx}=\text{dt}$
$\Rightarrow \text{x}\text{ dx}=\frac{-\text{dt}}{2}$
when $\text{x}\rightarrow0;\text{t}\rightarrow1$
and $\text{x}\rightarrow\pi;\text{ t}\rightarrow-\pi^2$
$\therefore\ \text{I}=\int\limits^\pi_0\frac{1}{\sqrt{1-\text{x}^2}}-\int\limits^{(1-\pi^2)}_1\frac{-\text{dt}}{2\sqrt{\text{t}}}$
$=\big[\sin^{-1}\text{x}\big]^{\pi}_0+\frac{2}{2}\big[\sqrt{\text{t}}\big]^{1-\pi^2}_1$
$=\big[0-0\big]+\Big[\sqrt{1-\pi^2}-\sqrt{1}\Big]$
$=\sqrt{1-\pi^2}-1$
View full question & answer→MCQ 1611 Mark
The antiderivative of every odd function is:
- A
- ✓
- C
- D
Sometimes even, sometimes odd
AnswerThe anti derivative of an odd function is even. Let f(x) be odd
eg = f(x) = x odd function
$\int\text{xdx}=\frac{\text{x}^2}{2}+\text{c}$
$\text{g}'(\text{x})=\frac{{\text{x}}^{2}}{\text{x}}+\text{c}$ is even.
View full question & answer→MCQ 1621 Mark
$\int\frac{\sin\text{x}+\cos\text{x}}{\sqrt{1+2\sin\text{x}}}\text{dx}=$
AnswerCorrect option: B. $\text{x}$
View full question & answer→MCQ 1631 Mark
If $\int\sin\text{xd}(\sec\text{x})=\text{f}(\text{x})-\text{g}(\text{x})+\text{c},$ then:
- A
$\text{f}(\text{x})=\sec\text{x}$
- ✓
$\text{f}(\text{x})=\tan\text{x}$
- C
$\text{g}(\text{x})=2\text{x}$
- D
$\text{g}=-\text{x}$
AnswerCorrect option: B. $\text{f}(\text{x})=\tan\text{x}$
$\int\sin \text{xd}(\sec\text{x})=\int\sin\text{x}\sec\text{x}\tan\text{xdx}$
$=\int\tan^2\text{xdx}$
$=\int(\sec^2\text{x - 1})\text{dx}$
$=\tan\text{x - x}+\text{c}$
$\Rightarrow\text{f}(\text{x})=\tan\text{x},\text{g}(\text{x})=\text{x}$
View full question & answer→MCQ 1641 Mark
$\int\frac{\sin\text{x} + \cos\text{x}}{\sqrt{(1 + \sin2\text{x})}}\text{dx}$ is:
- A
$\sin\text{x + c}$
- ✓
$\text{x + c}$
- C
$\cos\text{x + c}$
- D
$\tan\text{x + c}$
AnswerCorrect option: B. $\text{x + c}$
$\int\frac{\sin\text{x} + \cos\text{x}}{\sqrt{(1 + \sin2\text{x})}}\text{dx}$
$\int\frac{\sin\text{x} + \cos\text{x}}{\sqrt{(\sin\text{x}+\cos\text{x})^2}}\text{dx}$
$=\int1\text{dx}$
$=\text{x + c}$
View full question & answer→MCQ 1651 Mark
Choose the correct answer in Exercise:
The value of $\int^{1}_{0}\tan^{-1}\bigg(\frac{2\text{x}-1}{1+\text{x}-\text{x}^{2}}\bigg)\text{dx}$ is
Answer$\text{Let I}=\int^{1}\limits_{0}\tan^{-1}\bigg(\frac{2\text{x}-1}{1+\text{x}-\text{x}^{2}}\bigg)\text{dx}$
$\Rightarrow\text{I}=\int^{1}\limits_{0}\tan^{-1}\bigg(\frac{\text{x}-(1-\text{x})}{1+\text{x}(1-\text{x})}\bigg)\text{dx}$
$\Rightarrow\text{I}=\int^{1}\limits_{0}\Big[\tan^{-1}\text{x}-\tan^{-1}(1-\text{x)}\Big]\text{dx}$
$\Rightarrow\text{I}=\int^{\text{1}}\limits_{0}\Big[\tan^{-1}(1-\text{x)}-\tan^{-1}(1-1+\text{x)}\Big]\text{dx}$
$\Rightarrow\text{I}=\int^{1}\limits_{0}\Big[\tan^{-1}(1-\text{x)}-\tan^{-1}\text{(x)}\Big]\text{dx}$
$\Rightarrow\text{I}=\int^{1}_{0}\Big[\tan^{-1}(1-\text{x)}-\tan^{-1}\text{(x)}\Big]\text{dx}$
Adding (1) and (2), we obtain
$\Rightarrow2\text{I}=\int^{1}\limits_{0}\Big(\tan^{-1}\text{x)}+\tan^{-1}(1-\text{x)}-\tan^{-1}\text{x}\Big)\text{dx}$
$\Rightarrow2\text{I}=0$
$\Rightarrow\text{I}=0$
View full question & answer→MCQ 1661 Mark
$\int_{-2}^{2}|\text{x}|\text{dx}=$
View full question & answer→MCQ 1671 Mark
$\int|\text{x}|\text{dx}$ is equal to:
- A
$\frac{1}{2}\text{x}^2+\text{c}$
