3 Marks Question

Question 3013 Marks

Evalute the following integrals:
$\int\frac{\cos2\text{x}+\text{x}+1}{\text{x}^2+\sin2\text{x}+2\text{x}}\text{dx}$

Answer

Let $\text{I}=\int\frac{\cos2\text{x}+\text{x}+1}{\text{x}^2+\sin2\text{x}+2\text{x}}\text{dx}$
Putting $\text{x}^2+\sin2\text{x}+2\text{x}=\text{t}$
$\Rightarrow2\text{x}+2\cos2\text{x}+2=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(\text{x}+\cos2\text{x}+1)\text{dx}=\frac{\text{dt}}{2}$
$\therefore\text{I}=\frac{1}{2}\int\frac{1}{\text{t}}\text{dt}$
$=\frac{1}{2}\text{ln}|\text{t}|+\text{C}$
$=\frac{1}{2}\text{ln}|\text{x}^2+\sin2\text{x}+2\text{x}|+\text{C}$
$\big[\because\text{t}=\text{x}^2+\sin2\text{x}+2\text{x}\big]$

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Question 3023 Marks

Evaluate the following integrals:$\int\sqrt{\text{cosec}\text{x}-1}\text{ dx}$

Answer

$\int\sqrt{\text{cosec}\text{x}-1}\text{ dx}$
$=\int\sqrt{\frac{1}{\sin\text{x}}-1}\text{ dx}$
$=\int\frac{\sqrt{1-\sin\text{x}}}{\sqrt{\sin\text{x}}}\text{ dx}$
$=\int\frac{\sqrt{(1-\sin\text{x})(1+\sin\text{x})}}{\sqrt{\sin\text{x}(1+\sin\text{x})}}\text{ dx}$
$=\int\frac{\sqrt{1-\sin^2\text{x}}}{\sqrt{\sin^2\text{x}+\sin\text{x}}}\text{ dx}$
$=\int\frac{\cos\text{x}\text{ dx}}{\sqrt{\sin^2\text{x}+\sin\text{x}}}$
Let $\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x}\text{ dx}=\text{dt}$
Now, $=\int\frac{\cos\text{x}\text{ dx}}{\sqrt{\sin^2\text{x}+\sin\text{x}}}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}^2+\text{t}}}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}^2+\text{t}+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}}$
$=\int\frac{\text{dt}}{\sqrt{\big(\text{t}+\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}}$
$=\log\Big|\Big(\text{t}+\frac{1}{2}\Big)+\sqrt{\big(\text{t}+\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}\Big|+\text{C}$
$=\log\Big|\text{t}+\frac{1}{2}+\sqrt{\text{t}^2+\text{t}}\Big|+\text{C}$
$=\log\Big|\sin\text{x}+\frac{1}{2}\sqrt{\sin^2\text{x}+\sin\text{x}}\Big|+\text{C}$

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Question 3033 Marks

Evaluate the following integrals:$\int\frac{1}{\sqrt{5-4\text{x}-2\text{x}^2}}\text{ dx}$

Answer

$\int\frac{1}{\sqrt{5-4\text{x}-2\text{x}^2}}$
$=\int\frac{\text{dx}}{\sqrt{2\big[\frac{5}{2}-2\text{x}-\text{x}^2}\big]}$
$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{5}{2}-2\text{x}-\text{x}^2}}$
$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{5}{2}(\text{x}^2+2\text{x})}}$
$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{5}{2}-(\text{x}^2+2\text{x}+1-1)}}$
$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{5}{2}-(\text{x}+1)^2+1}}$
$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{7}{2}-(\text{x}+1)^2}}$
$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\Big(\frac{\sqrt7}{\sqrt3}\Big)^2-(\text{x}+1)^2}}$
$=\frac{1}{\sqrt2}\sin^{-1}\Big(\frac{(\text{x}+1)\sqrt2}{\sqrt7}\Big)+\text{C}$
$=\frac{1}{\sqrt2}\sin^{-1}\Big(\sqrt{\frac{2}{7}}(\text{x}+1)\Big)+\text{C}$

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Question 3043 Marks

$\int\frac{1+\cos4\text{x}}{\cot\text{x}-\tan\text{x}}\text{dx}$

Answer

$\int\Big(\frac{1+\cos4\text{x}}{\cot\text{x}-\tan\text{x}}\Big)\text{dx}$
$=\int\frac{(1+\cos4\text{x})}{\big(\frac{\cos\text{x}}{\sin\text{x}}-\frac{\sin\text{x}}{\cos\text{x}}\big)}\text{dx}$
$=\int\frac{2\cos^22\text{x}\times\sin\text{x}\cos\text{x}}{(\cos^2\text{x}-\sin^2\text{x})}\text{dx}$
$=\int\frac{\cos^22\text{x}\times2\sin\text{x}\cos\text{x}}{\cos2\text{x}}\text{dx}$
$=\int\cos2\text{x}\sin2\text{x}\text{ dx}$
$=\frac{1}{2}\int2\sin2\text{x}\cos2\text{x dx}$
$=\frac{1}{2}\int\sin4\text{x dx}$
$=\frac{1}{2}\Big[-\frac{\cos4\text{x}}{4}\Big]+\text{c}$
$=-\frac{1}{8}\cos4\text{x}+\text{c}$

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Question 3053 Marks

Evaluate the following integrals:$\int\frac{\log(\text{x}+2)}{(\text{x}+2)^2}\text{dx}$

