MCQ 11 Mark
The value of $\int\limits^{\pi}_0\frac{\text{x}\tan\text{x}}{\sec\text{x}+\cos\text{x}}\text{ dx}$ is:
- ✓
$\frac{\pi^2}{4}$
- B
$\frac{\pi^2}{2}$
- C
$\frac{3\pi^2}{2}$
- D
$\frac{\pi^2}{2}$
AnswerCorrect option: A. $\frac{\pi^2}{4}$
We have,
$\text{I}= \int\limits^\pi_0\frac{\text{x}\tan\text{x}}{\sec\text{x}+\cos\text{x}}\text{dx}\ ...(\text{i})$
$=\int\limits^\pi_0\frac{(\pi-\text{x)}\tan(\pi-\text{x})}{\sec(\pi-\text{x})+\cos(\pi-\text{x})}\text{dx}$
$=\int\limits^\pi_0\frac{(\pi-\text{x})\tan\text{x}}{\sec\text{x}+\cos\text{x}}\ ...(\text{ii})$
Adding (i) and (ii), we get
$2\text{I}=\int\limits^\pi_0\Big[\frac{\text{x}\tan\text{x}}{\sec\text{x}+\cos\text{x}}+\frac{(\pi-\text{x})\tan\text{x}}{\sec\text{x}+\cos\text{x}}\Big]\text{dx}$
$\Rightarrow\text{I}=\frac{1}{2}\int\limits^\pi_0\frac{\pi\tan\text{x}}{\sec\text{x}+\cos\text{x}}\text{dx}$
$=\frac{\pi}{2}\int\limits^\pi_0\frac{\sin\text{x}}{1+\cos^2\text{x}}\text{dx}$
Putting $\cos\text{x}=\text{t}$
$\Rightarrow-\sin\text{x}\text{ dx}=\text{dt}$
$\Rightarrow \sin\text{x}\text{ dx}=-\text{dt}$
When $\text{x}\rightarrow0;\text{ t}\rightarrow1$
and $\text{x}\rightarrow\pi;\text{ t}\rightarrow-1$
$\Rightarrow\text{I}=\frac{\pi}{2}\int\limits^{-1}_1\frac{-\text{dt}}{1+\text{t}^2}$
$=\frac{\pi}{2}\int\limits^1_{-1}\frac{\text{dt}}{1+\text{t}^2}$
$=\frac{\pi}{2}\big[\tan^{-1}\text{t}\big]^1_{-1}$
$=\frac{\pi}{2}\big[\tan^{-1}(1)-\tan^{-1}(-1)\big]$
$=\frac{\pi}{2}\Big[\frac{\pi}{4}-\Big(-\frac{\pi}{4}\Big)\Big]$
$=\frac{\pi}{2}\times\frac{\pi}{2}=\frac{\pi^2}{4}$
Hence, $\text{I}=\frac{\pi^2}{4}$
View full question & answer→MCQ 21 Mark
$\int\limits^1_0\frac{\text{d}}{\text{dx}}\Big\{\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\Big\}\text{dx}$ is equal to:
- A
$0$
- B
${\pi}$
- ✓
$\frac{\pi}{2}$
- D
$\frac{\pi}{4}$
AnswerCorrect option: C. $\frac{\pi}{2}$
We have,
$\text{I}=\int\limits^1_0\frac{\text{d}}{\text{dx}}\Big\{\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\Big\}\text{dx}$
We know since $\int\text{f}'(\text{x})=\text{f}(\text{x})$
$\text{f}(\text{x})=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ and $\text{f}'(\text{x})=\frac{\text{d}}{\text{dx}}\Big\{\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\Big\}$
Therefore, $\text{I}=\Big[\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\Big]^1_0$
$=\sin^{-1}(1)-\sin^{-1}(0)$
$=\frac{\pi}{2}$
View full question & answer→MCQ 31 Mark
$\int\limits^\sqrt{3}_1\frac{1}{1+\text{x}^2}\text{ dx}$ is equal to:
- ✓
$\frac{\pi}{12}$
- B
$\frac{\pi}{6}$
- C
$\frac{\pi}{4}$
- D
$\frac{\pi}{3}$
AnswerCorrect option: A. $\frac{\pi}{12}$
$\int\limits^\sqrt{3}_1\frac{1}{1+\text{x}^2}\text{dx}$
$=\big[\tan^{-1}\text{x}\big]^\sqrt{3}_1$
$=\frac{\pi}{3}-\frac{\pi}{4}$
$=\frac{\pi}{12}$
View full question & answer→MCQ 41 Mark