- B
$-\frac{\text{x}^2}{2}+\text{c}$
- C
$\text{x}|\text{x}|+\text{c}$
- ✓
$\frac{1}{2}\text{x}|\text{x}|+\text{c}$
AnswerCorrect option: D. $\frac{1}{2}\text{x}|\text{x}|+\text{c}$
View full question & answer→MCQ 1681 Mark
If $\text{g}'(\text{x})=\int\text{x}^\text{x}\log_\text{e}(\text{ex})\text{dx}$ then $\text{g}(\pi)$ equals:
AnswerCorrect option: D. $\pi^\pi$
$\text{g}'(\text{x})=\int\text{x}^{\text{x}}(1+\log{\text{e}^\text{x}})\text{dx}$
$=\int\text{d}(\text{x}^{\text{x}})$
$\text{g'}(\text{x})=\text{x}^{\text{x}}$
$\text{g'}({\pi})={\pi}^{\pi}$
View full question & answer→MCQ 1691 Mark
Choose the correct answer in Exercises:If $\frac{\text{d}}{\text{dx}}\text{f}\text{(x)}=4\text{x}^3-\frac{3}{\text{x}^4}$such that $\text{f}(2)=0.$Then $\text{f}\text{(x)}$ is
- ✓
$\text{x}^4+\frac{1}{\text{x}^3}-\frac{129}{8}$
- B
$\text{x}^3+\frac{1}{\text{x}^4}+\frac{129}{8}$
- C
$\text{x}^4+\frac{1}{\text{x}^3}+\frac{129}{8}$
- D
$\text{x}^3+\frac{1}{\text{x}^4}-\frac{129}{8}$
AnswerCorrect option: A. $\text{x}^4+\frac{1}{\text{x}^3}-\frac{129}{8}$
$\text{f(x)}=\int\bigg(4\text{x}^3-\frac{3}{\text{x}^4}\bigg)\text{ dx}$
$=4\int\text{x}^3\text{ dx}-3\int\frac{1}{\text{x}^4}\text{ dx} $
$=4.\frac{\text{x}^4}{4}-3\int\text{x}^{-4}\text{ dx} =\text{x}^4-3\frac{\text{x}^-3}{-3}+\text{c} $
$\Rightarrow\ \ \ \ \ \ \ \text{f(x)}=\text{x}^4+\frac{1}{\text{x}^3}+\text{c} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ....\text{(i)} $
$\Rightarrow\ \ \ \ \ \ \ \text{f(2)}=16+\frac{1}{8}+\text{c} \ \ \ \ \Rightarrow\ \ \ \ \ \ 0=\frac{128+1}{8}+\text{c} $
$\Rightarrow\ \ \ \ \ \ \ \text{c }+\frac{129}{8}=0 \ \ \ \ \Rightarrow\ \ \ \ \ \ \text{c }=\frac{-129}{8} $
Putting $\text{c}=\frac{-129}{8}$ in eq. (i),
$\text{f(x)}=\text{x}^4+\frac{1}{\text{x}^3}-\frac{129}{8} $
Therefore, option (A) is correct.
View full question & answer→MCQ 1701 Mark
$\int\log_{10}\text{xdx}=$
- A
$\log_\text{e}10.\text{x}\log_\text{e}(\frac{\text{x}}{\text{e}})+\text{c}$
- ✓
$\log_{10}\text{e}.\text{x}\log_\text{e}(\frac{\text{x}}{\text{e}})+\text{c}$
- C
$(\text{x}-1)\log_\text{e}\text{x}+\text{c}$
- D
$\frac{1}{\text{x}}+\text{c}$
AnswerCorrect option: B. $\log_{10}\text{e}.\text{x}\log_\text{e}(\frac{\text{x}}{\text{e}})+\text{c}$
View full question & answer→MCQ 1711 Mark
Choose the correct answer in Exercise:
$\int\frac{\cos2\text{x}}{(\sin\text{x}+\cos\text{x)}^{2}}$ is equal to
- 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
$\frac{1}{(\sin\text{x}+\cos\text{x)}^{2}}+\text{C}$
AnswerCorrect option: B. $\log|\sin\text{x}+\cos\text{x}|+\text{C}$
$\text{Let I}=\int\frac{\cos2\text{x}}{(\cos\text{x}+\sin\text{x)}^{2}}$
$\text{I}=\int\frac{(\cos^{2}\text{x}-\sin^{2}\text{x)}}{(\cos\text{x}+\sin\text{x)}^{2}}\text{dx}$
$=\int\frac{(\cos\text{x}+\sin\text{x})(\cos\text{x}-\sin\text{x)}}{(\cos\text{x}+\sin\text{x})^2}\text{dx}$
$=\int\frac{\cos\text{x}-\sin\text{x}}{\cos\text{x}+\sin\text{x}}\text{dx}$
$\text{Let}\cos\text{x}+\sin\text{x}=\text{t}\Rightarrow(\cos\text{x}-\sin\text{x)}\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|\cos\text{x}+\sin\text{x}|+\text{C}$
View full question & answer→MCQ 1721 Mark
$\int\frac{10\text{x}^9+10^\text{x}\log_\text{e}10}{10^\text{x}+\text{x}^{10}}\text{dx}$ is equal to:
- A
$10^\text{x}-\text{x}^{10}+\text{c}$
- B
$10^\text{x}+\text{x}^{10}+\text{c}$
- C
$(10^\text{x}-\text{x}^{10})^{-1}+\text{c}$
- ✓
$\log_\text{e}(10^\text{x}+\text{x}^{10})+\text{c}$
AnswerCorrect option: D. $\log_\text{e}(10^\text{x}+\text{x}^{10})+\text{c}$
View full question & answer→MCQ 1731 Mark
Choose the correct answer in Exercises:
The anti derivative of $\bigg(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\bigg)$equals
- A
$\frac{1}{3}\text{x}^{\frac{1}{3}}+2\text{x}^{\frac{1}{2}}+\text{c}$
- B
$\frac{2}{3}\text{x}^{\frac{2}{3}}+\frac{1}{2}\text{x}^2+\text{c}$
- ✓
$\frac{2}{3}\text{x}^{\frac{3}{2}}+2\text{x}^{\frac{1}{2}}+\text{c}$
- D
$\frac{3}{2}\text{x}^{\frac{3}{2}}+\frac{1}{2}\text{x}^{\frac{1}{2}}+\text{c}$
AnswerCorrect option: C. $\frac{2}{3}\text{x}^{\frac{3}{2}}+2\text{x}^{\frac{1}{2}}+\text{c}$
$\int\bigg(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\bigg)\text{ dx}=\int\bigg(\text{x}^{\frac{1}{2}}+\text{x}^{\frac{-1}{2}}\bigg)\text{ dx} $ $=\int\text{x}^\frac{1}{2}\text{ dx}+\int\text{x}^\frac{-1}{2}\text{ dx}=\frac{\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1} +\frac{\text{x}^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+\text{C} $ $=\frac{\text{x}^{\frac{3}{2}}}{\frac{3}{2}} +\frac{\text{x}^{\frac{1}{2}}}{\frac{1}{2}}+\text{C}$$=\frac{2}{3}\text{x}^\frac{3}{2}+2\text{x}^\frac{1}{2}+\text{C}$
View full question & answer→MCQ 1741 Mark
$\int\limits^{2\text{a}}_0\text{f}(\text{x})\text{dx}$ is equal to:
- A
$2\int\limits^{\text{a}}_0\text{f(x)}\text{dx}$
- B
$0$
- ✓
$\int\limits^{\text{a}}_0\text{f}(\text{x})\text{dx}+\int\limits^{\text{a}}_0\text{f}(2\text{a}-\text{x})\text{dx}$
- D
$\int\limits^{\text{a}}_0\text{f}(\text{x})\text{dx}+\int\limits^{2\text{a}}_0\text{f}(2\text{a}-\text{x})\text{dx}$
AnswerCorrect option: C. $\int\limits^{\text{a}}_0\text{f}(\text{x})\text{dx}+\int\limits^{\text{a}}_0\text{f}(2\text{a}-\text{x})\text{dx}$
$\int\limits^\text{a}_0\text{f}(\text{x})\text{dx}+\int\limits^\text{a}_0\text{f}(2\text{a}-\text{x})\text{dx}$
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,
$\int\limits^{2\text{a}}_0\text{f}(\text{x})\text{dx}=\int\limits^\text{a}_0\text{f}(\text{x})\text{dx}+\int\limits^\text{a}_0\text{f}(2\text{a}-\text{x})\text{dx}$
View full question & answer→MCQ 1751 Mark
The value of $\int\limits^{\pi}_0\frac{1}{5+3\cos\text{x}}\text{ dx}$ is:
- ✓
$\frac{\pi}{4}$
- B
$\frac{\pi}{8}$
- C
$\frac{\pi}{2}$
- D
$0$
AnswerCorrect option: A. $\frac{\pi}{4}$
$\int\limits^{\pi}_0\frac{1}{5+3\cos\text{x}}\text{ dx}$
$=\int\limits^{\pi}_0\frac{1}{5+3\frac{1-\tan^{2}\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}}}\text{ dx}$
$=\int\limits^{\pi}_0\frac{1+\tan^{2}\frac{\text{x}}{2}}{5+5\tan^{2}\frac{\text{x}}{2}+3-3\tan^{2}\frac{\text{x}}{2}}$
$=\int\limits^{\pi}_0\frac{\sec^2\frac{\text{x}}{2}}{8+2\tan^{2}\frac{\text{x}}{2}}\text{ dx}$
Let $\tan\frac{\text{x}}{2}=\text{t},$ then $\sec^2\frac{\text{x}}{2}\text{ dx}=2\text{dt}$
When $\text{x}=0,\text{ t}=0,\text{x}=\pi,\text{ t}=\infty$
Therefore the integral becomes
$\frac{1}{2}\int\limits^{\infty}_0\frac{\text{dt}}{4+\text{t}^2}$
$=\frac{1}{2}\Big[\tan^{-1}\frac{\text{t}}{2}\Big]^{\infty}_0$