Answer

Let $\text{I}=\int\frac{\log(\text{x}+2)}{(\text{x}+2)^2}\text{dx}$
Let $\frac{1}{\text{x}+2}=\text{t}$
$-\frac{1}{(\text{x}+2)^2}\text{dx = dt}$
$\text{I}=-\int\log\big(\frac{1}{\text{t}}\big)\text{dt}$
$=-\int\log\text{t}^{-1}\text{dt}$
$=-\int1\times\log\text{t dt}$
Using integration by parts,
$\text{I}=\log\text{t}\int\text{dt}-\int\big(\frac{1}{\text{t}}\int\text{dt}\big)\text{dt}$
$=\text{t}\log\text{t}-\int\Big(\frac{1}{\text{t}}\times\text{t}\Big)\text{dt}$
$=\text{t}\log\text{t}-\int\text{dt}$
$=\text{t}\log\text{t}-\text{t+C}$
$=\frac{1}{\text{x}+2}\big(\log(\text{x}+2)^{-1}-1\big)+\text{C}$
$\text{I}=\frac{-1}{\text{x}+2}-\frac{\log(\text{x}+2)}{\text{x}+2}+\text{C}$

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Question 3063 Marks

Evaluate the following integrals:$\int\frac{\text{x}}{\sqrt{\text{x}^4+\text{a}^4}}\text{ dx}$

Answer

$\int\frac{\text{x}\text{ dx}}{\sqrt{\text{x}^4+\text{a}^4}}$ $\int\frac{\text{x}\text{ dx}}{\sqrt{(\text{x}^2)^2+(\text{a}^2)^2}}$ Let $\text{x}^2=\text{t}$ $\Rightarrow2\text{x}\text{ dx}=\text{dt}$ $\Rightarrow\text{x}\text{ dx}=\frac{\text{dt}}{2}$Now, $\int\frac{\text{x}\text{ dx}}{\sqrt{(\text{x}^2)^2+(\text{a}^2)^2}}$
$=\frac{1}{2}\int\frac{\text{x}\text{ dx}}{\sqrt{{\text{t}^2+(\text{a}^2)^2}}}$
$=\frac{1}{2}\log\Big|\text{t}+\sqrt{\text{t}^2+\text{a}^4}\Big|+\text{C}$
$=\frac{1}{2}\log\Big|\text{x}^2+\sqrt{\text{x}^4+\text{a}^4}\Big|+\text{C}$

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Question 3073 Marks

Evaluate the following integrals:
$\int\frac{1}{\text{x}^2}\cos^2\Big(\frac{1}{\text{x}}\Big)\text{dx}$

Answer

$\int\frac{1}{\text{x}^2}\cos^2\Big(\frac{1}{\text{x}}\Big)\text{dx}$
Let $\frac{1}{\text{x}}=\text{t}$
$\Rightarrow-\frac{1}{\text{x}^2}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{\text{x}^2}\text{ dx}=-\text{dt}$
Now, $\int\frac{1}{\text{x}^2}\cos^2\Big(\frac{1}{\text{x}}\Big)\text{dx}$
$=-\int\cos^2\text{t}\ \text{dt}$
$=-\int\Big(\frac{1+\cos2\text{t}}{2}\Big)\text{dt}$
$=-\frac{1}{2}\int(1+\cos2\text{t})\text{dt}$
$=-\frac{1}{2}\Big[\text{t}+\frac{\sin2\text{t}}{2}\Big]+\text{C}$
$=-\frac{1}{2}\Bigg[\frac{1}{\text{x}}+\frac{\sin\big(\frac{2}{\text{x}}\big)}{2}\Bigg]+\text{C}$
$=-\frac{1}{2}\Big(\frac{1}{\text{x}}\Big)-\frac{1}{4}\sin\Big(\frac{2}{\text{x}}\Big)+\text{C}$

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Question 3083 Marks

Evaluate the following integrals:$\int\sin^{-1}(3\text{x}-4\text{x}^3)\text{dx}$

Answer

Let $\text{I}=\int\sin^{-1}(3\text{x}-4\text{x}^3)\text{dx}$
Let $\text{x}=\sin\theta$
$\text{dx}=\cos\theta \text{d}\theta$
$=\int\sin^{-1}(3\sin\theta-4\sin^3\theta)\cos\theta\text{d}\theta$
$=\int\sin^{-1}(\sin3\theta)\cos\theta\text{d}\theta$
$=\int3\theta\cos\theta\text{d}\theta$
$=3[\theta\int\cos\theta\text{d}\theta-\int(1\int\cos\theta\text{d}\theta)\text{d}\theta]$
$=3[\theta\sin\theta-\int\sin\theta\text{d}\theta]$
$=3[\theta\sin\theta+\cos\theta]+\text{C}$
$\text{I}=3\Big[\text{x}\sin^{-1}\text{x}+\sqrt{1-\text{x}^2}\Big]+\text{C}$

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Question 3093 Marks

Evaluate the following intregals:
$\int\frac{\text{x}}{\sqrt{\text{x}^2+\text{x}+1}}\text{dx}$

Answer

Let $\text{I}=\int\frac{\text{x}}{\sqrt{\text{x}^2+\text{x}+1}}\text{dx}$
Consider,
$\text{X}=\text{A}\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{x}+1)+\text{B}$
$\text{x}=\text{A}(2\text{x}+1)+\text{B}$
$\Rightarrow\text{x}=(2\text{A})\text{x}+\text{A}+\text{B}$
Equating coefficient of like terms
$2\text{A}=1$
$\Rightarrow\text{A}=\frac{1}{2}$
And
$\text{A}+\text{B}=0$
$\Rightarrow\frac{1}{2}+\text{B}=0$
$\Rightarrow\text{B}=-\frac{1}{2}$
$\therefore\text{I}=\int\frac{\big(\frac{1}{2}(2\text{x}+1)-\frac{1}{2}\big)}{\sqrt{\text{x}^2+\text{x}+1}}\text{dx}$

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MATHS 3 Marks Question