M.C.Q (1 Marks)

Question 511 Mark

$\int(1+2\text{x}+3\text{x}^2+4\text{x}^3+ ... )\text{dx }(\mid\text{x}\mid<1)$

$-(1+\text{x})^{-1}+\text{c}$
$(1-\text{x})^{-1}+\text{c}$
$-(1-\text{x})^{-2}+\text{c}$
None of these

Answer

$(1-\text{x})^{-1}+\text{c}$

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MCQ 521 Mark

Integrate the following functions with respect to t $\int(3\text{t}^2-2\text{t})\text{dt:}$

✓
$t^3 - t^2 + C$
B
$3t^3 - 2t^2 + C$
C
$\frac{\text{t}^{3}}{3}-\frac{\text{t}^{2}}{2}+\text{C}$
D
$t^3 - t^2$

Answer

Correct option: A.

$t^3 - t^2 + C$

$\int(3\text{t}^2-2\text{t})\text{dt}=\text{t}^3-\text{t}^2+\text{c}$

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Question 531 Mark

$\int\limits^\infty_0\frac{1}{1+\text{e}^\text{x}}\text{dx}$ equals:

$\log2-1$
$\log2$
$\log4-1$
$-\log2$

Answer

Solution:
$\int\limits^\frac{\pi^2}{4}_0\frac{\sin\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{dx}$
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
$\int\limits^\frac{\pi}{2}_02\sin\text{t}\text{ dt}$
$=-2\big[\cos\text{t}\big]^\frac{\pi}{2}_0$
$=2$

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Question 541 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:

$\text{x}^{\sin\text{x}}+\text{C}$
$\text{x}^{\sin\text{x}}\cos\text{x}+\text{C}$
$\frac{(\text{x}^{\sin\text{x}})^2}{2}+\text{C}$
None of these.

Answer

$\text{x}^{\sin\text{x}}+\text{C}$

Solution:
$\int\text{x}^{\sin\text{x}}\Big(\frac{\sin\text{x}}{\text{x}}+\cos\text{x}\cdot\log\text{x}\Big)\text{dx}$
Put $\text{x}^{\sin\text{x}}=\text{t}$
Taking $\log$ on both sides,
$\log\text{t}=\sin\text{x}\log\text{x}$
$\frac{1}{\text{t}}=\frac{\sin\text{x}}{\text{x}}+\cos\text{x}\log\text{x}$
$1=\int\text{t}\cdot\frac{\text{dt}}{\text{t}}$
$1=\text{t}+\text{C}$
$1=\text{x}^{\sin\text{x}}+\text{C}$

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Question 551 Mark

$\int_{0}^{1}\frac{\text{x}}{1+\text{x}}\text{dx}=$

$1-\log2$
$2$
$1+\log 2$
$\log2$

Answer

$1-\log2$

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Question 561 Mark

Choose the correct answer in Exercise:
$\int\frac{\text{dx}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$ is equal to

$\tan^{-1}(\text{e}^{\text{x}})+\text{C}$
$\tan^{-1}\text{(e}^{\text{x}})+\text{C}$
$\log(\text{e}^{\text{x}}-\text{e}^{\text{x}})+\text{C}$
$\log(\text{e}^{\text{x}}+\text{e}^\text{x})+\text{C}$

Answer

$\tan^{-1}(\text{e}^{\text{x}})+\text{C}$

$\text{Let I}=\int\frac{\text{dx}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}\text{dx}=\int\frac{\text{e}^{\text{x}}}{\text{e}^{2\text{x}}+1}\text{dx}$
Also, let $\text{e}^{\text{x}}=\text{t}\Rightarrow\text{e}^{\text{x}}\ \text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{1+\text{t}^{2}}$
$=\tan^{-1}\text{t}+\text{C}$
$=\tan^{-1}\text{(e}^{\text{x}})+\text{C}$

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Question 571 Mark

$\int\limits^\frac{\pi}{2}_0\frac{\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$ equals to:

$\pi$
$\frac{\pi}{2}$
$\frac{\pi}{3}$
$\frac{\pi}{4}$

Answer

$\frac{\pi}{4}$

Solution:
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 $\text{I}=\frac{\pi}{4}$

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Question 581 Mark

The value of $\int\frac{\cos\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$ is:

$2\cos\sqrt{\text{x}}+\text{C}$
$\sqrt{\frac{\cos\text{x}}{\text{x}}}+\text{C}$
$\sin\sqrt{\text{x}}+\text{C}$
$2\sin\sqrt{\text{x}}+\text{C}$

Answer

$2\sin\sqrt{\text{x}}+\text{C}$

Solution:
$\text{I}=\int\frac{\cos\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$
Put $\sqrt{\text{x}}=\text{t}$
$\frac{1}{2\sqrt{\text{x}}}\text{ dx}=\text{dt}$
$\frac{1}{\sqrt{\text{x}}}\text{ dx}=2\text{dt}$
$\text{I}=\int\cos\text{t }2\text{ dt}$
$\text{I}=2\sin\text{t}+\text{C}$
$\text{I}=2\sin\sqrt{\text{x}}+\text{C}$

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MCQ 591 Mark

$\int(3\text{x}^2-1)\text{dx:}$

✓
$x^3 - x$
B
$x^2 - x$
C
$x^3 - 1$
D
None

Answer

Correct option: A.

