3 Marks Question

Question 2513 Marks

Evaluate the following integrals:
$\int\frac{\text{x}^2}{\text{x}^6+\text{a}^6}\text{dx}$

Answer

$\int\frac{\text{x}^2}{\text{x}^6+\text{a}^6}\text{dx}$
$\Rightarrow\int\frac{\text{x}^2\text{dx}}{(\text{x}^3)^2+(\text{a}^3)^2}$
Let $\text{x}^3=\text{t}$
$\Rightarrow3\text{x}^3\text{dx = dt}$
$\Rightarrow\text{x}^2\text{dx}=\frac{\text{dt}}{3}$
Now $\int\frac{\text{x}^2}{\text{x}^6+\text{a}^6}\text{dx}$
$=\frac{1}{3}\int\frac{\text{dt}}{\text{t}^2+(a^3)^2}$
$=\frac{1}{3\text{a}^3}\tan^{-1}\Big(\frac{\text{t}}{\text{a}^3}\Big)+\text{C}$
$=\frac{1}{3\text{a}^3}\tan^{-1}\Big(\frac{\text{x}^3}{\text{a}^3}\Big)+\text{C}$

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Question 2523 Marks

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

Answer

$\int\sqrt{\text{e}^\text{x}-1}\text{ dx}$ Let $\text{I}=\int\frac{1}{(\text{x}+1)(\text{x}^2+2\text{x}+2)}\text{ dx}$ $=\int\frac{1}{(\text{x}+1)\big((\text{x}+1)^2+\text{1}\big)}\text{ dx}$Let $\text{x}+1=\tan\text{u}$
$\Rightarrow\text{dx}=\sec^2\text{u du}$ $\therefore\ \text{I}=\int\frac{\sec^2\text{u}}{\tan\text{u}(\tan^2\text{u}+1)}\text{ du}$ $=\int\frac{\cos\text{u}}{\sin\text{u}}\text{ du}$ $=\log|\sin\text{u}|+\text{C}$ $=\log\Big|\frac{\tan\text{u}}{\sec^2\text{u}}\Big|+\text{C}$ $=\log\bigg|\frac{\text{x}+1}{\sqrt{\text{x}^2+2\text{x}+2}}\bigg|+\text{C}$

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Question 2533 Marks

Evaluate the following integrals:
$\int\frac{1}{\sqrt{1-\text{x}^2}(\sin^{-1}\text{x})^2}\text{dx}$

Answer

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

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Question 2543 Marks

Evaluate the following integrals:
$\int\frac{\cos\text{x}}{1+\cos\text{x}}\text{dx}$

Answer

$\int\frac{\cos\text{x}}{1+\cos\text{x}}\text{dx}$
$=\int\frac{\cos\text{x}(1-\cos\text{x})}{(1+\cos\text{x})(1-\cos\text{x})}\text{dx}$
$=\int\frac{\cos\text{x}-\cos^2\text{x}}{1-\cos^2\text{x}}\text{dx}$
$=\int\frac{\cos\text{x}-\cos^2\text{x}}{\sin^2\text{x}}\text{dx}$
$=\int\frac{\cos\text{x}}{\sin^2\text{x}}-\frac{\cos^2\text{x}}{\sin^2\text{x}}\text{dx}$
$=\int(\cot\text{x}\text{ cosec x}-\cot^2\text{x})\text{dx}$
$=\int(\cot\text{x}\text{ cosec x}-\text{cosec}^2\text{x}+1)\text{dx}$
$=\int\cot\text{x}\text{ cosec x dx}-\int\text{cosec}^2\text{x dx}+\int1\text{dx}$
$=-\text{cosec x}+\cot\text{x}+\text{x}+\text{C}$

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Question 2553 Marks

$\int\frac{1}{\sqrt{\text{x}+1}+\sqrt{\text{x}}}\text{dx}$

Answer

Let $\text{I}=\int\frac{1}{\sqrt{\text{x}+1}+\sqrt{\text{x}}}\text{dx}.$ Then,
$\text{I}=\int\frac{1}{\sqrt{\text{x}+1}+\sqrt{\text{x}}}\times\frac{\sqrt{\text{x}+1}-\sqrt{\text{x}}}{\sqrt{\text{x}+1}-\sqrt{\text{x}}}\times\text{dx}$
$=\int\frac{\sqrt{\text{x+1}}-\sqrt{\text{x}}}{(\sqrt{\text{x}+1})^2-(\sqrt{\text{x}})^2}\times\text{dx}$
$=\int\frac{\sqrt{\text{x}+1}-\sqrt{\text{x}}}{\text{x}+1-\text{x}}\times\text{dx}$
$=\int(\sqrt{\text{x}+1}-\sqrt{\text{x}})\times\text{dx}$
$=\int(\text{x}+1)^{\frac{1}{2}}\text{dx}-\int\text{x}^{\frac{1}{2}}\text{dx}$
$=\frac{2}{3}(\text{x}+1)^{\frac{3}{2}}-\frac{2}{3}\text{x}^{\frac{3}{2}}+\text{c}$
$\therefore\text{I}=\frac{2}{3}(\text{x}+1)^{\frac{3}{2}}-\frac{2}{3}\text{x}^{\frac{3}{2}}+\text{c}$.

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Question 2563 Marks

Evaluate the following integrals:
$\int\frac{1+\cos\text{x}}{(\text{x}+\sin\text{x})^3}\text{dx}$

Answer

$\int\frac{(1+\cos\text{x})}{(\text{x}+\sin\text{x})^3}\text{dx}$
$\text{Let x}+\sin\text{x}=\text{t}$
$\Rightarrow(1+\cos\text{x})=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(1+\cos\text{x})\text{dx}={\text{dt}}$
$\text{Now,}\int\frac{(1+\cos\text{x})}{(\text{x}+\sin\text{x})^3}\text{dx}$
$=\int\frac{\text{dt}}{\text{t}^3}$
$=\int\text{t}^{-3}\text{dt}$
$=\frac{\text{t}^{-3+1}}{-3+1}+\text{C}$
$=\frac{-1}{2\text{t}^2}+\text{C}$
$=\frac{-1}{2(\text{x}+\sin\text{x})^2}+\text{C}$

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MATHS 3 Marks Question