$=\frac{1}{2}\Big(\frac{\pi}{2}-0\Big)$
$=\frac{\pi}{4}$
View full question & answer→MCQ 1761 Mark
$\int\frac{1}{7+5\cos\text{x}}\text{ dx}=$
- ✓
$\frac{1}{\sqrt{6}}\tan^{-1}\Big(\frac{1}{\sqrt{6}}\tan\frac{\text{x}}{2}\Big)+\text{C}$
- B
$\frac{1}{\sqrt{3}}\tan^{-1}\Big(\frac{1}{\sqrt{3}}\tan\frac{\text{x}}{2}\Big)+\text{C}$
- C
$\frac{1}{4}\tan^{-1}\Big(\tan\frac{\text{x}}{2}\Big)+\text{C}$
- D
$\frac{1}{7}\tan^{-1}\Big(\tan\frac{\text{x}}{2}\Big)+\text{C}$
AnswerCorrect option: A. $\frac{1}{\sqrt{6}}\tan^{-1}\Big(\frac{1}{\sqrt{6}}\tan\frac{\text{x}}{2}\Big)+\text{C}$
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 $\tan\frac{\text{x}}{2}=\text{t}$
$\frac{1}{2}\sec^2\Big(\frac{\text{x}}{2}\Big)\text{dx}=\text{dt}$
$\sec^2\Big(\frac{\text{x}}{2}\Big)\text{dx}=2\text{dt}$
$\therefore\ \text{I}=\frac{1}{2}\int\frac{2\text{ dt}}{\text{t}^2+(\sqrt{6})^2}$
$=\frac{1}{\sqrt{6}}\tan^{-1}\Big(\frac{\text{t}}{\sqrt{6}}\Big)+\text{C}$ $\Big(\because\int\frac{1}{\text{a}^2+\text{x}^2}=\frac{1}{\text{a}}\tan^{-1}\frac{\text{x}}{\text{a}}+\text{C}\Big)$
$=\frac{1}{\sqrt{6}}\tan^{-1}\bigg(\frac{\tan\frac{\pi}{2}}{\sqrt{16}}\bigg)+\text{C}$ $\Big(\because\text{t}=\tan\frac{\text{x}}{2}\Big)$
View full question & answer→MCQ 1771 Mark
$\int(1+2\text{x}+3\text{x}^2+4\text{x}^3+ ... )\text{dx }(\mid\text{x}\mid<1)$
- A
$-(1+\text{x})^{-1}+\text{c}$
- ✓
$(1-\text{x})^{-1}+\text{c}$
- C
$-(1-\text{x})^{-2}+\text{c}$
- D
AnswerCorrect option: B. $(1-\text{x})^{-1}+\text{c}$
View full question & answer→MCQ 1781 Mark
Integrate the following functions with respect to t $\int(3\text{t}^2-2\text{t})\text{dt:}$
AnswerCorrect option: A. $t^3 - t^2 + C$
View full question & answer→MCQ 1791 Mark
$\int\limits^\infty_0\frac{1}{1+\text{e}^\text{x}}\text{dx}$ equals:
- ✓
$\log2-1$
- B
$\log2$
- C
$\log4-1$
- D
$-\log2$
AnswerCorrect option: A. $\log2-1$
$\int\limits^\frac{\pi^2}{4}_0\frac{\sin\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{dx}$
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
$\int\limits^\frac{\pi}{2}_02\sin\text{t}\text{ dt}$
$=-2\big[\cos\text{t}\big]^\frac{\pi}{2}_0$
$=2$
View full question & answer→MCQ 1801 Mark
$\int\text{x}^{\sin\text{x}}\Big(\frac{\sin\text{x}}{\text{x}}+\cos\text{x}\cdot\log\text{x}\Big)\text{dx}$ is equal to:
- ✓
$\text{x}^{\sin\text{x}}+\text{C}$
- B
$\text{x}^{\sin\text{x}}\cos\text{x}+\text{C}$
- C
$\frac{(\text{x}^{\sin\text{x}})^2}{2}+\text{C}$
- D
AnswerCorrect option: A. $\text{x}^{\sin\text{x}}+\text{C}$
$\int\text{x}^{\sin\text{x}}\Big(\frac{\sin\text{x}}{\text{x}}+\cos\text{x}\cdot\log\text{x}\Big)\text{dx}$
Put $\text{x}^{\sin\text{x}}=\text{t}$
Taking $\log$ on both sides,
$\log\text{t}=\sin\text{x}\log\text{x}$
$\frac{1}{\text{t}}=\frac{\sin\text{x}}{\text{x}}+\cos\text{x}\log\text{x}$
$1=\int\text{t}\cdot\frac{\text{dt}}{\text{t}}$
$1=\text{t}+\text{C}$
$1=\text{x}^{\sin\text{x}}+\text{C}$
View full question & answer→MCQ 1811 Mark
Choose the correct answer in Exercise:
$\int\frac{\text{dx}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$ is equal to
- ✓
$\tan^{-1}(\text{e}^{\text{x}})+\text{C}$
- B
$\tan^{-1}\text{(e}^{\text{x}})+\text{C}$
- C
$\log(\text{e}^{\text{x}}-\text{e}^{\text{x}})+\text{C}$