$x^3 - x$

$\int(3\text{x}^2-1)\text{dx:}$
$=3\frac{\text{x}^3}{3}-\text{x}$
$= x^3- x$

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Question 601 Mark

$\int\log_{10}\text{xdx}=$

$\log_\text{e}10.\text{x}\log_\text{e}(\frac{\text{x}}{\text{e}})+\text{c}$
$\log_{10}\text{e}.\text{x}\log_\text{e}(\frac{\text{x}}{\text{e}})+\text{c}$
$(\text{x}-1)\log_\text{e}\text{x}+\text{c}$
$\frac{1}{\text{x}}+\text{c}$

Answer

$\log_{10}\text{e}.\text{x}\log_\text{e}(\frac{\text{x}}{\text{e}})+\text{c}$

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Question 611 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:

$10^\text{x}-\text{x}^{10}+\text{c}$
$10^\text{x}+\text{x}^{10}+\text{c}$
$(10^\text{x}-\text{x}^{10})^{-1}+\text{c}$
$\log_\text{e}(10^\text{x}+\text{x}^{10}+\text{c}$

Answer

$\log_\text{e}(10^\text{x}+\text{x}^{10}+\text{c}$

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Question 621 Mark

$\int\text{e}^{\text{x}}\{\text{f(x)}+\text{f}'(\text{x})\}\text{dx}=$

$\text{e}^{\text{x}}\text{f(x)}+\text{C}$
$\text{e}^{\text{x}}+\text{f(x)}$
$2\text{e}^{\text{x}}\text{f(x)}$
$\text{e}^{\text{x}}-\text{f(x)}$

Answer

$\text{e}^{\text{x}}\text{f(x)}+\text{C}$

Solution:
$\int\text{e}^{\text{x}}\{\text{f(x)}+\text{f}'(\text{x})\}\text{dx}=\text{e}^{\text{x}}\text{f(x)}+\text{C}$

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Question 631 Mark

$\int|\text{x}|\text{dx}$ is equal to:

$\frac{1}{2}\text{x}^2+\text{c}$
$-\frac{\text{x}^2}{2}+\text{c}$
$\text{x}|\text{x}|+\text{c}$
$\frac{1}{2}\text{x}|\text{x}|+\text{c}$

Answer

$\frac{1}{2}\text{x}|\text{x}|+\text{c}$

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Question 641 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:

$\text{a}=\frac{1}{3},\text{ b}=1$
$\text{a}=-\frac{1}{3},\text{ b}=1$
$\text{a}=-\frac{1}{3},\text{ b}=-1$
$\text{a}=\frac{1}{3},\text{ b}=-1$

Answer

$\text{a}=\frac{1}{3},\text{ b}=-1$

Solution:
$\text{I}=\int\frac{\text{x}^3}{\sqrt{1+\text{x}^2}}\text{ dx}$
$1+\text{x}^2=\text{t}$
$2\text{xdx}=\text{dt}$
$\text{x dx}=\frac{\text{dt}}{2}$
$\text{I}=\int\frac{\text{x}^2}{\sqrt{1+\text{x}^2}}\text{x dx}$
$\text{I}=\int\frac{\text{t}-1}{\sqrt{\text{t}}}\frac{\text{dt}}{2}$
$\text{I}=\frac{1}{2}\Big(\frac{2}{3}\text{t}^{\frac{3}{2}}-2\sqrt{\text{t}}\Big)+\text{C}$
$\text{I}=\frac{1}{3}(1+\text{x}^{2})^{\frac{3}{2}}-\sqrt{1+\text{x}^2}+\text{C}$
$\text{a}=\frac{1}{3},\text{ b}=-1$

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Question 651 Mark

$\int\frac{\sin\text{x}}{3+4\cos^2\text{x}}\text{ dx}=$

$\log(3+4\cos^2\text{x})+\text{C}$
$\frac{1}{2\sqrt{3}}\tan^{-1}\Big(\frac{\cos\text{x}}{\sqrt{3}}\Big)+\text{C}$
$-\frac{1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\cos\text{x}}{\sqrt{3}}\Big)+\text{C}$
$\frac{1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\cos\text{x}}{\sqrt{3}}\Big)+\text{C}$

Answer

$-\frac{1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\cos\text{x}}{\sqrt{3}}\Big)+\text{C}$

Solution:
$\text{I}=\int\frac{\sin\text{x}}{3+4\cos^2\text{x}}\text{ dx}$
Put $\cos\text{x}=\text{t}$
$-\sin\text{x dx}=\text{dt}$
$\sin\text{x dx}=-\text{dt}$
$\text{I}=\int\frac{-\text{dt}}{3+4\text{t}^2}$
$\text{I}=\frac{-1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\text{t}}{\sqrt{3}}\Big)+\text{C}$
$\text{I}=\frac{-1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\cos\text{x}}{\sqrt{3}}\Big)+\text{C}$

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Question 661 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:

$\text{a}=\frac{1}{3},\text{b}=1$
$\text{a}=\frac{-1}{3},\text{b}=1$
$\text{a}=\frac{-1}{3},\text{b}=-1$
$\text{a}=\frac{1}{3},\text{b}=-1$

Answer

$\text{a}=\frac{1}{3},\text{b}=-1$

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MCQ 671 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:

A
$4a^2$
✓
$0$
C
$2a^2$
D
none of these

Answer

Correct option: B.