$\int\limits^{\frac{\pi}{2}}_0\sin2\text{x }\log\tan\text{x dx}$ is equal to:
- A
$\pi$
- B
$\frac{\pi}{2}$
- ✓
$0$
- D
$2\pi$
Answer$\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin2\text{x }\log\tan\text{x dx}\ ....(\text{i})$
$\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin(\pi-2\text{x})\log\tan\big(\frac{\pi}{2}-\text{x}\big)\text{dx}$
$\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin2\text{x}\log\cot\text{x dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin2\text{x}\big(\log\tan\text{x}+\log\cot\text{x}\big)\text{dx}$
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin2\text{x}\big(\log\tan\text{x}\cot\text{x}\big)\text{dx}$
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\sin\text{x}(\log1)\text{dx}$
$\text{I}=0$
View full question & answer→MCQ 51 Mark
The value of the integral $\int\limits^{\frac{\pi}{2}}_0\frac{\sqrt{\cos\text{x}}}{\sqrt{\cos\text{x}}+\sqrt{\sin\text{x}}}\text{ dx}$ is:
- A
$0$
- B
$\frac{\pi}{2}$
- ✓
$\frac{\pi}{4}$
- D
AnswerCorrect option: C. $\frac{\pi}{4}$
$\text{ I}=\int\limits^\frac{\pi}{2}_0\frac{\sqrt{\cos\text{x}}}{\sqrt\cos\text{x}+\sqrt{\sin\text{x}}}\text{dx}\ ...(\text{i})$
$=\int\limits^\frac{\pi}{2}_0\frac{\sqrt{\cos\big(\frac{\pi}{2}-\text{x}\big)}}{\sqrt{\cos\big(\frac{\pi}{2}-\text{x}\big)+\sqrt{\sin\big(\frac{\pi}{2}-\text{x}\big)}}}\text{dx}$
$=\int\limits^\frac{\pi}{2}_0\frac{\sqrt{\sin\text{x}}}{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}\text{dx}$
$=\int\limits^\frac{\pi}{2}_0\frac{\sqrt{\sin{\text{x}}}}{\sqrt{\cos\text{x}}+\sqrt{\sin\text{x}}}\text{dx}...(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^\frac{\pi}{2}_0\bigg[\frac{\sqrt{\cos\text{x}}}{\sqrt{\cos\text{x}}+\sqrt{\sin\text{x}}}+\frac{\sqrt{\sin\text{x}}}{\sqrt{\cos\text{x}}+\sqrt{\sin\text{x}}}\bigg]\text{dx}$
$=\int\limits^\frac{\pi}{2}_0\text{dx}$
$=[\text{x}]^\frac{\pi}{2}_0=\frac{\pi}{2}$
Hence, $\text{I}=\frac{\pi}{4}$
View full question & answer→MCQ 61 Mark
The value of $\int\limits^\frac{\pi}{4}_0\cos\text{x}\text{ e}^{\sin\text{x}}\text{ dx}$ is:
AnswerLet, $\text{I}=\int\limits^\frac{\pi}{4}_0\cos\text{x}\text{ e}^{\sin\text{x}}\text{dx}$
Let $\sin\text{x}=\text{t},$ then $\cos\text{x}\text{ dx}=\text{dt}$
when $\text{x}=0,\text{t}=0$ and $\text{x}=\frac{\pi}{2},\text{t}=1$
Therefore the integrel becomes
$\text{I}=\int\limits^1_0\text{e}^\text{t}\text{dt}$
$=\big[\text{e}^\text{t}\big]^1_0$
$=\text{e}-1$
View full question & answer→MCQ 71 Mark
$\int\limits^1_0\sqrt{\text{x}(1-\text{x})}\text{ dx}$ equals:
- A
$\frac{\pi}{2}$
- B
$\frac{\pi}{4}$
- C
$\frac{\pi}{6}$
- ✓
$\frac{\pi}{8}$
AnswerCorrect option: D. $\frac{\pi}{8}$
$\text{I}=\int\limits^1_0\sqrt{\text{x}(1-\text{x})}\text{ dx} $
$\text{I}= \int\limits^1_0\sqrt{\text{x}-\text{x}^2}\text{dx}$
$\text{I}=\int\limits^1_0\sqrt{\frac{1}{4}+\text{x}-\text{x}^2+\frac{1}{4}}\text{dx}$