- D
$\log(\text{e}^{\text{x}}+\text{e}^\text{x})+\text{C}$
AnswerCorrect option: A. $\tan^{-1}(\text{e}^{\text{x}})+\text{C}$
$\text{Let I}=\int\frac{\text{dx}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}\text{dx}=\int\frac{\text{e}^{\text{x}}}{\text{e}^{2\text{x}}+1}\text{dx}$
Also, let $\text{e}^{\text{x}}=\text{t}\Rightarrow\text{e}^{\text{x}}\ \text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{1+\text{t}^{2}}$
$=\tan^{-1}\text{t}+\text{C}$
$=\tan^{-1}\text{(e}^{\text{x}})+\text{C}$
View full question & answer→MCQ 1821 Mark
$\int\limits^\frac{\pi}{2}_0\frac{\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$ equals to:
- A
$\pi$
- B
$\frac{\pi}{2}$
- C
$\frac{\pi}{3}$
- ✓
$\frac{\pi}{4}$
AnswerCorrect option: D. $\frac{\pi}{4}$
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 $\text{I}=\frac{\pi}{4}$
View full question & answer→MCQ 1831 Mark
The value of $\int\frac{\cos\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$ is:
- A
$2\cos\sqrt{\text{x}}+\text{C}$
- B
$\sqrt{\frac{\cos\text{x}}{\text{x}}}+\text{C}$
- C
$\sin\sqrt{\text{x}}+\text{C}$
- ✓
$2\sin\sqrt{\text{x}}+\text{C}$
AnswerCorrect option: D. $2\sin\sqrt{\text{x}}+\text{C}$
$\text{I}=\int\frac{\cos\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$
Put $\sqrt{\text{x}}=\text{t}$
$\frac{1}{2\sqrt{\text{x}}}\text{ dx}=\text{dt}$
$\frac{1}{\sqrt{\text{x}}}\text{ dx}=2\text{dt}$
$\text{I}=\int\cos\text{t }2\text{ dt}$
$\text{I}=2\sin\text{t}+\text{C}$
$\text{I}=2\sin\sqrt{\text{x}}+\text{C}$
View full question & answer→MCQ 1841 Mark
$\int(3\text{x}^2-1)\text{dx:}$
- ✓
$x^3 - x$
- B
$x^2 - x$
- C
$x^3 - 1$
- D
AnswerCorrect option: A. $x^3 - x$
View full question & answer→MCQ 1851 Mark
$\int\text{e}^{\text{x}}\{\text{f(x)}+\text{f}'(\text{x})\}\text{dx}=$
- ✓
$\text{e}^{\text{x}}\text{f(x)}+\text{C}$
- B
$\text{e}^{\text{x}}+\text{f(x)}$
- C
$2\text{e}^{\text{x}}\text{f(x)}$
- D
$\text{e}^{\text{x}}-\text{f(x)}$
AnswerCorrect option: A. $\text{e}^{\text{x}}\text{f(x)}+\text{C}$
$\int\text{e}^{\text{x}}\{\text{f(x)}+\text{f}'(\text{x})\}\text{dx}=\text{e}^{\text{x}}\text{f(x)}+\text{C}$
View full question & answer→MCQ 1861 Mark
$\int\frac{\text{x}^3}{\sqrt{1+\text{x}^2}}\text{ dx}=\text{a}(1+\text{x}^2)^{\frac{3}{2}}+\text{b}\sqrt{1+\text{x}^2}+\text{C},$ then:
- A
$\text{a}=\frac{1}{3},\text{ b}=1$
- B
$\text{a}=-\frac{1}{3},\text{ b}=1$
- C
$\text{a}=-\frac{1}{3},\text{ b}=-1$
- ✓
$\text{a}=\frac{1}{3},\text{ b}=-1$
AnswerCorrect option: D. $\text{a}=\frac{1}{3},\text{ b}=-1$
$\text{I}=\int\frac{\text{x}^3}{\sqrt{1+\text{x}^2}}\text{ dx}$
$1+\text{x}^2=\text{t}$
$2\text{xdx}=\text{dt}$
$\text{x dx}=\frac{\text{dt}}{2}$
$\text{I}=\int\frac{\text{x}^2}{\sqrt{1+\text{x}^2}}\text{x dx}$
$\text{I}=\int\frac{\text{t}-1}{\sqrt{\text{t}}}\frac{\text{dt}}{2}$
$\text{I}=\frac{1}{2}\Big(\frac{2}{3}\text{t}^{\frac{3}{2}}-2\sqrt{\text{t}}\Big)+\text{C}$
$\text{I}=\frac{1}{3}(1+\text{x}^{2})^{\frac{3}{2}}-\sqrt{1+\text{x}^2}+\text{C}$
$\text{a}=\frac{1}{3},\text{ b}=-1$
View full question & answer→MCQ 1871 Mark
$\int\frac{\sin\text{x}}{3+4\cos^2\text{x}}\text{ dx}=$
- A
$\log(3+4\cos^2\text{x})+\text{C}$
- B
$\frac{1}{2\sqrt{3}}\tan^{-1}\Big(\frac{\cos\text{x}}{\sqrt{3}}\Big)+\text{C}$