$0$

$\int\limits^1_0(\text{a}-\text{x})^2\text{ f}(\text{x})\text{dx}$
$=\text{a}^2\int\limits^1_0\text{f}(\text{x})\text{dx}+\int\limits^1_0\text{x}^2\text{f}(\text{x})\text{dx}-2\text{a}\int\limits^1_0\text{x}\text{f}(\text{x})\text{dx}$
$=\text{a}^2\times1+\text{a}^2-2\text{aa}$ (As per given values)
$=2\text{a}^2-2\text{a}^2$
$=0$

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Question 681 Mark

$\int\tan^{-1}\sqrt{\text{xdx}}$ is equal to:

$(\text{x}+1)\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\text{c}$
$\text{x}\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\text{c}$
$\sqrt{\text{x}}-\text{x}\tan^{-1}\sqrt{\text{x}}+\text{c}$
$\sqrt{\text{x}}-(\text{x}+1)\tan^{-1}\sqrt{\text{x}}+\text{c}$

Answer

$(\text{x}+1)\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\text{c}$

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Question 691 Mark

$\int\text{x}^2\sin\text{x}^3\text{dx}=$

$\frac{1}{3}\cos\text{x}^3+\text{c}$
$-\frac{1}{3}\cos\text{x}+\text{c}$
$\frac{-1}{3}\cos\text{x}^3+\text{c}$
$\frac{1}{2}\sin^2\text{x}^3+\text{c}$

Answer

$\frac{-1}{3}\cos\text{x}^3+\text{c}$

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Question 701 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:

A = 5, B = 3
A = 5, B = -3
A = -5, B = 3
A = -5, B = -3

Answer

A = 5, B = -3

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Question 711 Mark

$\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{1}{1+\sqrt{\cot\text{x}}}\text{ dx}$ is:

$\frac{\pi}{3}$
$\frac{\pi}{6}$
$\frac{\pi}{12}$
$\frac{\pi}{2}$

Answer

$\frac{\pi}{12}$

Solution:
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, $\text{I}=\frac{\pi}{12}$

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Question 721 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 =

$-\frac{1}{16}$
$\frac{1}{8}$
$\frac{1}{16}$
$-\frac{1}{8}$

Answer

$\frac{1}{16}$

Solution:
$\int\frac{\cos8\text{x}+1}{\tan2\text{x}-\cot2\text{x}}\text{ dx}$
$=\int\frac{2\cos^24\text{x}}{\frac{\sin2\text{x}}{\cos2\text{x}}-\frac{\cos2\text{x}}{\sin2\text{x}}}\text{ dx}$
$=\int\frac{2\cos^24\text{x}}{\sin^22\text{x}-\cos^22\text{x}}\times\sin2\text{x}\cos2\text{x dx}$
$=\int-\frac{\cos^24\text{x}\sin4\text{x}}{\cos4\text{x}}\text{ dx}$
$=\frac{-1}{2}\int\sin8\text{x dx}$
$=\frac{\cos8\text{x}}{16}+\text{C}$
$\text{a}=\frac{1}{16}$

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MCQ 731 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
$\frac{1}{2}\text{e}^{\text{x}^2}+\text{C}$

Answer

Correct option: A.

$\frac{1}{3}\text{e}^{\text{x}^3}+\text{C}$

Let $\text{I}=\int\text{x}^2\text{e}^{\text{x}^3}\text{dx}$
Also, let $x^3 = t$
$\Rightarrow 3x^2 dx = dt$
$\Rightarrow\ \text{I}=\frac{1}{3}\int\text{e}^\text{t}\text{dt}$
$=\frac{1}{3}(\text{e}^\text{t})+\text{C}$
$=\frac{1}{3}\text{e}^{\text{x}^3}+\text{C}$

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Question 741 Mark

$\int\frac{1}{\cos\text{x}+\sqrt{3}\sin\text{x}}\text{ dx}$ is equal to:

$\log\tan\Big(\frac{\pi}{3}+\frac{\pi}{2}\Big)+\text{C}$
$\log\tan\Big(\frac{\pi}{2}-\frac{\pi}{3}\Big)+\text{C}$
$\frac{1}{2}\log\tan\Big(\frac{\pi}{2}+\frac{\pi}{3}\Big)+\text{C}$
None of these.

