MCQ 511 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
$\frac{1}{10\text{x}}\Big(\frac{1}{\text{x}^2}+4\Big)^{-5}+\text{C}$
- ✓
$\frac{1}{10}\Big(\frac{1}{\text{x}^2}+4\Big)^{-5}+\text{C}$
AnswerCorrect option: D. $\frac{1}{10}\Big(\frac{1}{\text{x}^2}+4\Big)^{-5}+\text{C}$
Let $\text{I}=\int\frac{\text{x}^9}{(4\text{x}^2+1)^6}\text{ dx}$
$=\int\frac{\text{x}^{9}}{\text{x}^{\frac{1}{2}}\big(4+\frac{1}{\text{x}^2}\big)^6}\text{ dx}$
$=\int\frac{\text{x}^{\frac{1}{3}}}{\big(4+\frac{1}{\text{x}^2}\big)^6}\text{ dx}$
Let $\Big(4+\frac{1}{\text{x}^2}\Big)=\text{t}$
On differentiating both sides, we get
$-\frac{2}{\text{x}^3}\text{dx}=\text{dt}$
$\therefore\ \text{I}=-\frac{1}{2}\int\frac{1}{(\text{t}^{6})}\text{dt}$
$=-\frac{1}{2}\Big(-\frac{1}{5}\Big)\text{t}^{-5}+\text{C}$
$=\frac{1}{10}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{C}$
Therefore, $\int\frac{\text{x}^9}{(4\text{x}^2+1)^6}\text{ dx}=\frac{1}{10}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{C}$
Hence, the correct option is (d)
View full question & answer→MCQ 521 Mark
$\int\limits_{2}^{2} \mid\text{x}\mid\text{dx}=$
View full question & answer→MCQ 531 Mark
$\int\sin^{-1}\text{xdx}$ is equal to:
- A
$\cos^{-1}\text{x}+\text{c}$
- B
$\text{x}\sin^{-1}\text{x}+\sqrt{1-\text{x}^2}+\text{c}$
- C
$\frac{1}{\sqrt{1-\text{x}^2}}+\text{c}$
- ✓
$\text{x}\sin^{-1}\text{x}-\sqrt{1-\text{x}^2}+\text{c}$
AnswerCorrect option: D. $\text{x}\sin^{-1}\text{x}-\sqrt{1-\text{x}^2}+\text{c}$
View full question & answer→MCQ 541 Mark
The number of integral solutions (x, y) of the equations $\text{x}{\sqrt{\text{y}}}+\text{y}\sqrt{\text{x}}=20$ and $\text{x}{\sqrt{\text{x}}}+\text{y}\sqrt{\text{y}}=65$ is:
AnswerBy trial and error method Put y = 1, x = 16 in the same way x = 1, y = 16 The equation gets satisfied.
View full question & answer→MCQ 551 Mark
$\int\limits^\frac{\pi}{2}_0\frac{1}{2+\cos\text{x}}\text{ dx}$ equals:
- A
$\frac{1}{3}\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$
- ✓
$\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$
- C
${\sqrt{3 }}\tan^{-1}\big({\sqrt{3}}\big)$
- D
$2{\sqrt{3 }}\tan^{-1}{\sqrt{3}}$
AnswerCorrect option: B. $\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$
We have,
$=\int\limits^\frac{\pi}{2}_0\frac{1}{2+\cos\text{x}}\text{ dx}$
$=\int\limits^\frac{\pi}{2}_0\frac{1}{2+\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}}\text{ dx}$
$=\int\limits^\frac{\pi}{2}_0\frac{1+\tan^2\frac{\text{x}}{2}}{2+2\tan^2\frac{\text{x}}{2}+1-\tan^2\frac{\text{x}}{2}}\text{ dx}$
$=\int\limits^\frac{\pi}{2}_0\frac{\sec^2\frac{\text{x}}{2}}{3+\tan^2\frac{\text{x}}{2}}\text{dx}$
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 $\text{x}\rightarrow\frac{\pi}{2};\text{ t}\rightarrow1$
$\therefore\ \text{I}=\int\limits^1_0\frac{2}{3+\text{t}^2}\text{ dt}$
$=2\int\limits^1_0\frac{1}{(\sqrt{3})^2+\text{t}^2}\text{dt}$
$=\frac{2}{\sqrt{3}}\Big[\tan^{-1}\frac{\text{t}}{\sqrt{3}}\Big]^1_0$
$=\frac{2}{\sqrt{3}}\Big[\tan^{-1}\frac{1}{\sqrt{3}}-\tan^{-1}\frac{0}{\sqrt{3}}\Big]$
$=\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$
View full question & answer→MCQ 561 Mark
Given that $\int\limits^{\infty}_0\frac{\text{x}^2}{(\text{x}^2+\text{a}^2)(\text{x}^2+\text{b}^2)(\text{x}^2+\text{c}^2)}\text{ dx}=\frac{\pi}{2(\text{a}+\text{b})(\text{b}+\text{c})(\text{c}+\text{a})},$ the value of $\int\limits^\infty_0\frac{1}{(\text{x}^2+4)(\text{x}^2+9)},$ is:
- A
$\frac{\pi}{60}$
- ✓
$\frac{\pi}{20}$
- C
$\frac{\pi}{40}$
- D
$\frac{\pi}{80}$
AnswerCorrect option: B. $\frac{\pi}{20}$
$\int\limits^\infty_0\frac{1}{\big(\text{x}^2+4\big)\big(\text{x}^2+9\big)}\text{dx}$
$=\frac{1}{5}\int\limits^\infty_0\frac{1}{\big(\text{x}^2+4\big)}-\frac{1}{\big(\text{x}^2+9\big)}\text{dx}$
$=\frac{1}{5}\bigg[\frac{1}{2\tan^{-1}{}}\frac{\text{x}}{2}-\frac{1}{3}\tan^{-1}\frac{\text{x}}{3}\bigg]^\infty_0$
$=\frac{1}{5}\bigg[\frac{1}{2}\times\frac{\pi}{2}-\frac{1}{3}\times\frac{\pi}{2}\bigg]$
$=\frac{1}{5}\times\frac{\pi}{12}$
$=\frac{\pi}{60}$
View full question & answer→MCQ 571 Mark
What is the value of $\int_{0}^{1}\frac{\text{d}}{\text{dx}}\{\sin^{-1}(\frac{2\text{x}}{1+\text{x}^2})\}\text{dx}?$
- A
$0$
- B
$\pi$
- C
$-\pi$
- ✓
$\frac{\pi}{2}$
AnswerCorrect option: D. $\frac{\pi}{2}$
View full question & answer→MCQ 581 Mark
The anti derivative of $\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)$ 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. $\frac{2}{3}\text{x}^\frac{3}{2}+2\text{x}^\frac{1}{2}+\text{c}$
$\int\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)\text{dx}$