$\text{I}=\int\limits^1_0\sqrt{\frac{1}{4}-\Big(\text{x}^2-\text{x}+\frac{1}{4}\Big)}\text{dx}$
$\text{I}=\int\limits^1_0\sqrt{\Big(\frac{1}{2}\Big)^2-\Big(\text{x}-\frac{1}{2}\Big)^2}\text{dx}$
$\text{I}=\Bigg[\frac{\text{x}-\frac{1}{2}}{2}\sqrt{\text{x}(1-\text{x})}+\frac{1}{2}\times\frac{1}{4}\sin^{-1}(2\text{x}-1)\Bigg]^1_0$
$\text{I}=0+\frac{1}{8}\big(\sin^{-1}(1)-\sin^{-1}(-1)\big)$
$\text{I}= \frac{1}{8}\Big(\frac{\pi}{2}-\Big(\frac{\pi}{2}\Big)\Big)$
$\text{I}= \frac{\pi}{8}$
View full question & answer→MCQ 81 Mark
$\int\limits^{\pi}_0\frac{1}{1+\sin\text{x}}\text{ dx}$ equals:
- A
$0$
- B
$\frac{1}{2}$
- ✓
$2$
- D
$\frac{3}{2}$
Answer$\int\limits^{\pi}_0\frac{1}{1+\sin\text{x}}\text{ dx}$
$=\int\limits^\pi_0\frac{1}{1+\sin\text{x}}\times\frac{1-\sin\text{x}}{1-\sin\text{x}}\text{dx}$
$= \int\limits^\pi_0\frac{1-\sin\text{x}}{1-\sin^2\text{x}}\text{dx}$
$= \int\limits^\pi_0\frac{1-\sin\text{x}}{\cos^2\text{x}}\text{dx}$
$=\int\limits^\pi_0(\sec^2\text{x}-\sec\text{x}\tan\text{x})\text{dx}$
$=\big[\tan\text{x}-\sec\text{x}\big]^\pi_0$
$=0+1-0+1$
$=2$
View full question & answer→MCQ 91 Mark
The value of $\int\limits^\frac{\pi}{2}_0\log\Big(\frac{4+3\sin\text{x}}{4+3\cos\text{x}}\Big)\text{dx}$ is:
AnswerLet $\text{I}=\int\limits^\frac{\pi}{2}_0\log\Big(\frac{4+3\sin\text{x}}{4+3\cos\text{x}}\Big)\text{dx}\ ...(\text{i})$
$=\int\limits^\frac{\pi}{2}_0\log\Bigg[\frac{4+3\sin\big(\frac{\pi}{2}-\text{x}\big)}{4+3\cos\big(\frac{\pi}{2}-\text{x}\big)}\Bigg]\text{dx}$
$=\int\limits^\frac{\pi}{2}_0\log\Big(\frac{4+3\cos\text{x}}{4+3\sin\text{x}}\Big)\text{dx}\ ....(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^\frac{\pi}{2}_0\Big[\log\Big(\frac{4+3\sin\text{x}}{4+3\cos\text{x}}\Big)+\log\Big(\frac{4+3\cos\text{x}}{4+3\sin\text{x}}\Big)\Big]\text{dx}$
$=\int\limits^\frac{\pi}{2}_0\log1\text{ dx}=0$
Hence, $\text{I}=0$
View full question & answer→MCQ 101 Mark
$\int\limits^\text{e}_1\log\text{x}\text{ dx}=$
Answer$\int\limits^\text{e}_1\log\text{x}\text{ dx}$
$=\int\limits^\text{e}_1\log\text{x}\text{ x}^0\text{dx}$
$=\big[\text{x}\log\text{x}\big]^\text{e}_1-\int\limits^\text{e}_1\frac{1}{\text{x}}\text{dx}$
$=\big[\text{x}\log\text{x}\big]^\text{e}_1-\big[\text{x}\big]^\text{e}_1$
$=(\text{e}-0)-(\text{e}-1)$
$= \text{e}-\text{e}+1$
$=1$
View full question & answer→MCQ 111 Mark
The value of $\int\limits^1_0\tan^{-1}\Big(\frac{2\text{x}-1}{1+\text{x}-\text{x}^2}\Big)\text{ dx},$ is:
AnswerLet, $\text{I}=\int\limits^1_0\tan^{-1}\frac{2\text{x}-1}{1+\text{x}-\text{x}^2}\text{ dx}\ ....(\text{i})$
$=\int\limits^1_0\tan^{-1}\frac{2(1-\text{x})-1}{1+(1-\text{x})-(1-\text{x})^2}\text{ dx}$
$=\int\limits^1_0\tan^{-1}\frac{1-2\text{x}}{2-\text{x}-1-\text{x}^2+2\text{x}}\text{ dx}$
$=\int\limits^1_0\tan^{-1}\frac{1-2\text{x}}{1+\text{x}-\text{x}^2}\text{ dx}$