Answer

$\frac{1}{2}\log\tan\Big(\frac{\pi}{2}+\frac{\pi}{3}\Big)+\text{C}$

Solution:
$\text{I}=\int\frac{1}{\cos\text{x}+\sqrt{3}\sin\text{x}}\text{ dx}$
$\text{I}=\frac{1}{2}\int\frac{2}{\frac{\cos\text{x}}{2}+\frac{\sqrt{3}}{2}\sin\text{x}}\text{ dx}$
$\text{I}=\frac{1}{2}\int\frac{1}{\cos\big(\text{x}-\frac{\pi}{6}\big)}\text{ dx}$
$\text{I}=\frac{1}{2}\int\sec\Big(\text{x}-\frac{\pi}{6}\Big)\text{dx}$
$\text{I}=\frac{1}{2}\ln\Big|\tan\Big(\frac{\text{x}}{2}+\frac{\pi}{3}\Big)\Big|+\text{C}$

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Question 751 Mark

If $\int\text{x}\sin\text{x dx}=-\text{x}\cos\text{x}+\text{a},$ then a is equal to:

$\sin\text{x}+\text{C}$
$\cos\text{x}+\text{C}$
$\text{C}$
none of these.

Answer

$\sin\text{x}+\text{C}$

Solution:
$\int\text{x}\sin\text{x dx}=-\text{x}\cos\text{x}+\text{a}$
$\text{I}=\int\text{x}\sin\text{x dx}$
$\text{I}=\text{x}\int\sin\text{x dx}-\int\Big(\frac{\text{dx}}{\text{dx}}\int\sin\text{x dx}\Big)\text{dx}$
$\text{I}=-\text{x}\cos\text{x}+\int\cos\text{x dx}$
$\text{I}=\text{x}\cos\text{x}+\sin\text{x}+\text{C}$
$\text{a}=\sin\text{x}+\text{C}$

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Question 761 Mark

$\int\frac{-1}{\text{y}^2}\text{dy}$ is:

$\frac{1}{\text{y}}$
1 - y
y
1 + y

Answer

$\frac{1}{\text{y}}$

Solution:
$\int\frac{-1}{{\text{y}}^2}\text{dy}$
$=-\int\text{y}^{-2}\text{dy}$
$=\text{y}^{-1}=\frac{1}{\text{y}}$

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Question 771 Mark

Integrate the following functions with respect to x: $\int\frac{\text{dx}}{4\text{x}+5}$

$\frac{1}{4}\text{ In }(4\text{x}+5)+\text{c}$
$\frac{1}{4}\text{ In }(4\text{x}+5)-\text{c}$
$\frac{-1}{4}\text{ In }(4\text{x}+5)-\text{c}$
$4\text{ In }(4\text{x}-5)-\text{c}$

Answer

$\frac{1}{4}\text{ In }(4\text{x}+5)+\text{c}$

Solution:
$\int\frac{\text{dx}}{4\text{x}+5}=\frac{1}{4}\int\frac{\text{dx}}{\text{x}}$ where x = 4x + 5
$=\frac{1}{4}\text{ In }\text{x}+\text{c}_{1}=\frac{1}{4}\text{ In }(4\text{x}+5)+\text{c}_{2}$

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Question 781 Mark

What is the value of $\int\limits_{-1}^{1} \sin^3\text{x}\cos^2\text{xdx:}$

0
1
$\frac{1}{2}$
2

Answer

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Question 791 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:

$\sin(\text{b}-\text{a)}\log\Big|\frac{\sin(\text{x}-\text{b})}{\sin(\text{x}-\text{a})}\Big|+\text{C}$
$\text{cosec}(\text{b}-\text{a)}\log\Big|\frac{\sin(\text{x}-\text{a})}{\sin(\text{x}-\text{b})}\Big|+\text{C}$
$\text{cosec}(\text{b}-\text{a)}\log\Big|\frac{\sin(\text{x}-\text{b})}{\sin(\text{x}-\text{a})}\Big|+\text{C}$
$\sin(\text{b}-\text{a)}\log\Big|\frac{\sin(\text{x}-\text{b})}{\sin(\text{x}-\text{a})}\Big|+\text{C}$

Answer

$\text{cosec}(\text{b}-\text{a)}\log\Big|\frac{\sin(\text{x}-\text{b})}{\sin(\text{x}-\text{a})}\Big|+\text{C}$

Solution:
Let $\text{I}=\frac{\text{dx}}{\sin(\text{x}-\text{a})\sin(\text{x}-\text{b})}$
$=\frac{1}{\sin(\text{b}-\text{a})}\int\frac{\sin(\text{b}-\text{a})}{\sin(\text{x}-\text{a})\sin(\text{x}-\text{b})}\text{dx}$
$=\frac{1}{\sin(\text{b}-\text{a})}\int\frac{\sin(\text{x}-\text{a}-\text{x}+\text{b})}{\sin(\text{x}-\text{a})\sin(\text{x}-\text{b})}\text{dx}$
$=\frac{1}{\sin(\text{b}-\text{a})}\int\frac{\sin\big\{(\text{x}-\text{a})-(\text{x}+\text{b})\big\}}{\sin(\text{x}-\text{a})\sin(\text{x}-\text{b})}\text{dx}$
$=\frac{1}{\sin(\text{b}-\text{a})}\int\frac{\sin(\text{x}-\text{a})\cos(\text{x}-\text{b})-\cos(\text{x}-\text{a})\sin(\text{x}-\text{b})}{\sin(\text{x}-\text{a})\sin(\text{x}-\text{b})}\text{dx}$
$=\frac{1}{\sin(\text{b}-\text{a})}\int\frac{\sin(\text{x}-\text{a})\cos(\text{x}-\text{b})}{\sin(\text{x}-\text{a})\sin(\text{x}-\text{b})}-\frac{\cos(\text{x}-\text{a})\sin(\text{x}-\text{b})}{\sin(\text{x}-\text{a} )\sin(\text{x}-\text{b})}\text{dx}$
$=\frac{1}{\sin(\text{b}-\text{a})}\int\frac{\cos(\text{x}-\text{b})}{\sin(\text{x}-\text{a})}-\frac{\cos(\text{x}-\text{a})}{\sin(\text{x}-\text{a} )}\text{dx}$
$=\frac{1}{\sin(\text{b}-\text{a})}\big[\log\sin|(\text{x}-\text{b})|-\log|\sin(\text{x}-\text{b})|\big]+\text{C}$
$=\text{cosec}(\text{b}-\text{a})\log\Big|\frac{\sin(\text{x}-\text{b})}{\sin(\text{x}-\text{a})}\Big|+\text{C}$