$=\int\text{x}^\frac{1}{2}\text{dx}+\int\text{x}^-\frac{1}{2}\text{dx}$
we know that $\int\text{x}^\text{n}\text{dx}=\frac{\text{x}^\text{n}+1}{\text{n}+1}$
$=\frac{\text{x}^\frac{3}{2}}{\frac{3}{2}}+\frac{\text{x}^\frac{1}{2}}{\frac{1}{2}}+\text{c}$
$=\frac{2}{3}\text{x}^\frac{3}{2}+2\text{x}^\frac{1}{2}+\text{c}$
View full question & answer→MCQ 591 Mark
$\int\limits^{\frac{\pi}{2}}_0\text{x}\sin\text{x dx}$ is equal to:
- A
$\frac{\pi}{4}$
- B
$\frac{\pi}{2}$
- C
$\pi$
- ✓
$1$
AnswerWe have,
$\text{I}=\int\limits^{\frac{\pi}{2}}_0\text{x}\sin\text{x dx}$
$=\big[-\text{x}\cos\text{x}\big]^{\frac{\pi}{2}}_0-\int\limits^{\frac{\pi}{2}}_01(-\cos\text{x})\text{dx}$
$=\big[-\text{x}\cos\text{x}\big]^{\frac{\pi}{2}}_0+\int\limits^{\frac{\pi}{2}}_0\cos\text{x dx}$
$=-\big[\text{x}\cos\text{x}\big]^{\frac{\pi}{2}}_0+\big[\sin\text{x}\big]^{\frac{\pi}{2}}_0$
$=-\big[0-0\big]+\big[1-0\big]$
$=1$
View full question & answer→MCQ 601 Mark
$\int(\frac{\cos2\theta-1}{\cos2\theta+1})\text{d}\theta=$
- A
$\tan\theta-\theta+\text{c}$
- B
$\theta+\tan\theta+\text{c}$
- ✓
$\theta-\tan\theta+\text{c}$
- D
$-\theta-\cot\theta+\text{c}$
AnswerCorrect option: C. $\theta-\tan\theta+\text{c}$
$\theta-\tan\theta+\text{c}$
View full question & answer→MCQ 611 Mark
If $\int\limits_{0}^{\frac{\pi}{2}}\sin\text{x}\cos\text{xdx}$ is equal to:
- ✓
$\frac{1}{2}$
- B
$\frac{1}{4}$
- C
$2$
- D
$1$
AnswerCorrect option: A. $\frac{1}{2}$
$\int\limits_{0}^{\frac{\pi}{2}}\sin\text{x}\cos\text{xdx}$
$\sin\text{x}=\text{t}\Rightarrow\cos\text{xdx}=\text{dt}$
$\text{x}\Rightarrow0\Rightarrow\frac{\pi}{2}$
$\int\limits_{0}^{\frac{\pi}{2}}\text{tdt}$
$\Rightarrow\frac{\text{t}^2}{2}\mid^1_0$
$\Rightarrow\frac{1}{2}-0=\frac{1}{2}$
View full question & answer→MCQ 621 Mark
The value of $\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\big(\text{x}^3+\text{x}\cos\text{x}+\tan^5\text{x}+1\big)\text{dx},$ is:
Answer$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\big(\text{x}^3+\text{x}\cos\text{x}+\tan^5\text{x}+1\big)\text{dx}$
$=\Big[\frac{\text{x}^4}{4}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}+\Big[\text{x}\sin\text{x}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}-\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin\text{x}\text{dx}\\+\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\tan^3\text{x}(\sec^2\text{x}-1)\text{dx}+\Big[\text{x}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}$
$=\frac{\pi^4}{64}-\frac{\pi^4}{64}+\frac{\pi}{2}-\frac{\pi}{2}-\Big[-\cos\text{x}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}\\+\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\tan^3\text{x}\sec^2\text{x}\text{ dx}-\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\tan^3\text{x}\text{ dx}+\frac{\pi}{2}+\frac{\pi}{2}$
$=+0+\Big[\frac{\tan^4\text{x}}{4}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}-\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\tan\text{x}\sec^2\text{x}\text{ dx}-\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\tan\text{x}\text{ dx}$
$=\pi-\Big[\frac{\tan^2\text{x}}{2}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}-\Big[-\log\big(\cos\text{x}\big)\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}$
$=\pi$
View full question & answer→MCQ 631 Mark
Choose the correct answer in exercise:
$\int\frac{\text{x dx}}{(\text{x}-1)(\text{x}-2)}$ equals
- A
$\text{log}\Bigg|\frac{(\text{x}-1)^2}{\text{x}-2}\Bigg|+\text{C}$
- ✓
$\text{log}\Bigg|\frac{(\text{x}-2)^2}{\text{x}-1}\Bigg|+\text{C}$
- C
$\text{log}\Bigg|\Bigg(\frac{\text{x}-1}{\text{x}-2}\Bigg)^2\Bigg|+\text{C}$
- D
$\text{log}|(\text{x}-1)(\text{x}-2)|+\text{c}$
AnswerCorrect option: B. $\text{log}\Bigg|\frac{(\text{x}-2)^2}{\text{x}-1}\Bigg|+\text{C}$
$\text{log}\Bigg|\frac{(\text{x}-2)^2}{\text{x}-1}\Bigg|+\text{C}$Let, $\frac{\text{x}}{(\text{x}-1)(\text{x}-2)}=\frac{\text{A}}{\text{x}-1}+\frac{\text{B}}{\text{x}-2}\dots(\text{i})$
⇒ x = A(x - 2) + B(x -1)
⇒ x = Ax - 2A + Bx - B
Comparing coefficients of x:
A + B = 1 ……….(ii)
Comparing constants
–2A – B = 0 ……….(iii)
On solving eq. (ii) and (iii), we get
A = –1, B = 2
Putting these values of A and B in eq. (i),
$\frac{\text{x}}{(\text{x}-1)(\text{x}-2)}=\frac{-1}{\text{x}-1}+\frac{2}{\text{x}-2}$
$\Rightarrow\ \int\frac{\text{x}}{(\text{x}-1)(\text{x}-2)}\text{dx}=-\int\frac{1}{\text{x}-1}\text{dx}+2\int\frac{1}{\text{x}-2}\text{dx}$
$=-\text{log}|\text{x}-1|+2\text{log}|\text{x}-2|+\text{c}$
$=\text{log}|(\text{x}-2)^2|-\text{log}|\text{x}-1|+\text{c}=\text{log}\Bigg|\frac{(\text{x}-2)^2}{\text{x}-1}\Bigg|+\text{c}$