- ✓
$-\frac{1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\cos\text{x}}{\sqrt{3}}\Big)+\text{C}$
- D
$\frac{1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\cos\text{x}}{\sqrt{3}}\Big)+\text{C}$
AnswerCorrect option: C. $-\frac{1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\cos\text{x}}{\sqrt{3}}\Big)+\text{C}$
$\text{I}=\int\frac{\sin\text{x}}{3+4\cos^2\text{x}}\text{ dx}$
Put $\cos\text{x}=\text{t}$
$-\sin\text{x dx}=\text{dt}$
$\sin\text{x dx}=-\text{dt}$
$\text{I}=\int\frac{-\text{dt}}{3+4\text{t}^2}$
$\text{I}=\frac{-1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\text{t}}{\sqrt{3}}\Big)+\text{C}$
$\text{I}=\frac{-1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\cos\text{x}}{\sqrt{3}}\Big)+\text{C}$
View full question & answer→MCQ 1881 Mark
If $\int\frac{\text{x}^3\text{dx}}{\sqrt{1+\text{x}^2}}=\text{a}(1+\text{x}^2)^{\frac{3}{2}}+\text{b}\sqrt{1+\text{x}^2}+\text{c,}$ then:
- A
$\text{a}=\frac{1}{3},\text{b}=1$
- B
$\text{a}=\frac{-1}{3},\text{b}=1$
- C
$\text{a}=\frac{-1}{3},\text{b}=-1$
- ✓
$\text{a}=\frac{1}{3},\text{b}=-1$
AnswerCorrect option: D. $\text{a}=\frac{1}{3},\text{b}=-1$
View full question & answer→MCQ 1891 Mark
If $\int\limits^1_0\text{f}(\text{x})\text{dx}=1,\int\limits^1_0\text{x}\text{f}(\text{x})\text{dx}=\text{a},\int\limits^1_0\text{x}^2\text{f}(\text{x})\text{dx}=\text{a}^2,$ then $\int\limits^1_0(\text{a}-\text{x})^2\text{f(x)}\text{dx}$ equals:
Answer$\int\limits^1_0(\text{a}-\text{x})^2\text{ f}(\text{x})\text{dx}$
$=\text{a}^2\int\limits^1_0\text{f}(\text{x})\text{dx}+\int\limits^1_0\text{x}^2\text{f}(\text{x})\text{dx}-2\text{a}\int\limits^1_0\text{x}\text{f}(\text{x})\text{dx}$
$=\text{a}^2\times1+\text{a}^2-2\text{aa} ($As per given values$)$
$=2\text{a}^2-2\text{a}^2$
$=0$
View full question & answer→MCQ 1901 Mark
$\int_{a}^{b}\text{x}^2\text{dx}=$
- A
$\frac{1}{2}\tan\frac{\text{x}}{2}+\text{k}$
- B
$2\tan\frac{\text{x}}{2}+\text{k}$
- ✓
$\tan\frac{\text{x}}{2}+\text{k}$
- D
$\tan^2\frac{\text{x}}{2}+\text{k}$
AnswerCorrect option: C. $\tan\frac{\text{x}}{2}+\text{k}$
View full question & answer→MCQ 1911 Mark
$\int\tan^{-1}\sqrt{\text{xdx}}$ 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. $(\text{x}+1)\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\text{c}$
View full question & answer→MCQ 1921 Mark
$\int\text{x}^2\sin\text{x}^3\text{dx}=$
- A
$\frac{1}{3}\cos\text{x}^3+\text{c}$
- B
$-\frac{1}{3}\cos\text{x}+\text{c}$
- ✓
$\frac{-1}{3}\cos\text{x}^3+\text{c}$
- D
$\frac{1}{2}\sin^2\text{x}^3+\text{c}$
AnswerCorrect option: C. $\frac{-1}{3}\cos\text{x}^3+\text{c}$
View full question & answer→MCQ 1931 Mark
$\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{1}{1+\sqrt{\cot\text{x}}}\text{ dx}$ is:
- A
$\frac{\pi}{3}$
- B
$\frac{\pi}{6}$
- ✓
$\frac{\pi}{12}$
- D
$\frac{\pi}{2}$
AnswerCorrect option: C. $\frac{\pi}{12}$
Let, $\text{I}=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\frac{1}{1+\sqrt{\cot\text{x}}}\text{dx}\ ...{\text{(i)}}$
$=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\frac{1}{\sqrt{\cot\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}}\text{dx}$ $\bigg[\text{using}\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx}=\int\limits^\text{b}_\text{a}\text{f}\big(\text{a}+\text{b}-\text{x}\big)\text{dx}\bigg]$