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Question 801 Mark

$\int_{a}^{b}\text{x}^2\text{dx}=$

$\frac{1}{2}\tan\frac{\text{x}}{2}+\text{k}$
$2\tan\frac{\text{x}}{2}+\text{k}$
$\tan\frac{\text{x}}{2}+\text{k}$
$\tan^2\frac{\text{x}}{2}+\text{k}$

Answer

$\tan\frac{\text{x}}{2}+\text{k}$

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Question 811 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:

$\frac{1}{3\ln\text{x}}$
$\frac{1}{3\ln\text{x}}-\frac{1}{2\ln\text{x}}$
$\big(\ln\text{x}\big)^{-1}\text{x}(\text{x}-1)$
$\frac{3\text{x}^2}{\ln\text{x}}$

Answer

$\big(\ln\text{x}\big)^{-1}\text{x}(\text{x}-1)$

Solution:
$\text{f}'(\text{x})=\frac{1}{\log_\text{e}\text{x}^3}(3\text{x}^2)-\frac{1}{\log_\text{e}\text{x}^2}(2\text{x})$
$=\frac{3\text{x}^2}{3\ln\text{ x}}-\frac{2\text{x}}{2\ln\text{ x}}$
$=\frac{\text{x}^2}{\ln\text{ x}^{-1}}-\frac{\text{x}}{\ln\text{ x}}$
$=\frac{1}{\ln\text{ x}}\text{x}(\text{x}-1)$
$=\big(\ln\text{ x}\big)^{-1}\text{x}(\text{x}-1)$

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Question 821 Mark

$\int\log_{10}\text{xdx}=$

$\log_{\text{e}}10.\text{x}\log_{\text{e}}\big(\frac{\text{x}}{\text{e}}\big)+\text{c}$
$\log_{10}\text{e.x}\log_{\text{e}}\big(\frac{\text{x}}{\text{e}}\big)+\text{c}$
$\log_{10}\text{e.x}\log_{\text{e}}\big(\frac{\text{x}}{\text{e}}\big)+\text{n}$
$\log_{10}\text{e.x}\log_{\text{e}}\big(\frac{\text{x}}{\text{n}}\big)+\text{n}$

Answer

$\log_{10}\text{e.x}\log_{\text{e}}\big(\frac{\text{x}}{\text{e}}\big)+\text{c}$

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Question 831 Mark

$\int(\frac{\cos2\theta-1}{\cos2\theta+1})\text{d}\theta=$

$\tan\theta-\theta+\text{c}$
$\theta+\tan\theta+\text{c}$
$\theta-\tan\theta+\text{c}$
$-\theta-\cot\theta+\text{c}$

Answer

$\theta-\tan\theta+\text{c}$

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Question 841 Mark

Choose the correct answer in Exercise:
$\text{If}\ \text{f}(\text{x})=\int^{\text{x}}_{0}\text{t}\sin\text{t}\ \text{dt}$, then $\text{f}\text{(x)}$ is

$\cos\text{x}+\text{x}\sin\text{x}$
$\text{x}\sin\text{x}$
$\text{x}\cos\text{x}$
$\sin\text{x}+\text{x}\cos\text{x}$

Answer

$\text{x}\sin\text{x}$

$\text{f}\text{(x)}=\int\limits_{0}^{\text{x}}\text{t}\sin\text{t}\ \text{dt}$
$\therefore\text{f}\text{'(x)}=\text{x}\sin\text{x}\ $ $\big[$$\therefore$ of first fundamental theorem$\big]$
$\text{f}\text{(x)}=\int\limits_{0}^{\text{x}}\text{t}\sin\text{t}\ \text{dt}$
$\therefore\text{f}\text{'(x)}=\text{x}\sin\text{x}\ \ $ $\big[$$\therefore$ of first fundamental theorem$\big]$

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MCQ 851 Mark

Choose the correct answer in exercise:
$\int\frac{\text{dx}}{\text{x}(\text{x}^2+1)}$ equals

✓
$\text{log}|\text{x}|-\frac{1}{2}\text{log}(\text{x}^2+1)+\text{C}$
B
$\text{log}|\text{x}|+\frac{1}{2}\text{log}(\text{x}^2+1)+\text{C}$
C
$-\text{log}|\text{x}|+\frac{1}{2}\text{log}(\text{x}^2+1)+\text{C}$
D
$\text{log}|\text{x}|+\frac{1}{2}\text{log}(\text{x}^2+1)+\text{C}$

Answer

Correct option: A.