View full question & answer→MCQ 641 Mark
Evaluate: $\int\frac{1-\cos\text{x}}{cos\text{x}(1+cos\text{x})}\text{dx}.$
- ✓
$\log|\text{sec}\text{x}+\tan\text{x}|-2\tan(\frac{\text{x}}{2})+\text{c}$
- B
$\log|\text{sec}\text{x}-\tan\text{x}|-2\tan(\frac{\text{x}}{2})+\text{c}$
- C
$\log|\text{sec}\text{x}+\tan\text{x}|+2\tan(\frac{\text{x}}{2})+\text{c}$
- D
$\text{None of these}$
AnswerCorrect option: A. $\log|\text{sec}\text{x}+\tan\text{x}|-2\tan(\frac{\text{x}}{2})+\text{c}$
$\log|\text{sec}\text{x}+\tan\text{x}|-2\tan(\frac{\text{x}}{2})+\text{c}$
View full question & answer→MCQ 651 Mark
$\int\limits^1_0\frac{\text{x}}{(1-\text{x})^{54}}\text{ dx}=$
- A
$\frac{15}{16}$
- B
$\frac{3}{16}$
- C
$-\frac{3}{16}$
- ✓
$-\frac{16}{3}$
AnswerCorrect option: D. $-\frac{16}{3}$
$\text{I}=\int\limits^1_0\frac{\text{x}}{(1-\text{x})^{\frac{5}{4}}}\text{ dx}$
Put, 1 - x = t ⇒ x =1 -t
⇒ dx = -dt
$\text{I}=\int\limits^0_1\frac{(1-\text{t})(-\text{dt})}{\text{t}^{\frac{5}{4}}}$
$\text{I}=\int\limits^1_0\Big(\text{t}^\frac{5}{4}-\text{t}^\frac{-1}{4}\Big)\text{dt}$
$\text{I}=\Bigg[\frac{\text{t}^{-\frac{5}{4}}}{\frac{-1}{4}}-\frac{\text{t}^\frac{3}{4}}{\frac{3}{4}}\Bigg]^1_0$
$\text{I}=-4-\frac{4}{3}$
$\text{I}=\frac{-16}{3}$ View full question & answer→MCQ 661 Mark
Choose the correct option from given four options:
$\int\limits^{\frac{\pi}{2}}_0\sqrt{1-\sin2\text{x}}\text{ dx}$ is equal to:
- A
$2\sqrt{2}$
- B
$\big(\sqrt{2}+1)$
- C
$2$
- ✓
$2\big(\sqrt{2}-1)$
AnswerCorrect option: D. $2\big(\sqrt{2}-1)$
Let $\text{I}=\int\limits^{\frac{\pi}{2}}_0\sqrt{1-\sin2\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\sqrt{(\cos\text{x}-\sin\text{x})^2}\text{ dx}+\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{4}}\sqrt{(\sin\text{x}-\cos\text{x})^2}\text{ dx}$
$=\Big[\sin\text{x}+\cos\text{x}\Big]^{\frac{\pi}{4}}_0+\Big[-\cos\text{x}-\sin\text{x}\Big]^{\frac{\pi}{2}}_{\frac{\pi}{4}}$
$=\frac{1}{\sqrt{2}}+=\frac{1}{\sqrt{2}}-0-1+\Big(-0-1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\Big)$
$=2\sqrt{2}-2=2(\sqrt{2}-1)$
View full question & answer→MCQ 671 Mark
If $\int\frac{1}{5+4\sin\text{x}}\text{ dx}=\text{A}\tan^{-1}\Big(\text{B}\tan\frac{\pi}{2}+\frac{4}{3}\Big)+\text{C},$ then:
- ✓
$\text{A}=\frac{2}{3},\text{B}=\frac{5}{3}$
- B
$\text{A}=\frac{1}{3},\text{B}=\frac{2}{3}$
- C
$\text{A}=-\frac{2}{3},\text{B}=\frac{5}{3}$
- D
$\text{A}=\frac{1}{3},\text{B}=-\frac{5}{3}$
AnswerCorrect option: A. $\text{A}=\frac{2}{3},\text{B}=\frac{5}{3}$
$\int\frac{1}{5+4\sin\text{x}}\text{ dx}=\text{A}\tan^{-1}\Big(\text{B}\tan\frac{\pi}{2}+\frac{4}{3}\Big)+\text{C}\ ...(\text{i})$
Considering the LHS of eq. (1)
Putting $\sin\text{x}=\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}$
$\Rightarrow\int\frac{1}{5+\frac{8\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}}\text{dx}$
$\Rightarrow\frac{\Big(1+\tan^2\frac{\text{x}}{2}\Big)}{5\Big(1+\tan^2\frac{\text{x}}{2}\Big)+8\tan\frac{\text{x}}{2}}\text{dx}$
$\Rightarrow\frac{\sec^2\frac{\text{x}}{2}}{5\tan^2\frac{\text{x}}{2}+8\tan\frac{\text{x}}{2}+5}\text{dx}\dots(2)$
Let $\tan\frac{\text{x}}{2}=\text{t}$
$\Rightarrow\sec^2\frac{\text{x}}{2}\times\frac{1}{2}\text{dx}=\text{dt}$
$\Rightarrow\sec^2\Big(\frac{\text{x}}{2}\Big)\text{dx}=\text{2dt}$
$\Rightarrow\frac{2}{5}\int\frac{\text{dt}}{\text{t}^2+\frac{8}{5}\text{t}+1}$
$\Rightarrow\frac{2}{5}\int\frac{\text{dt}}{\text{t}^2+\frac{8}{5}\text{t}+\big(\frac{4}{5}\big)^2-\big(\frac{4}{5}\big)^2+1}$
$\Rightarrow\frac{2}{5}\int\frac{\text{dt}}{\big(\text{t}+\frac{4}{5}\big)^2+1-\frac{16}{25}}$
$\Rightarrow\frac{2}{5}\int\frac{\text{dt}}{\big(\text{t}+\frac{4}{5}\big)^2+\big(\frac{3}{5}\big)^2}$
$\Rightarrow\frac{2}{5}\times\frac{2}{3}\tan^{-1}\bigg(\frac{\text{t}+\frac{4}{5}}{\frac{3}{5}}\bigg)+\text{C}$
$\Rightarrow\frac{2}{3}\tan^{-1}\Big(\frac{5\text{t}+4}{3}\Big)+\text{C}$
$\Rightarrow\frac{2}{3}\tan^{-1}\Big(\frac{5}{3}\tan\frac{\text{x}}{2}+\frac{4}{3}\Big)+\text{C}\ ...\text{(ii)}$ $\Big(\because\text{t}=\tan\frac{\text{x}}{2}\Big)$
Comparing eq. (ii) with the RHS of eq. (i) we get,
$\therefore\ \text{A}=\frac{2}{3},\text{ B}=\frac{5}{3}$
View full question & answer→MCQ 681 Mark
$\int\frac{\text{dx}}{2\sqrt{\text{x}}(1+\text{x})}=$
- A
$\frac{1}{2}\tan(\sqrt{\text{x}})+\text{c}$
- ✓
$\tan^{-1}(\sqrt{\text{x}})+\text{c}$
- C
$2\tan^{-1}(\sqrt{\text{x}})+\text{c}$
- D
AnswerCorrect option: B. $\tan^{-1}(\sqrt{\text{x}})+\text{c}$
$\tan^{-1}(\sqrt{\text{x}})+\text{c}$
View full question & answer→MCQ 691 Mark