$=-\int\limits^1_0\tan^{-1}\frac{2\text{x}-1}{1+\text{x}-\text{x}^2}\text{ dx}\ ....(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^1_0\tan^{-1}\frac{2\text{x}-1}{1+\text{x}-\text{x}^2}\text{ dx}-\int\limits^1_0\tan^{-1}\frac{2\text{x}-1}{1+\text{x}-\text{x}^2}\text{ dx}$
Hence, $\text{I}=0$
View full question & answer→MCQ 121 Mark
$\int\limits^3_0\frac{3\text{x}+1}{\text{x}^2+9}\text{ dx}=$
- ✓
$\frac{\pi}{12}+\log\big(2\sqrt{2}\big)$
- B
$\frac{\pi}{2}+\log\big(2\sqrt{2}\big)$
- C
$\frac{\pi}{6}+\log\big(2\sqrt{2}\big)$
- D
$\frac{\pi}{3}+\log\big(2\sqrt{2}\big)$
AnswerCorrect option: A. $\frac{\pi}{12}+\log\big(2\sqrt{2}\big)$
We have,
$\text{I}=\int\limits^3_0\frac{3\text{x}+1}{\text{x}^2+9}\text{ dx}$
$\text{I}=\int\limits^3_0\frac{3\text{x}}{\text{x}^2+9}\text{ dx}+\int\limits^3_0\frac{1}{\text{x}^2+9}\text{ dx}$
$\text{I}_1=\int\limits^3_0\frac{3\text{x}}{\text{x}^2+9}\text{ dx}$ and $\text{I}_2=\int\limits^3_0\frac{1}{\text{x}^2+9}\text{ dx}$
Putting $\text{x}^2+9=\text{t}$ in $I_1$
$\Rightarrow 2\text{x}\text{ dx}=\text{dt}$
$\Rightarrow \text{x}\text{ dx}=\frac{\text{dt}}{2}$
when $\text{x}\rightarrow0;\text{t}\rightarrow9$
and $\text{x}\rightarrow3;\text{t}\rightarrow18$
$\therefore\text{I}=\int\limits^{18}_9\frac{3\text{ dt}}{2\text{ t}}+\int\limits^3_0\frac{1}{\text{x}^2+9}\text{ dx}$
$=\frac{3}{2}\int\limits^{18}_9\frac{\text{dt}}{\text{t}}+\int\limits^3_0\frac{1}{\text{x}^2+3^2}\text{ dx}$
$=\frac{3}{2}\big[\log(\text{t})\big]^{18}_9+\frac{1}{3}\Big[\tan^{-1}\Big(\frac{\pi}{3}\Big)\Big]^3_0$
$=\frac{3}{2}\big[\log18-\log9\big]+\frac{1}{3}\Big(\frac{\pi}{4}-0\Big)$
$=\frac{3}{2}\Big[\log\frac{18}{9}\Big]+\frac{\pi}{12}$
$=\frac{3}{2}\big[\log2\big]+\frac{\pi}{12}$
$=\log(\sqrt{8})+\frac{\pi}{12}$
$=\log(2\sqrt{2})+\frac{\pi}{12}$
$=\frac{\pi}{12}+\log(2\sqrt{2})$
View full question & answer→MCQ 131 Mark
$\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\frac{1}{\sin2\text{x}}\text{ dx}$ is equal to:
- A
$\log_\text{e}{3}$
- ✓
$\log_\text{e}\sqrt{3}$
- C
$\frac{1}{2}\log(-1)$
- D
$\log(-1)$
AnswerCorrect option: B. $\log_\text{e}\sqrt{3}$
$\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\frac{1}{\sin2\text{x}}\text{ dx}$
$=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\text{cosec }2\text{x}\text{ dx}$
$=\frac{1}{2}\int\limits^\frac{\pi}{3}_\frac{\pi}{6}2\text{cosec }2\text{x}\text{ dx}$
$=\frac{-1}{2}\big[\log(\text{cosec}2\text{x}+\cot2\text{x})\big]^\frac{\pi}{3}_\frac{\pi}{6}$
$=\frac{-1}{2}\big[-2\log\sqrt{3}\big]$
$=\log\sqrt{3}$