$=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\frac{1}{1+\sqrt{\tan\text{x}}}\text{dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\bigg[\frac{1}{1+\sqrt{\cot\text{x}}}+\frac{1}{1+\sqrt{\tan\text{x}}}\bigg]\text{dx}$
$=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\frac{2+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}{\big(1+\sqrt{\cot\text{x}}\big)+\big(1+\sqrt{\tan\text{x}}\big)}\text{ dx}$
$=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\Bigg[\frac{2+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}{2+\sqrt{\cot\text{x}+\sqrt{\tan\text{x}}}}\Bigg]\text{dx}$
$=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\text{dx}$
$=\big[\text{x}\big]^\frac{\pi}{3}_\frac{\pi}{6}$
$=\frac{\pi}{3}-\frac{\pi}{6}$
$=\frac{\pi}{6}$
Hence, $\text{I}=\frac{\pi}{12}$
View full question & answer→MCQ 1941 Mark
If $\int\frac{\cos8\text{x}+1}{\tan2\text{x}-\cot2\text{x}}\text{ dx}=\text{a}\cos8\text{x}+\text{C},$ then a =
- A
$-\frac{1}{16}$
- B
$\frac{1}{8}$
- ✓
$\frac{1}{16}$
- D
$-\frac{1}{8}$
AnswerCorrect option: C. $\frac{1}{16}$
$\int\frac{\cos8\text{x}+1}{\tan2\text{x}-\cot2\text{x}}\text{ dx}$
$=\int\frac{2\cos^24\text{x}}{\frac{\sin2\text{x}}{\cos2\text{x}}-\frac{\cos2\text{x}}{\sin2\text{x}}}\text{ dx}$
$=\int\frac{2\cos^24\text{x}}{\sin^22\text{x}-\cos^22\text{x}}\times\sin2\text{x}\cos2\text{x dx}$
$=\int-\frac{\cos^24\text{x}\sin4\text{x}}{\cos4\text{x}}\text{ dx}$
$=\frac{-1}{2}\int\sin8\text{x dx}$
$=\frac{\cos8\text{x}}{16}+\text{C}$
$\text{a}=\frac{1}{16}$
View full question & answer→MCQ 1951 Mark
Choose the correct answer in Exercise: $\int\text{x}^2\text{e}^{\text{x}^3}\text{dx}$ equals
- ✓
$\frac{1}{3}\text{e}^{\text{x}^3}+\text{C}$
- B
$\frac{1}{3}\text{e}^{\text{x}^2}+\text{C}$
- C
$\frac{1}{2}\text{e}^{\text{x}^3}+\text{C}$
- D
$\frac{1}{2}\text{e}^{\text{x}^2}+\text{C}$
AnswerCorrect option: A. $\frac{1}{3}\text{e}^{\text{x}^3}+\text{C}$
View full question & answer→MCQ 1961 Mark
$\int\frac{1}{\cos\text{x}+\sqrt{3}\sin\text{x}}\text{ dx}$ is equal to:
- A
$\log\tan\Big(\frac{\pi}{3}+\frac{\pi}{2}\Big)+\text{C}$
- B
$\log\tan\Big(\frac{\pi}{2}-\frac{\pi}{3}\Big)+\text{C}$
- ✓
$\frac{1}{2}\log\tan\Big(\frac{\pi}{2}+\frac{\pi}{3}\Big)+\text{C}$
- D
AnswerCorrect option: C. $\frac{1}{2}\log\tan\Big(\frac{\pi}{2}+\frac{\pi}{3}\Big)+\text{C}$
$\text{I}=\int\frac{1}{\cos\text{x}+\sqrt{3}\sin\text{x}}\text{ dx}$
$\text{I}=\frac{1}{2}\int\frac{2}{\frac{\cos\text{x}}{2}+\frac{\sqrt{3}}{2}\sin\text{x}}\text{ dx}$
$\text{I}=\frac{1}{2}\int\frac{1}{\cos\big(\text{x}-\frac{\pi}{6}\big)}\text{ dx}$
$\text{I}=\frac{1}{2}\int\sec\Big(\text{x}-\frac{\pi}{6}\Big)\text{dx}$
$\text{I}=\frac{1}{2}\ln\Big|\tan\Big(\frac{\text{x}}{2}+\frac{\pi}{3}\Big)\Big|+\text{C}$
View full question & answer→MCQ 1971 Mark
Choose the correct option from given four options:
$\frac{\text{dx}}{\sin(\text{x}-\text{a})\sin(\text{x}-\text{b})}$ is equal to:
- A
$\sin(\text{b}-\text{a)}\log\Big|\frac{\sin(\text{x}-\text{b})}{\sin(\text{x}-\text{a})}\Big|+\text{C}$
- B
$\text{cosec}(\text{b}-\text{a)}\log\Big|\frac{\sin(\text{x}-\text{a})}{\sin(\text{x}-\text{b})}\Big|+\text{C}$