Evaluate: $\int\sec^{\frac{4}{3}}\text{x}\text{cosec}^{\frac{8}{3}}\text{xdx}.$
- A
$\frac{3}{5}\tan^{\frac{-5}{3}}\text{x}-3\tan^{\frac{1}{3}}\text{x}+\text{c}$
- ✓
$-\frac{3}{5}\tan^{\frac{-5}{3}}\text{x}+3\tan^{\frac{1}{3}}+\text{c}$
- C
$-\frac{3}{5}\tan^{\frac{-05}{3}}\text{x}-3\tan^{\frac{1}{3}}+\text{c}$
- D
$\text{None of these}$
AnswerCorrect option: B. $-\frac{3}{5}\tan^{\frac{-5}{3}}\text{x}+3\tan^{\frac{1}{3}}+\text{c}$
$-\frac{3}{5}\tan^{\frac{-5}{3}}\text{x}+3\tan^{\frac{1}{3}}+\text{c}$
View full question & answer→MCQ 701 Mark
$\int1.\text{dx}=$
AnswerCorrect option: A. $\text{x}+\text{k}$
$\text{x}+\text{k}$
View full question & answer→MCQ 711 Mark
If $x > a, \int\frac{\text{dx}}{\text{x}^2-\text{a}^2}=$
- ✓
$\frac{2}{2\text{a}}\text{log }\frac{\text{x-a}}{\text{x+a}}+\text{k}$
- B
$\frac{2}{2\text{a}}\text{log }\frac{\text{x+a}}{\text{x-a}}+\text{k}$
- C
$\frac{1}{\text{a}}\text{log}(\text{x}^2-\text{a}^2)+\text{k}$
- D
$\log(\text{x}+\sqrt{\text{x}^2-\text{a}^2}+\text{k})$
AnswerCorrect option: A. $\frac{2}{2\text{a}}\text{log }\frac{\text{x-a}}{\text{x+a}}+\text{k}$
View full question & answer→MCQ 721 Mark
$\frac{\text{x}}{\text{dx}}\int\text{f(x)}\text{dx}$ is equal to:
View full question & answer→MCQ 731 Mark
$\int\cos^{-1}(\frac{1}{\text{x}})\text{dx}$ equals:
- A
$\text{x}\sec^{-1}\text{x}+\log|\text{x}+\sqrt{\text{x}^2-1}|+\text{c}$
- ✓
$\text{x}\sec^{-1}\text{x}-\log|\text{x}+\sqrt{\text{x}^2-1}|+\text{c}$
- C
$-\text{x}\sec^{-1}\text{x}-\log|\text{x}+\sqrt{\text{x}^2-1}|+\text{c}$
- D
$\text{None of these}$
AnswerCorrect option: B. $\text{x}\sec^{-1}\text{x}-\log|\text{x}+\sqrt{\text{x}^2-1}|+\text{c}$
$\text{x}\sec^{-1}\text{x}-\log|\text{x}+\sqrt{\text{x}^2-1}|+\text{c}$
View full question & answer→MCQ 741 Mark
Choose the correct option from given four options:
If $\frac{\text{x}^3\text{dx}}{\sqrt{1+\text{x}^2}}=\text{a}(1+\text{x}^2)^{\frac{3}{2}}+\text{b}\sqrt{1+\text{x}^2}+\text{C},$ 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. $\text{a}=\frac{1}{3},\text{b}=-1$
Given $\text{I}=\int\frac{\text{x}^3}{\sqrt{1+\text{x}^2}}\text{dx}=\text{a}(1+\text{x}^2)^{\frac{3}{2}}+\text{b}\sqrt{1+\text{x}^2}+\text{C}$
$\text{I}=\int\frac{\text{x}^3}{\sqrt{1+\text{x}^2}}\text{dx}$
Put $1+\text{x}^2=\text{t}^2$
$\Rightarrow\ 2\text{x dx}=2\text{t dt}$
$\therefore\ \text{I}\int\frac{\text{t}(\text{t}^2-1)}{\text{t}}\text{dt}$ $=\frac{\text{t}^3}{3}-\text{t}+\text{C}=\frac{1}{3}(1+\text{x}^2)^{\frac{3}{2}}-\sqrt{1+\text{x}^2}+\text{C}$
$\therefore\ \text{a}=\frac{1}{3}$ and $\text{b}=-1$
View full question & answer→MCQ 751 Mark
$\int\Big(\frac{4\text{e}^\text{x}-25}{2\text{e}^{\text{x}}-5}\Big)\text{dx}=\text{Ax}+\text{B}\log\mid{2\text{e}^\text{x}-5}\mid+\text{ c}$ then:
View full question & answer→MCQ 761 Mark
$\text{I}=\int\frac{(\text{x+a})^3}{\text{x}^3}\text{dx}$ is equal to:
- ✓
$\text{x}+3\text{a}\log\text{x}-\frac{3\text{a}^2}{\text{x}}-\frac{\text{a}^3}{2\text{x}^2}+\text{c}$
- B
$\text{x}^{2}+3\text{a}\log\text{x}-\frac{3\text{a}^2}{\text{x}}-\frac{\text{a}^3}{2\text{x}^2}+\text{c}$
- C
$\text{x}^{3}+3\text{a}\log\text{x}-\frac{2\text{a}^2}{\text{x}}-\frac{\text{3a}^3}{2\text{x}^2}+\text{c}$
- D
${1}+2\text{a}\log\text{x}-\frac{2\text{a}^2}{\text{x}}-\frac{\text{3a}^3}{2\text{x}^2}+\text{c}$
AnswerCorrect option: A. $\text{x}+3\text{a}\log\text{x}-\frac{3\text{a}^2}{\text{x}}-\frac{\text{a}^3}{2\text{x}^2}+\text{c}$
$\text{I}=\int\frac{(\text{x+a})^3}{\text{x}^3}\text{dx}$
$=\int\frac{\text{x}^3+\text{a}^3+3\text{ax}^2+3\text{a}^2\text{x}}{\text{x}^3}\text{dx}$
$=\int\Big(1+\frac{\text{a}^3}{\text{x}^3}+\frac{3\text{a}}{\text{x}}+\frac{3\text{a}^2}{\text{x}^2}\Big)\text{dx}$
$=\text{x}-\frac{\text{a}^3}{2\text{x}^2}+3\text{a}\log\text{x}-\frac{3\text{a}^2}{\text{x}}+\text{c}$
$\text{x}+3\text{a}\log\text{x}-\frac{3\text{a}^2}{\text{x}}-\frac{\text{a}^3}{2\text{x}^2}+\text{c}$
View full question & answer→MCQ 771 Mark
If $\int\limits^\alpha_0\frac{1}{1+4\text{x}^2}\text{ dx}=\frac{\pi}{8},$ then a equals:
- A
$\frac{\pi}{2}$
- ✓
$\frac{1}{2}$
- C
$\frac{\pi}{4}$
- D
$1$
AnswerCorrect option: B. $\frac{1}{2}$
$\int\limits^\alpha_0\frac{1}{1+4\text{x}^2}\text{ dx}=\frac{\pi}{8}$
$\Rightarrow\int\limits^\alpha_0\frac{1}{1+(2\text{x)}^2}\text{ dx}=\frac{\pi}{8}$
$\Rightarrow \frac{1}{2}\big[\tan^-2\text{x}\big]^\alpha_0=\frac{\pi}{8}$
$\Rightarrow \frac{1}{2}\tan^{-1}2\alpha=\frac{\pi}{8}$
$\Rightarrow 2\alpha=\tan\frac{\pi}{4}$
$\Rightarrow2\alpha=1$
$\therefore\alpha=\frac{1}{2}$
View full question & answer→MCQ 781 Mark
Choose the correct answer in Exercise.