View full question & answer→MCQ 141 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 151 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 161 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 171 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 181 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 191 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 201 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 211 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 221 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 231 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 241 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 251 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,
$-\int\limits^{\frac{\pi}{2}}_0\log(\sin\theta)\text{ d}\theta-\int\limits^{\frac{\pi}{2}}_0\log\cos\theta\text{ d}\theta$
$-2\int\limits^{\frac{\pi}{2}}_0\log\sin\theta\text{ d}\theta=-2\times\text{I}$
$=-2\times-\frac{\pi}{2}\log2$
$\Big[\text{Where I}=-\frac{\pi}{2}\log2\Big]$
$=\pi\log2$
View full question & answer→MCQ 261 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 271 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 281 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 291 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 301 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 311 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 321 Mark
$\int\limits^\infty_0\frac{1}{1+\text{e}^\text{x}}\text{dx}$ equals:
- A
$\log2-1$
- B
$\log2$
- C
$\log4-1$
- ✓
$-\log2$
AnswerCorrect option: D. $-\log2$
Let, $\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\cos\text{x}}{(2+\sin\text{x})(1+\sin\text{x})}\text{dx}$Let $\sin\text{x}$ then $\cos\text{x}\text{ dx}=\text{dt}$
When $\text{x}=0,\text{t}=0,\text{x}=\frac{\pi}{2},\text{t}=1$
Therefore the integral becomes
$\text{I}=\int\limits^1_0\frac{\text{dt}}{(2+\text{t})(1+\text{t})}$
$=\int\limits^1_0\Big[\frac{-1}{2+\text{t}}+\frac{1}{1+\text{t}}\Big]\text{dt}$
$=\big[-\log(2+\text{t})+\log(1+\text{t})\big]^1_0$
$= \big[\log(1+\text{t})-\log(2+\text{t})\big]^1_0$
$=\log2-\log3-\log1+\log2$
$=\log\frac{4}{3}$
View full question & answer→MCQ 331 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 341 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 351 Mark
$\int\limits^\infty_0\frac{1}{1+\text{e}^\text{x}}\text{dx}$ equals:
- A
$\log2-1$
- ✓
$\log2$
- C
$\log4-1$
- D
$-\log2$
AnswerCorrect option: B. $\log2$
We have,
$\text{I}=\int\limits^\infty_0\frac{1}{1+\text{e}^\text{x}}\text{ dx}$
Putting $\text{e}^\text{x}=\text{t}$
$\Rightarrow \text{e}^\text{x}\text{ dx}=\text{dt}$
$\Rightarrow \text{dx} = \frac{\text{dt}}{\text{t}}$
When $\text{x}\rightarrow0;\text{ t}\rightarrow1$
and $\text{x}\rightarrow\infty;\text{ t}\rightarrow\infty$
$\therefore\text{I}=\int\limits^\infty_1\frac{1}{\text{t}(1+\text{t})}\text{dt}$
$=\int\limits^\infty_1\frac{1}{\text{t}+\text{t}^2}\text{dt}$
$=\int\limits^\infty_1\frac{1}{\big(\text{t}+\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}\text{dt}$
$=\frac{1}{2\times\frac{1}{2}}\Bigg[\log\Bigg|\frac{\text{t}+\frac{1}{2}-\frac{1}{2}}{\text{t}+\frac{1}{2}+\frac{1}{2}}\Bigg|\Bigg]^\infty_1$