- ✓
$\text{cosec}(\text{b}-\text{a)}\log\Big|\frac{\sin(\text{x}-\text{b})}{\sin(\text{x}-\text{a})}\Big|+\text{C}$
- D
$\sin(\text{b}-\text{a)}\log\Big|\frac{\sin(\text{x}-\text{b})}{\sin(\text{x}-\text{a})}\Big|+\text{C}$
AnswerCorrect option: C. $\text{cosec}(\text{b}-\text{a)}\log\Big|\frac{\sin(\text{x}-\text{b})}{\sin(\text{x}-\text{a})}\Big|+\text{C}$
Let $\text{I}=\frac{\text{dx}}{\sin(\text{x}-\text{a})\sin(\text{x}-\text{b})}$
$=\frac{1}{\sin(\text{b}-\text{a})}\int\frac{\sin(\text{b}-\text{a})}{\sin(\text{x}-\text{a})\sin(\text{x}-\text{b})}\text{dx}$
$=\frac{1}{\sin(\text{b}-\text{a})}\int\frac{\sin(\text{x}-\text{a}-\text{x}+\text{b})}{\sin(\text{x}-\text{a})\sin(\text{x}-\text{b})}\text{dx}$
$=\frac{1}{\sin(\text{b}-\text{a})}\int\frac{\sin\big\{(\text{x}-\text{a})-(\text{x}+\text{b})\big\}}{\sin(\text{x}-\text{a})\sin(\text{x}-\text{b})}\text{dx}$
$=\frac{1}{\sin(\text{b}-\text{a})}\int\frac{\sin(\text{x}-\text{a})\cos(\text{x}-\text{b})-\cos(\text{x}-\text{a})\sin(\text{x}-\text{b})}{\sin(\text{x}-\text{a})\sin(\text{x}-\text{b})}\text{dx}$
$=\frac{1}{\sin(\text{b}-\text{a})}\int\frac{\sin(\text{x}-\text{a})\cos(\text{x}-\text{b})}{\sin(\text{x}-\text{a})\sin(\text{x}-\text{b})}-\frac{\cos(\text{x}-\text{a})\sin(\text{x}-\text{b})}{\sin(\text{x}-\text{a} )\sin(\text{x}-\text{b})}\text{dx}$
$=\frac{1}{\sin(\text{b}-\text{a})}\int\frac{\cos(\text{x}-\text{b})}{\sin(\text{x}-\text{a})}-\frac{\cos(\text{x}-\text{a})}{\sin(\text{x}-\text{a} )}\text{dx}$
$=\frac{1}{\sin(\text{b}-\text{a})}\big[\log\sin|(\text{x}-\text{b})|-\log|\sin(\text{x}-\text{b})|\big]+\text{C}$
$=\text{cosec}(\text{b}-\text{a})\log\Big|\frac{\sin(\text{x}-\text{b})}{\sin(\text{x}-\text{a})}\Big|+\text{C}$
View full question & answer→MCQ 1981 Mark
If $\int\text{x}\sin\text{x dx}=-\text{x}\cos\text{x}+\text{a},$ then a is equal to:
- ✓
$\sin\text{x}+\text{C}$
- B
$\cos\text{x}+\text{C}$
- C
$\text{C}$
- D
AnswerCorrect option: A. $\sin\text{x}+\text{C}$
$\int\text{x}\sin\text{x dx}=-\text{x}\cos\text{x}+\text{a}$
$\text{I}=\int\text{x}\sin\text{x dx}$
$\text{I}=\text{x}\int\sin\text{x dx}-\int\Big(\frac{\text{dx}}{\text{dx}}\int\sin\text{x dx}\Big)\text{dx}$
$\text{I}=-\text{x}\cos\text{x}+\int\cos\text{x dx}$
$\text{I}=\text{x}\cos\text{x}+\sin\text{x}+\text{C}$
$\text{a}=\sin\text{x}+\text{C}$
View full question & answer→MCQ 1991 Mark
$\int\frac{-1}{\text{y}^2}\text{dy}$ is:
AnswerCorrect option: A. $\frac{1}{\text{y}}$
$\int\frac{-1}{{\text{y}}^2}\text{dy}$
$=-\int\text{y}^{-2}\text{dy}$
$=\text{y}^{-1}=\frac{1}{\text{y}}$
View full question & answer→MCQ 2001 Mark
Integrate the following functions with respect to x: $\int\frac{\text{dx}}{4\text{x}+5}$
- ✓
$\frac{1}{4}\text{ In }(4\text{x}+5)+\text{c}$
- B
$\frac{1}{4}\text{ In }(4\text{x}+5)-\text{c}$
- C
$\frac{-1}{4}\text{ In }(4\text{x}+5)-\text{c}$
- D
$4\text{ In }(4\text{x}-5)-\text{c}$
AnswerCorrect option: A. $\frac{1}{4}\text{ In }(4\text{x}+5)+\text{c}$
$\int\frac{\text{dx}}{4\text{x}+5}=\frac{1}{4}\int\frac{\text{dx}}{\text{x}}$ where x = 4x + 5
$=\frac{1}{4}\text{ In }\text{x}+\text{c}_{1}=\frac{1}{4}\text{ In }(4\text{x}+5)+\text{c}_{2}$
View full question & answer→