$\int\sqrt{1+\text{x}^2}\text{dx}$ is equal to
- ✓
$\frac{\text{x}}{2}\sqrt{1+\text{x}^2}+\frac{1}{2}\text{log}\Bigg|\Big(\text{x}+\sqrt{1+\text{x}^2}\Big)\Bigg|+\text{C}$
- B
$\frac{2}{3}(1+\text{x}^2)^{\frac{3}{2}}+\text{C}$
- C
$\frac{2}{3}\text{x}(1+\text{x}^2)^{\frac{3}{2}}+\text{C}$
- D
$\frac{\text{x}^2}{2}\sqrt{1+\text{x}^2}+\frac{1}{2}\text{x}^2\text{log}\Bigg|\text{x}+\sqrt{1+\text{x}^2}\Bigg|+\text{C}$
AnswerCorrect option: A. $\frac{\text{x}}{2}\sqrt{1+\text{x}^2}+\frac{1}{2}\text{log}\Bigg|\Big(\text{x}+\sqrt{1+\text{x}^2}\Big)\Bigg|+\text{C}$
$\frac{\text{x}}{2}\sqrt{1+\text{x}^2}+\frac{1}{2}\text{log}\Bigg|\Big(\text{x}+\sqrt{1+\text{x}^2}\Big)\Bigg|+\text{C}$
$\int\sqrt{1+\text{x}^2}\text{dx}=\int\sqrt{\text{x}^2+1^2}\text{dx}=\frac{\text{x}}{2}\sqrt{\text{x}^2+1^2}+\frac{1^2}{2}\text{log}\Big|\text{x}+\sqrt{\text{x}^2+1^2}\Big|+\text{C}$
$=\frac{\text{x}}{2}\sqrt{\text{x}^2+1}+\frac{1}{2}\text{log}\Big|\text{x}+\sqrt{\text{x}^2+1}\Big|+\text{C}$
$\Bigg[\therefore\ \int\sqrt{\text{x}^2+\text{a}^2}\text{dx}=\frac{\text{x}}{2}\sqrt{\text{x}^2+\text{a}^2}+\frac{\text{a}^2}{2}\text{log}\Big|\text{x}+\sqrt{\text{x}^2+\text{a}^2}\Big|\Bigg]+\text{c}$
View full question & answer→MCQ 791 Mark
$\int\frac{2}{(\text{e}^{\text{x}}+\text{e}^{-\text{x}})^2}\text{ dx}=$
- ✓
$\frac{-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+\text{C}$
- B
$-\frac{1}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+\text{C}$
- C
$\frac{-1}{(\text{e}^{\text{x}}+1)^2}+\text{C}$
- D
$\frac{1}{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}+\text{C}$
AnswerCorrect option: A. $\frac{-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+\text{C}$
$\text{I}=\int\frac{2}{(\text{e}^{\text{x}}+\text{e}^{-\text{x}})^2}\text{ dx}$
$\text{I}=\int\frac{2\text{e}^{2\text{x}}}{(\text{e}^{2\text{x}}+1)^2}\text{ dx}$
Put $\text{t}=\text{e}^{2\text{x}}+1$
$\text{dt}=2\text{e}^{2\text{x}}\text{ dx}$
$\text{I}=\int\frac{\text{dt}}{\text{t}^2}$
$\text{I}=\frac{-1}{\text{t}}+\text{C}$
$\text{I}=\frac{-1}{\text{e}^{2\text{x}}+1}+\text{C}$
$\text{I}=\frac{-\frac{1}{\text{e}^{\text{x}}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+\text{C}$
$\text{I}=\frac{-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+\text{C}$
View full question & answer→MCQ 801 Mark
$\int\frac{\text{e}^{\text{x}}(1+\text{x})}{\cos^2(\text{x}\text{e}^{\text{x}})}\text{ dx}=$
- A
$2\log_\text{e}\cos(\text{xe}^{\text{x}})+\text{C}$
- B
$\sec(\text{xe}^{\text{x}})+\text{C}$
- ✓
$\tan(\text{xe}^{\text{x}})+\text{C}$
- D
$\tan(\text{x}+\text{e}^{\text{x}})+\text{C}$
AnswerCorrect option: C. $\tan(\text{xe}^{\text{x}})+\text{C}$
$\text{I}=\int\frac{\text{e}^{\text{x}}(1+\text{x})}{\cos^{2}(\text{xe}^{\text{x}})}\text{ dx}$
Put $\text{xe}^{\text{x}}=\text{t}$
$\text{e}^{\text{x}}(1+\text{x})\text{dx}=\text{dt}$
$\text{I}=\int\frac{\text{dt}}{\cos^2\text{t}}$
$=\text{I}=\int\sec^2\text{t dt}$
$\text{t}=\tan\text{t}+\text{C}$
$\text{I}=\tan(\text{xe}^{\text{x}})+\text{C}$
View full question & answer→MCQ 811 Mark
The symbol dx represents:
- ✓
The differential of the variable x
- B
The variable of integration is x
- C
Integral with respect to x
- D
AnswerCorrect option: A. The differential of the variable x
The differential of the variable x
View full question & answer→MCQ 821 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. $2(\sin\text{x}+\text{x}\cos\theta)+\text{c}$
$2(\sin\text{x}+\text{x}\cos\theta)+\text{c}$
View full question & answer→MCQ 831 Mark
Choose the correct answer in Exercise:
The value of the integral $\int^{1}_{\frac{1}{3}}\frac{(\text{x}-\text{x}^{3})^{\frac{1}{3}}}{\text{x}^{4}}\text{dx}\ \text{is}$
Answer$\text{Let}\ \text{I}=\int\limits_{\frac{1}{3}}^{1}\frac{(\text{x}-\text{x}^{3})^{\frac{1}{3}}}{\text{x}^{4}}\text{dx}=\int\limits_{3}^{1}\frac{\text{x}\bigg(\frac{1}{\text{x}^{2}}-1\bigg)^{\frac{1}{3}}}{\text{x}^{4}}\text{dx} $
$=\int\limits_{\frac{1}{3}}^{1}\frac{\bigg(\frac{1}{\text{x}^{2}}-1\bigg)^{\frac{1}{3}}}{\text{x}^{3}}\text{dx}=\int\limits_{\frac{1}{3}}^{1}\bigg(\frac{1}{\text{x}^{2}}-1\bigg)^{\frac{1}{3}}.\frac{1}{\text{x}^{3}}\text{dx}$
$\text{put}\frac{1}{\text{x}^{2}}=\text{t},\ \therefore-\frac{2}{\text{x}^{3}}\text{dx}=\text{dt}\Rightarrow\frac{1}{\text{x}^{3}}\text{dx}=-\frac{1}{2}\text{dt}$
$\text{when}\ \text{x}=\frac{1}{3},\text{t}=9$
$\text{when}\ \text{x}=1,\text{t}=1$
$\therefore\ \ \text{I}=-\frac{1}{2}\int^{1}_{9}\big(\text{t}-1\big)^{\frac{1}{3}}\text{dt}=-\frac{1}{2}\Bigg[\frac{\big(\text{t}-1\big)^{\frac{4}{3}}}{\frac{4}{3}}\Bigg]^{1}_{0}$
$=-\frac{3}{8}\bigg[\big(\text{t}-1\big)^{\frac{4}{3}}\bigg]^{1}_{9}=-\frac{3}{8}\bigg[\big(1-1\big)^{\frac{4}{3}}-\big(9-1\big)^{\frac{4}{3}}\bigg]$