$\text{log}|\text{x}|-\frac{1}{2}\text{log}(\text{x}^2+1)+\text{C}$

Let $\text{I}=\int\frac{1}{\text{x}(\text{x}^2+1)}\text{dx}=\int\frac{2\text{x}}{2\text{x}^2(\text{x}^2+1)}\text{dx}=\int\frac{2\text{x}}{2\text{x}^2(\text{x}^2+1)}\text{dx}\dots(\text{i})$
Putting $x^2 = t$
$\Rightarrow 2x\ dx = dt$
Putting this value in eq. $(i),$
$\text{I}=\int\frac{\text{dt}}{2\text{t}(\text{t}+1)}=\frac{1}{2}\int\frac{(\text{t}+1)-\text{t}}{\text{t}(\text{t}+1)}\text{dt}$
$\text{I}=\frac{1}{2}\int\Bigg(\frac{\text{t}+1}{\text{t}(\text{t}+1)}-\frac{\text{t}}{\text{t}(\text{t}+1)}\Bigg)\text{dt}=\frac{1}{2}\int\Bigg(\frac{1}{\text{t}}-\frac{1}{\text{t}+1}\Bigg)\text{dt}=\frac{1}{2}\Bigg[\int\frac{1}{\text{t}}\text{dt}-\int\frac{1}{\text{t}+1}\text{dt}\Bigg]$
$=\frac{1}{2}[\text{log}|\text{t}|-\text{log}|\text{t}+1|]+\text{c}$
$=\frac{1}{2}[\text{log}|\text{x}^2|-\text{log}(\text{x}^2+1)]+\text{c}$
$=\text{log}|\text{x}|-\frac{1}{2}\text{log}|\text{x}^2+1|+\text{c}$

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Question 861 Mark

$\int\frac{1}{1-\cos\text{x}-\sin\text{x}}\text{ dx}=$

$\log\big|1+\cot\frac{\pi}{2}\big|+\text{C}$
$\log\big|1-\tan\frac{\pi}{2}\big|+\text{C}$
$\log\big|1-\cot\frac{\pi}{2}\big|+\text{C}$
$\log\big|1+\tan\frac{\pi}{2}\big|+\text{C}$

Answer

$\log\big|1-\cot\frac{\pi}{2}\big|+\text{C}$

Solution:
$\int\frac{\text{dx}}{1-\cos\text{x}-\sin\text{x}}$
Put $\text{t}=\tan\frac{\text{x}}{2}$
$\text{dx}=\frac{2\text{dt}}{1+\text{t}^2}$
$\cos\text{x}=\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}=\frac{1-\text{t}^2}{1+\text{t}^2}$
$\sin\text{x}=\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}=\frac{2\text{t}}{1+\text{t}^2}$
Put in the question
$\text{I}=\int\frac{\frac{2\text{dt}}{1+\text{t}^2}}{1-\frac{1-\text{t}^2}{1+\text{t}^2}-\frac{2\text{t}}{1+\text{t}^2}}$
$\text{I}=\int\frac{\text{dt}}{\text{t}^2-1}$
$\text{I}=\int\frac{\text{dt}}{\text{t}^2-\text{t}+\frac{1}{4}-\frac{1}{4}}$
$\text{I}=\ln\Big|1-\cot\frac{\text{x}}{2}\Big|+\text{C}$

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Question 871 Mark

Choose the correct option from given four options:
$\int\tan^{-1}\sqrt{\text{x}}\text{ dx}$ is equal to:

$(\text{x}+1)\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\text{C}$
$\text{x}\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\text{C}$
$\sqrt{\text{x}}-\text{x}\tan^{-1}\sqrt{\text{x}}+\text{C}$
$\sqrt{\text{x}}-(\text{x}+1)\tan^{-1}\sqrt{\text{x}}+\text{C}$

Answer

$(\text{x}+1)\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\text{C}$

Solution:
Let $\text{I}=\int1\cdot\tan^{-1}\sqrt{\text{x}}\text{ dx}$
$=\text{x}\tan^{-1}\sqrt{\text{x}}-\int\text{x}\cdot\frac{1}{1+\text{x}}\frac{1}{2\sqrt{\text{x}}}\text{dx}$
$=\text{x}\tan^{-1}\sqrt{\text{x}}-\frac{1}{2}\int\sqrt{\text{x}}\cdot\frac{1}{1+\text{x}}\text{dx}$
Put $\text{x}=\text{t}^2\Rightarrow\text{dx}=2\text{t dt}$
$\therefore$ we get;
$=\text{x}\tan^{-1}\sqrt{\text{x}}-\int\frac{\text{t}^2}{1+\text{t}^2}\text{dt}$
$=\text{x}\tan^{-1}\sqrt{\text{x}}-\int\Big(1-\frac{1}{1+\text{t}^2}\Big)\text{dt}$
$=\text{x}\tan^{-1}\sqrt{\text{x}}-\text{t}+\tan^{-1}\text{t}+\text{C}$
$=\text{x}\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\tan^{-1}\text{t}+\text{C}$
$=\text{x}\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\tan^{-1}\sqrt{\text{x}}+\text{C}$
$=(\text{x}+1)\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\text{C}$