$=\Big[\log\Big|\frac{\text{t}}{\text{t+1}}\Big|\Big]^\infty_1$
$=\Bigg[\log\Bigg|\frac{\frac{\text{t}}{\text{t}}}{\frac{\text{t}}{\text{t}}+\frac{1}{\text{t}}}\Bigg|\Bigg]^\infty_1$
$=\Bigg[\log\Bigg|\frac{1}{1+\frac{1}{\text{t}}}\Bigg|\Bigg]^\infty_1$
$=\log\frac{1}{1+0}-\log\frac{1}{1+1}$
$=\log(1)-\log(\frac{1}{2})$
$=0-(-\log2)$
$=\log2$
View full question & answer→MCQ 361 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 371 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 381 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 391 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 401 Mark
The derivative of $\text{f(x)}=\int\limits^{\text{x}^3}_{\text{x}^2}\frac{1}{\log_{\text{e}}\text{t}}\text{ dt},(\text{x}>0),$ is:
- A
$\frac{1}{3\ln\text{x}}$
- B
$\frac{1}{3\ln\text{x}}-\frac{1}{2\ln\text{x}}$
- ✓
$\big(\ln\text{x}\big)^{-1}\text{x}(\text{x}-1)$
- D
$\frac{3\text{x}^2}{\ln\text{x}}$
AnswerCorrect option: C. $\big(\ln\text{x}\big)^{-1}\text{x}(\text{x}-1)$
$\text{f}'(\text{x})=\frac{1}{\log_\text{e}\text{x}^3}(3\text{x}^2)-\frac{1}{\log_\text{e}\text{x}^2}(2\text{x})$
$=\frac{3\text{x}^2}{3\ln\text{ x}}-\frac{2\text{x}}{2\ln\text{ x}}$
$=\frac{\text{x}^2}{\ln\text{ x}^{-1}}-\frac{\text{x}}{\ln\text{ x}}$
$=\frac{1}{\ln\text{ x}}\text{x}(\text{x}-1)$
$=\big(\ln\text{ x}\big)^{-1}\text{x}(\text{x}-1)$
View full question & answer→MCQ 411 Mark
$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin|\text{x}|\text{dx}$ is equal to:
Answer$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin|\text{x}|\text{dx}$
$=-\int\limits^0_{-\frac{\pi}{2}}\sin\text{x}\text{ dx}+\int\limits^\frac{\pi}{2}_0\sin\text{x}\text{ dx}$
$=-\big[-\cos\text{x}\big]^0_{-\frac{\pi}{2}}+\big[-\cos\text{x}\big]^\frac{\pi}{2}_0$
$=1-0-0+1$
$=2$
View full question & answer→MCQ 421 Mark
$\lim\limits_{\text{n}\rightarrow\infty}\Big\{\frac{1}{2\text{n}+1}+\frac{1}{2\text{n}+2}+\ .....+\frac{1}{2\text{n}+\text{n}}\Big\}$ is equal to:
- A
$\ln\Big(\frac{1}{3}\Big)$
- B
$\ln\Big(\frac{2}{3}\Big)$
- ✓
$\ln\Big(\frac{3}{2}\Big)$
- D
$\ln\Big(\frac{4}{3}\Big)$
AnswerCorrect option: C. $\ln\Big(\frac{3}{2}\Big)$
$\lim\limits_{\text{n}\rightarrow\infty}\Big\{\frac{1}{2\text{n}+1}+\frac{1}{2\text{n}+2}+\ .....\ +\frac{1}{2\text{n}+\text{n}}\Big\}$
$=\lim\limits_{\text{n}\rightarrow\infty}\sum\limits^\text{n}_{\text{r}=1}\frac{1}{2\text{n}+\text{r}}$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{1}{\text{n}}\sum\limits^\text{n}_{\text{r}=1}\frac{1}{2+\frac{\text{r}}{\text{n}}}$
let $\frac{\text{r}}{\text{n}}=\text{x}$
$=\int\limits^\infty_0\frac{1}{2+\text{x}}\text{dx}$
$=\Big[\log\big(2+\text{x}\big)\Big]^\infty_0$
$=\log3-\log2$
$=\log\frac{3}{2}$
$=\ln\Big(\frac{3}{2}\Big)$
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