$=-\frac{3}{8}\bigg[{0}-(8)^{\frac{4}{3}}\bigg]^{1}_{9}=\frac{3}{8}(2)^{3^{\frac{4}{3}}}=\frac{3}{8}\text{x}\ 2^{4}=\frac{3}{8}\text{x}16=6$
$\text{Let}\ \text{I}=\int^{1}_{\frac{1}{3}}\frac{\big(\text{x}-\text{x}^{3}\big)^{\frac{1}{3}}}{\text{x}^{4}}\text{dx}=\int^{1}_{3}\frac{\text{x}\bigg(\frac{1}{\text{x}^{2}}-1\bigg)^{\frac{1}{3}}}{\text{x}^{4}}\text{dx}$
$=\int^{1}_{\frac{1}{3}}\frac{\bigg(\frac{1}{\text{x}^{2}}-1\bigg)^{\frac{1}{3}}}{\text{x}^{3}}\text{dx}=\int^{1}_{\frac{1}{3}}\bigg(\frac{1}{\text{x}^{2}}-1\bigg)^{\frac{1}{3}}.\frac{1}{\text{x}^{3}}\text{dx}$
$\text{put}\frac{1}{\text{x}^{2}}=\text{t},\ \therefore-\frac{2}{\text{x}^{3}}\text{dx}=\text{dt}\Rightarrow\frac{1}{\text{x}^{3}}\text{dx}=-\frac{1}{2}\text{dt}$
$\text{when}\ \text{x}=\frac{1}{3},\text{t}=9$
$\text{when}\ \text{x}=1,\text{t}=1$
$\therefore\ \ \text{I}=-\frac{1}{2}\int^{1}_{9}\big(\text{t}-1\big)^{\frac{1}{3}}\text{dt}=-\frac{1}{2}\bigg[\frac{\text{(t}-1)^{\frac{4}{3}}}{\frac{4}{3}}\bigg]^{1}_{0}$
$=-\frac{3}{8}\bigg[(\text{t}-1)^{\frac{4}{3}}\bigg]^{1}_{9}=-\frac{3}{8}\bigg[(1-1)^{\frac{4}{3}}-(9-1)^{\frac{4}{3}}\bigg]$
$=-\frac{3}{8}\bigg[0-(8)^{\frac{4}{3}}\bigg]=-\frac{3}{8}(2)^{3^{\frac{4}{3}}}=\frac{3}{8}\times2^{4}=\frac{3}{8}\times16=6$
View full question & answer→MCQ 841 Mark
$\int\frac{(1+\text{log x})^2}{1+\text{x}^2}\text{dx}=$
- ✓
$\frac{1}{3}(1+\text{log})^3+\text{c}$
- B
$\frac{1}{2}(1+\text{log})^2+\text{c}$
- C
$\log(\text{log }1+\text{x})+2$
- D
$\text{None of these}$
AnswerCorrect option: A. $\frac{1}{3}(1+\text{log})^3+\text{c}$
$\frac{1}{3}(1+\text{log})^3+\text{c}$
View full question & answer→MCQ 851 Mark
Choose the correct answer in Exercise:$\int^\frac{2}{3}_{0}\frac{\text{dx}}{4+9\text{x}^{2}}\text{equals}$
- A
$\frac{\pi}{6}$
- B
$\frac{\pi}{12}$
- ✓
$\frac{\pi}{24}$
- D
$\frac{\pi}{4}$
AnswerCorrect option: C. $\frac{\pi}{24}$
$\frac{\pi}{24}$ $\int\frac{\text{dx}}{4+9\text{x}^{2}}=\int\frac{\text{dx}}{(2)^{2}+(3\text{x})^{2}}$$\ \text{put}\ 3\text{x}=\text{t}\Rightarrow3\text{dx}=\text{dt}$
$\therefore\int\frac{\text{dx}}{(2)^{2}+(3\text{x})^{2}}=\frac{1}{3}\int\frac{\text{dt}}{(2)^{2}+\text{t}^{2}}$
$=\frac{1}{3}\bigg[\frac{1}{2}\tan^{-1}\frac{t}{2}\bigg]$
$=\frac{1}{6}\tan^{-1}\bigg(\frac{3\text{x}}{2}\bigg)$
$=\text{F}\text{(x)}$
By second fundamental theorem of calculus, we obtain $\int\limits_{0}^{\frac{2}{3}}\frac{\text{dx}}{4+9\text{x}^{2}}=\text{F}\bigg(\frac{2}{3}\bigg)-\text{F}(0)$ $=\frac{1}{6}\tan^{-1}\bigg(\frac{3}{2}.\frac{2}{3}\bigg)-\frac{1}{6}\tan^{-1}0$ $=\frac{1}{6}\tan^{-1}1-0$ $=\frac{1}{6}\times\frac{\pi}{4}$ $=\frac{\pi}{24}$
View full question & answer→MCQ 861 Mark
The value of the integral $\int\limits^2_{-2}\big|1-\text{x}^2\big|\text{dx}$ is:
AnswerWe have,
$\text{I}=\int\limits^2_{-2}\big|1-\text{x}^2\big|\text{dx}$
$\big|1-\text{x}^2\big|=\begin{cases}-\big(1-\text{x}^{2}\big), & -2<\text{x}<-1\\\big(1-\text{x}^2\big), & -1<\text{x}<1 \\-\big(1-\text{x}^2\big),&1<\text{x}<2\end{cases}$
$\therefore\ \text{I}=\int\limits^{-1}_{-2}\big|1-\text{x}^2\big|\text{dx}+\int\limits^1_{-1}\big|1-\text{x}^2\big|\text{dx}+\int\limits^2_1\big|1-\text{x}^2\big|\text{dx}$
$=\int\limits^{-1}_{-2}-\big(1-\text{x}^2)\text{dx}+\int\limits^1_{-1}(1-\text{x}^2)\text{dx}+\int\limits^2_1-(1-\text{x}^2)\text{dx}$
$=-\int\limits^{-1}_{-2}(1-\text{x}^2)\text{dx}+\int\limits^1_{-1}(1-\text{x}^2)\text{dx}-\int\limits^2_1(1-\text{x}^2)\text{dx}$
$=\Big[\text{x}-\frac{\text{x}^3}{3}\Big]^{-1}_{-2}+\Big[\text{x}-\frac{\text{x}^3}{3}\Big]^1_{-1}-\Big[\text{x}-\frac{\text{x}^3}{3}\Big]^2_1$
$=\Big[-1+\frac{1}{3}+2-\frac{8}{3}\Big]+\Big[1-\frac{1}{3}+1-\frac{1}{3}\Big]-\Big[2-\frac{8}{3}-1+\frac{1}{3}\Big]$
$=-\Big[1-\frac{7}{3}\Big]+\Big[2-\frac{2}{3}\Big]-\Big[1-\frac{7}{3}\Big]$
$=-1+\frac{7}{3}+2-\frac{2}{3}-1+\frac{7}{3}$
$=4$
View full question & answer→MCQ 871 Mark
Value of $\int\frac{\text{dx}}{\sqrt{2\text{x}-\text{x}^2}}$
- ✓
$\sin^{-1}(\text{x}-1)+\text{c}$
- B
$\sin^{-1}(1+\text{x})+\text{c}$
- C
$\sin^{-1}(1+\text{x}^2)+\text{c}$
- D
$-\sqrt{2\text{x}-\text{x}^2}+\text{c}$
AnswerCorrect option: A. $\sin^{-1}(\text{x}-1)+\text{c}$
$\sin^{-1}(\text{x}-1)+\text{c}$
View full question & answer→MCQ 881 Mark
$\int\frac{\text{x}^4+1}{\text{x}^2+1}\text{ dx}$ is equal to:
- A
$\frac{\text{x}^3}{3}+\text{x}+\tan^{-1}\text{x}+\text{c}$
- B
$\frac{\text{x}^3}{3}-\text{x}+\tan\text{x}+\text{c}$
- C
$\frac{\text{x}^3}{3}+\text{x}+2\tan^{-1}\text{x}+\text{c}$
- ✓
$\frac{\text{x}^3}{3}-\text{x}+2\tan^{-1}\text{x}+\text{c}$
AnswerCorrect option: D. $\frac{\text{x}^3}{3}-\text{x}+2\tan^{-1}\text{x}+\text{c}$
$\frac{\text{x}^3}{3}-\text{x}+2\tan^{-1}\text{x}+\text{c}$
View full question & answer→MCQ 891 Mark