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Question 881 Mark

Choose the correct answer in Exercise: The value of $\int^{\frac{\pi}{2}}\limits_{\frac{-\pi}{2}}\text{(x}^{3}+\text{x}\cos\text{x}+\tan^{5}\text{x}+1)\text{dx}\ $is

0
2
$\pi$
1

Answer

$\pi$

$\text{Let}\ \text{I}=\int^{\frac{\pi}{2}}\limits_{\frac{-\pi}{2}}\text{(x}^{3}+\text{x}\cos\text{x}+\tan^{5}\text{x}+1)\text{dx}=\int^{\frac{\pi}{2}}\limits_{\frac{-\pi}{2}}\text{x}^{3}\text{dx}+\int^{\frac{\pi}{2}}\limits_{\frac{-\pi}{2}}\text{x}\cos\text{x}\ \text{dx}+\int^{\frac{\pi}{2}}\limits_{\frac{-\pi}{2}}\tan^{5}\text{x}\ \text{dx}+\int^{\frac{\pi}{2}}\limits_{\frac{-\pi}{2}}1\text{dx}$
$\Rightarrow\ \ \text{I}=0+0+0+\text{(x)}^{\frac{\pi}{2}}_{\frac{-\pi}{2}}=\bigg(\frac{-\pi}{2}\bigg)=\pi$

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Question 891 Mark

$\int\limits^\text{e}_1\log\text{x}\text{ dx}=$

1
e - 1
e + 1
0

Answer

Solution:
$\int\limits^\text{e}_1\log\text{x}\text{ dx}$
$=\int\limits^\text{e}_1\log\text{x}\text{ x}^0\text{dx}$
$=\big[\text{x}\log\text{x}\big]^\text{e}_1-\int\limits^\text{e}_1\frac{1}{\text{x}}\text{dx}$
$=\big[\text{x}\log\text{x}\big]^\text{e}_1-\big[\text{x}\big]^\text{e}_1$
$=(\text{e}-0)-(\text{e}-1)$
$= \text{e}-\text{e}+1$
$=1$

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Question 901 Mark

$\int|\text{x}|^3\text{ dx}$ is equal to:

$\frac{-\text{x}^4}{4}+\text{C}$
$\frac{|\text{x}|^4}{4}+\text{C}$
$\frac{\text{x}^4}{4}+\text{C}$
none of these.

Answer

none of these.

Solution:
$\int|\text{x}|^3\text{ dx}$
$|\text{x}|=\begin{cases}\text{x},\text{ x}\geq0\\-\text{x},\text{ x}<0\end{cases}$
Case I:
When $\text{x}\geq0$
$\therefore\ \int|\text{x}|^3\text{ dx}$
$=\int\text{x}^3\text{ dx}$
$=\frac{\text{x}^4}{4}+\text{C}$
Case II:
$\text{x}<0$
$\int|\text{x}|^3\text{ dx}$
$=-\int\text{x}^3\text{ dx}$
$=\frac{-\text{x}^4}{4}+\text{C}$

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Question 911 Mark

$\int\limits^1_{-1}|1-\text{x}|\text{dx}$ is equal to:

-2
2
0
4

Answer

Solution:
$\int\limits^1_{-1}|1-\text{x}|\text{dx}$
$=\int\limits^0_{-1}(1-\text{x})\text{dx}+\int\limits^1_0(1-\text{x})\text{dx}$
$=\Big[\text{x}-\frac{\text{x}^2}{2}\Big]^0_{-1}+\Big[\text{x}-\frac{\text{x}^2}{2}\Big]^1_0$
$=0+1+\frac{1}{2}+1-\frac{1}{2}-0$
$=2$

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Question 921 Mark

Integration of $\frac{1}{1+(\log_\text{e}\text{x})^2}$ with respect to $\log_\text{e}\text{x}$ is:

$\frac{\tan^{-1}(\log_\text{e}\text{x})}{\text{x}}+\text{C}$
$\tan^{-1}(\log_\text{e}\text{x})+\text{C}$
$\frac{\tan^{-1}\text{x}}{\text{x}}+\text{C}$
none of these

Answer

$\tan^{-1}(\log_\text{e}\text{x})+\text{C}$

Solution:
$\frac{1}{1+(\log_\text{e}\text{x})^2}\text{ d}(\log_\text{e}\text{x})$
Put $\log_\text{e}\text{x}=\text{t}$
$\int\frac{\text{dt}}{1+\text{t}^2}=\tan^{-1}\text{t}+\text{C}$
$\tan^{-1}(\log_\text{e}\text{x})+\text{C}$

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Question 931 Mark

Choose the correct option from given four options:
$\int\frac{\text{x}+\sin\text{x}}{1+\cos\text{x}}\text{dx}$ is equal to:

$\log|1+\cos\text{x}|+\text{C}$
$\log|\text{x}+\sin\text{x}|+\text{C}$
$\text{x}-\tan\frac{\text{x}}{2}+\text{C}$
$\text{x}\cdot\tan\frac{\text{x}}{2}+\text{C}$

Answer

$\text{x}\cdot\tan\frac{\text{x}}{2}+\text{C}$

Solution:
$\text{I}=\int\frac{\text{x}+\sin\text{x}}{1+\cos\text{x}}\text{dx}$
$=\int\frac{\text{x}}{1+\cos\text{x}}\text{dx}+\int\frac{\sin\text{x}}{1+\cos\text{x}}\text{dx}$
$=\int\frac{\text{x}}{2\cos^2\frac{\text{x}}{2}}\text{dx}+\int\frac{2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}\text{dx}$
$=\int\text{x}\sec^2\frac{\text{x}}{2}\text{dx}+\int\tan\frac{\text{x}}{2}\text{dx}$
$=\frac{1}{2}\Big[\text{x}\cdot2\tan\frac{\text{x}}{2}-\int2\tan\frac{\text{x}}{2}\text{dx}\Big]+\int\tan\frac{\text{x}}{2}\text{dx}=\text{x}\cdot\tan\frac{\text{x}}{2}+\text{C}$

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Question 941 Mark

If $\int\frac{3\text{x}+4}{\text{x}^3-2\text{x}-4}\text{dx}=\log|\text{x}-2|+\text{k}\log\text{f(x)}+\text{c},$ then:

f(x) = |x² + 2x + 2|
f(x) = x² + 2x + 2
$\text{k}=-\frac{1}{2}$
All of these

Answer

All of these

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Question 951 Mark

What is the value of $\int_{1}^{\text{e}}\frac{1+\log\text{x}}{\text{x}}\text{dx}?$

$\frac{3}{2}$
$\frac{1}{2}$
$\text{e}$
$\frac{1}{\text{e}}$

Answer

$\frac{3}{2}$

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Question 961 Mark

What is the value of $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\text{dx}}{\sin2\text{x}}?$

$\frac{1}{2}\log(-1)$
$\log(-1)$
$\log3$
$\log\sqrt{3}$

Answer

$\log3$

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MCQ 971 Mark

Solve: $\int\frac{\text{x}^2+1}{\text{x}^2+1}\text{dx}=$

A
$1 + C$
B
$x^2 + C$
✓
$x + C$
D
$0$

Answer

Correct option: C.

$x + C$

Now, $\int\frac{\text{x}^2+1}{\text{x}^2+1}\text{dx}$
$=\int\text{dx}$
$= x + C \ [$Where $C$ is integrating constant$]$

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Question 981 Mark

The value of $\int\frac{\sin\text{x}+\cos\text{x}}{\sqrt{1-\sin2\text{x}}}\text{ dx}$ is equal to:

$\sqrt{\sin2\text{x}}+\text{C}$
$\sqrt{\cos2\text{x}}+\text{C}$
$\pm(\sin\text{x}-\cos\text{x})+\text{C}$
$\pm\log(\sin\text{x}-\cos\text{x})+\text{C}$

Answer

$\pm\log(\sin\text{x}-\cos\text{x})+\text{C}$

Solution:
Let $\text{I}=\int\frac{\sin\text{x}+\cos\text{x}}{\sqrt{1-\sin2\text{x}}}\text{ dx}$
$=\int\frac{(\sin\text{x}+\cos\text{x})\text{dx}}{\sqrt{\sin^2\text{x}+\cos^2\text{x}-2\sin\text{x}\cos\text{x}}}$
$=\int\frac{(\sin\text{x}+\cos\text{x})}{\sqrt{(\sin\text{x}-\cos\text{x})^2}}$
$=\int\frac{(\sin\text{x}+\cos\text{x})\text{dx}}{|\sin\text{x}-\cos\text{x}|}$
$=\pm\int\Big(\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}-\cos\text{x}}\Big)\text{dx}$
Let $\sin\text{x}-\cos\text{x}=\text{t}$
$(\cos\text{x}+\sin\text{x})\text{dx}=\text{dt}$
$\therefore\ \text{I}=\pm\int\frac{\text{dt}}{\text{t}}$
$=\ln|\text{t}|+\text{C}$
$=\pm\ln(\sin\text{x}-\cos\text{x})+\text{C}$ $(\because\text{t}=\sin\text{x}-\cos\text{x})$

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Question 991 Mark

Evaluate: $\int\sec^2(7-4\text{x})\text{dx}.$

$-\frac{1}{4}\tan(7-4\text{x})+\text{c}$
$\frac{1}{4}\tan(7-4\text{x})+\text{c}$
$\frac{1}{4}\tan(7+4\text{x})+\text{c}$
$-\frac{1}{4}\tan(7\text{x}-4)+\text{c}$

Answer

$-\frac{1}{4}\tan(7-4\text{x})+\text{c}$

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Question 1001 Mark

$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\sin^9\text{xdx}=$

-1
0
1
None of these

Answer

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MATHS M.C.Q (1 Marks)