If $\int\text{f(x)}\text{dx}=2\text{ f(x)}^3+\text{c}$ and $\text{f(x)}\neq0$ then f(x) is:
AnswerCorrect option: D. $\sqrt{\frac{\text{x}}{3}}$
$\sqrt{\frac{\text{x}}{3}}$
View full question & answer→MCQ 901 Mark
Choose the correct answer in Exercise:
$\int\frac{\text{dx}}{\sqrt{9\text{x}-4\text{x}^2}}\text{equals}$
- A
$\frac{1}{9}\sin^{-1}\bigg(\frac{9\text{x}-8}{8}\bigg)+\text{C}$
- ✓
$\frac{1}{2}\sin^{-1}\bigg(\frac{8\text{x}-9}{9}\bigg)+\text{C}$
- C
$\frac{1}{3}\sin^{-1}\bigg(\frac{9\text{x}-8}{8}\bigg)+\text{C}$
- D
$\frac{1}{2}\sin^{-1}\bigg(\frac{9\text{x}-8}{9}\bigg)+\text{C}$
AnswerCorrect option: B. $\frac{1}{2}\sin^{-1}\bigg(\frac{8\text{x}-9}{9}\bigg)+\text{C}$
$\text{Let I}=\int\frac{\text{dx}}{\sqrt{9\text{x}-4\text{x}^2}}$
$=\int\frac{1}{\sqrt{-4\text{x}^2+9\text{x}}}\text{ dx}$
$=\int\frac{1}{-4\bigg(\text{x}^2-\frac{9}{4}\text{x}\bigg)}\text{ dx}$
$=\int\frac{1}{-4\Bigg[\text{x}^2-\frac{9}{4}\text{x}+\bigg(\frac{9}{8}\bigg)^2-\bigg(\frac{9}{8}\bigg)^2\Bigg]}\text{ dx}$
$=\int\frac{1}{-4\Bigg[\bigg(\text{x}-\frac{9}{8}\bigg)^2+\bigg(\frac{9}{8}\bigg)^2\Bigg]}\text{ dx}$
$=\int\frac{1}{4\Bigg[\bigg(\frac{9}{8}\bigg)^2-\bigg(\text{x}-\frac{9}{8}\bigg)^2\Bigg]}\text{ dx}$
$=\frac{1}{2}\int\frac{1}{\Bigg[\bigg(\frac{9}{8}\bigg)^2-\bigg(\text{x}-\frac{9}{8}\bigg)^2\Bigg]}\text{ dx}$
$=\frac{1}{2}\sin^{-1}\frac{\text{x}-\frac{9}{8}}{\frac{9}{8}}+\text{C}$
$=\frac{1}{2}\sin^{-1}\bigg(\frac{8\text{x}-9}{9}\bigg)+\text{C}$
Therefore, option (B) is correct.
View full question & answer→MCQ 911 Mark
If $\frac{\text{dy}}{\text{dx}}=\text{x}^{-3}$ then $y:$
- ✓
$\frac{-1}{2\text{x}^{2}}+\text{c}$
- B
$\frac{-\text{x}^{-4}}{4}+\text{c}$
- C
$\frac{2}{\text{x}^{2}}+\text{c}$
- D
$\frac{\text{x}^{-2}}{2}+\text{c}$
AnswerCorrect option: A. $\frac{-1}{2\text{x}^{2}}+\text{c}$
View full question & answer→MCQ 921 Mark
$\int_{0}^{1}\text{x}(1-\text{x})^{99}$ is equal to:
- A
$\frac{1}{10010}$
- ✓
$\frac{1}{10100}$
- C
$\frac{1}{1010}$
- D
$\frac{11}{10100}$
AnswerCorrect option: B. $\frac{1}{10100}$
$\frac{1}{10100}$
View full question & answer→MCQ 931 Mark
$\int\frac{9\text{x}}{9\text{x}^2+1}=$
- A
$\frac{1}{3}\tan^{-1}(2\text{x})+\text{C}$
- B
$\frac{1}{3}\tan^{-1}\text{x}+\text{C}$
- ✓
$\frac{1}{3}\tan^{-1}(3\text{x})+\text{C}$
- D
$\frac{1}{3}\tan^{-1}(6\text{x})+\text{C}$
AnswerCorrect option: C. $\frac{1}{3}\tan^{-1}(3\text{x})+\text{C}$
$\frac{1}{3}\tan^{-1}(3\text{x})+\text{C}$
View full question & answer→MCQ 941 Mark
Choose the correct answer in Exercise:
$\int\text{e}^\text{x}\sec\text{x}(1+\tan\text{x})\text{dx}$ equals
- A
$\text{e}^\text{x}\cos\text{x}+\text{C}$
- ✓
$\text{e}^\text{x}\sec\text{x}+\text{C}$
- C
$\text{e}^\text{x}\sin\text{x}+\text{C}$
- D
$\text{e}^\text{x}\tan\text{x}+\text{C}$
AnswerCorrect option: B. $\text{e}^\text{x}\sec\text{x}+\text{C}$
$\text{e}^\text{x}\sec\text{x}+\text{C}$
$\int\text{e}^\text{x}\sec\text{x}(1+\tan\text{x})\text{dx}$
Let $\text{I}=\int\text{e}^\text{x}\sec\text{x}(1+\tan\text{x})\text{dx}=\int\text{e}^\text{x}(\sec\text{x}+\sec\text{x}\tan\text{x})\text{dx}$
Also, let $\sec\text{x}=\text{f}(\text{x})\Rightarrow \ \sec\text{x}\tan\text{x}=\text{f}'(\text{x})$
It is known that, $\int\text{e}^\text{x}\{\text{f}(\text{x})+\text{f}'(\text{x})\}\text{dx}=\text{e}^\text{x}\text{f}(\text{x})+\text{C}$
$\therefore\ \text{I}=\text{e}^\text{x}\sec\text{x}+\text{C}$
View full question & answer→MCQ 951 Mark
$\int\frac{1}{1+\tan\text{x}}\text{ dx}=$
- A
$\log_\text{e}(\text{x}+\sin\text{x})+\text{C}$
- B
$\log_\text{e}(\sin\text{x}+\cos\text{x})+\text{C}$
- C
$2\sec^2\frac{\text{x}}{2}+\text{C}$
- ✓
$\frac{1}{2}\big[\text{x}+\log(\sin\text{x}+\cos\text{x})\big]+\text{C}$
AnswerCorrect option: D. $\frac{1}{2}\big[\text{x}+\log(\sin\text{x}+\cos\text{x})\big]+\text{C}$
$\text{I}=\int\frac{1}{1+\tan\text{x}}\text{ dx}$
$=\int\frac{\cos\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$
Numerator can be written as,
$\cos\text{x}=\text{A}(\sin\text{x}+\cos\text{x})+\text{B}\frac{\text{d}(\sin\text{x}+\cos\text{x})}{\text{dx}}$
$\cos\text{x}=(\text{A}-\text{B})\sin\text{x}+(\text{A}+\text{B})\cos\text{x}$
$\text{A}-\text{B}=0$ and $\text{A}+\text{B}=1$
$\text{A}=\frac{1}{2}=\text{B}$
$\text{I}=\int\frac{\big[\frac{1}{2}(\sin\text{x}+\cos\text{x})+\frac{1}{2}(\cos\text{x}-\sin\text{x})\big]\text{dx}}{\sin\text{x}+\cos\text{x}}$
$\text{I}=\frac{1}{2}\int\Big(1+\frac{\cos\text{x}-\sin\text{x}}{\sin\text{x}+\cos\text{x}}\Big)\text{dx}$
$\text{I}=\frac{1}{2}\big[1+\int(\sin\text{x}+\cos\text{x})\big]+\text{C}$
View full question & answer→MCQ 961 Mark
$\int\sqrt{\frac{\text{x}}{1-\text{x}}}\text{ dx}$ is equal to:
- A
$\sin^{-1}\sqrt{\text{x}}+\text{C}$
- B
$\sin^{-1}\Big\{\sqrt{\text{x}}-\sqrt{\text{x}(1-\text{x})}\Big\}+\text{C}$
- C
$\sin^{-1}\Big\{\sqrt{\text{x}(1-\text{x})}\Big\}+\text{C}$
- ✓
$\sin^{-1}\sqrt{\text{x}}-\sqrt{\text{x}(1-\text{x})}+\text{C}$
AnswerCorrect option: D. $\sin^{-1}\sqrt{\text{x}}-\sqrt{\text{x}(1-\text{x})}+\text{C}$
$\text{I}=\int\sqrt{\frac{\text{x}}{1-\text{x}}}\text{ dx}$
$\text{I}=\int\sqrt{\frac{\text{x}}{1-\text{x}}\cdot\frac{\text{x}}{\text{x}}}\text{ dx}$
$\text{I}=\int\frac{\text{x dx}}{\sqrt{\text{x}-\text{x}^2}}$
Consider,
$\text{x}=\text{A}\frac{\text{d}(\text{x}-\text{x}^2)}{\text{dx}}+\text{B}$
$\text{x}=\text{A}(1-2\text{x})+\text{B}$
$\text{x}=-2\text{Ax}+\text{A}+\text{B}$
$-2\text{A}=1$
$\text{A}=\frac{-1}{2}$
$\text{I}=\int\frac{\frac{-1}{2}(1-2\text{x})+\frac{1}{2}}{\sqrt{\text{x}-\text{x}^2}}\text{ dx}$
$\text{I}=\int\Big(\frac{-1}{2}\frac{1-2\text{x}}{\sqrt{\text{x}-\text{x}^2}}+\frac{1}{2\sqrt{\text{x}-\text{x}^2}}\Big)\text{dx}$
$\text{I}=\frac{-1}{2}\times2\sqrt{\text{x}-\text{x}^2}+\frac{1}{2}\int\frac{1}{\sqrt{\text{x}-\text{x}^2}}\text{ dx}$
Second term after completing square method you will get as
$\text{I}=-\sqrt{\text{x}-\text{x}^2}+\sin^{-1}\sqrt{\text{x}}+\text{C}$
View full question & answer→MCQ 971 Mark
$\int\limits^1_{-1}|1-\text{x}|\text{dx}$ is equal to:
Answer$\int\limits^1_{-1}|1-\text{x}|\text{dx}$
$=\int\limits^0_{-1}(1-\text{x})\text{dx}+\int\limits^1_0(1-\text{x})\text{dx}$
$=\Big[\text{x}-\frac{\text{x}^2}{2}\Big]^0_{-1}+\Big[\text{x}-\frac{\text{x}^2}{2}\Big]^1_0$
$=0+1+\frac{1}{2}+1-\frac{1}{2}-0$
$=2$
View full question & answer→MCQ 981 Mark
$\int\sec^2\text{x}.\text{cosec}^2\text{xdx}=$
- ✓
$\tan\text{x}-\cot\text{x+c}$
- B
$\tan\text{x}+\cot\text{x+c}$
- C
$-\tan\text{x}+\cot\text{x+c}$
- D
$\sec\text{x}\tan\text{x+c}$
AnswerCorrect option: A. $\tan\text{x}-\cot\text{x+c}$
$\int\sec^2\text{x}.\text{cosec}^2\text{xdx}$
$=\int\frac{{1}}{{\cos^2\text{x.}\sin^2\text{x}}}\text{dx}$
$=\int\frac{{{{\cos^2\text{x.}\sin^2\text{x}}}}}{{\cos^2\text{x.}\sin^2\text{x}}}\text{dx}$
$=\int\frac{{{{\cos^2\text{x.}}}}}{{\cos^2\text{x}\sin^2\text{x}}}+\frac{{{{\sin^2\text{x}}}}}{{\cos^2\text{x.}\sin^2\text{x}}}\text{dx}$
$=\int(\text{cosec}^2\text{x}+\sec^2\text{x})\text{dx}$
$=-\cot\text{x}+\tan\text{x}+\text{c}$
$=\tan\text{x}-\cot\text{x+c}$
View full question & answer→MCQ 991 Mark
The value of the integral $\int\limits^\infty_0\frac{\text{x}}{(1+\text{x})(1+\text{x}^2)}\text{dx}$ is:
- A
$\frac{\pi}{2}$
- ✓
$\frac{\pi}{4}$
- C
$\frac{\pi}{6}$
- D
$\frac{\pi}{3}$
AnswerCorrect option: B. $\frac{\pi}{4}$
We have,
$\text{I}=\int\limits^\infty_0\frac{\text{x}}{(1+\text{x})(1+\text{x}^2)}\text{ dx}$
Putting $\text{x}=\tan\theta$
$\Rightarrow \text{dx}=\sec^2\theta\text{ d}\theta$
when $\text{x}\rightarrow0;\theta\rightarrow0$
and $\text{x}\rightarrow\infty;\theta\rightarrow\frac{\pi}{2}$
Now, integral becomes
$\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\tan\theta}{(1+\tan\theta)\sec^2\theta}\sec^2\theta\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\tan\theta}{1+\tan\theta}\text{d}\theta$
$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\frac{\sin\theta}{\cos\theta}}{1+\frac{\sin\theta}{\cos\theta}}\text{d}\theta$
$\Rightarrow \text{I}=\int\limits^\frac{\pi}{2}_0\frac{\sin\theta}{\sin\theta+\cos\theta}\text{d}\theta\ ...(\text{i})$
$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\sin\big(\frac{\pi}{2}-\theta\big)}{\sin\big(\frac{\pi}{2}-\theta\big)+\cos\big(\frac{\pi}{2}-\theta\big)}\text{ d}\theta$ $\bigg[\therefore\ \int\limits^\text{a}_0\text{f}(\text{x)}\text{dx}=\int\limits^\text{a}_0\text{f}(\text{a}-\text{x})\text{dx}\bigg]$
$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\cos\theta}{\cos\theta+\sin\theta}\text{d}\theta$
$\Rightarrow \text{I}=\int\limits^\frac{\pi}{2}_0\frac{\cos\theta}{\sin\theta+\cos\theta}\text{d}\theta\ ....(\text{ii})$
Adding (i) and (ii), we get
$2\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\sin\theta+\cos\theta}{\sin\theta+\cos\theta}\text{d}\theta$
$\Rightarrow 2\text{I}=\int\limits^\frac{\pi}{2}_0\text{d}\theta$
$\Rightarrow 2\text{I}=\frac{\pi}{2}$
$\Rightarrow \text{I}=\frac{\pi}{4}$
View full question & answer→MCQ 1001 Mark
If $\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}$ then y =
- ✓
$\text{ln }\text{x}+\text{c}$
- B
$\text{x}+\text{c}$
- C
$\frac{-1}{\text{x}^2}+\text{c}$
- D
$\frac{1}{\text{x}^2}+\text{c}$
AnswerCorrect option: A. $\text{ln }\text{x}+\text{c}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}$
$\text{dy}=\frac{\text{dx}}{\text{x}}$
$\int\text{dy}=\int\frac{\text{dx}}{\text{x}}=\log_{\text{e}}\text{x+c}$
$\therefore\text{y}=\text{ln }\text{x + c}$
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