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Maths 5 Marks Questions

Indefinite Integrals · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 321 questions.

Questions · Page 7 of 7

5 Marks Questions

Question 3015 Marks
$\int\sin\text{x}\sqrt{1-\cos2\text{x}}\text{ dx}$
Answer
Let I $=\int\sin\text{x}\sqrt{1-\cos2\text{x}}\text{ dx}.$ Then,
$\text{I}=\int\sin\text{x}\times\sqrt{2\sin^2\text{x}}\times\text{dx}$
$=\int\sin\text{x}\times\sqrt{2}\times\sin\text{x dx}$
$=\sqrt{2}\int\sin^2\text{x}\times\text{dx}$
$=\frac{\sqrt{2}}{2}\int2\sin^2\text{x}\times\text{dx}$
$=\frac{\sqrt{2}}{2}\Big[\text{x}-\frac{\sin2\text{x}}{2}\Big]+\text{C}$
$=\frac{\sqrt{2}\text{x}}{2}-\frac{\sqrt{2}}{4}\times\sin2\text{x}+\text{C}$
$=\frac{1}{\sqrt{2}}\times\text{x}-\frac{\sin2\text{x}}{2\sqrt{2}}+\text{C}$
$\therefore\text{I}=\frac{1}{\sqrt{2}}\times\text{x}-\frac{\sin2\text{x}}{2\sqrt{2}}+\text{C}$
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Question 3025 Marks
Evaluate the following integrals:
$\int\frac{\text{e}^{\text{m}\sin^{-1}\text{x}}}{\sqrt{1-\text{x}^2}}\text{ dx}$
Answer
Let $\text{I}=\int\frac{\text{e}^{\text{m}\sin^{-1}\text{x}}}{\sqrt{1-\text{x}^2}}\text{ dx}\ ....(1)$ Let $\text{m}\sin^{-1}\text{x}=\text{t}$ then, $\text{d}\big(\text{m}\sin^{-1}\text{x}\big)=\text{dt}$ $\Rightarrow\text{m}\frac{1}{\sqrt{1-\text{x}^2}}\text{ dx}=\text{dt}$ $\Rightarrow\frac{\text{dx}}{\sqrt{1-\text{x}^2}}=\frac{\text{dt}}{\text{m}}$Putting, $\text{m}\sin^{-1}\text{x}=\text{t}$ and $\frac{\text{dx}}{\sqrt{1-\text{x}^2}}=\frac{\text{dt}}{\text{m}}$ in equation (1),
We get,
$\text{I}=\int\text{e}^\text{t}\frac{\text{dt}}{\text{m}}$
$=\frac{1}{\text{m}}\text{e}^\text{t}+\text{C}$
$=\frac{1}{\text{m}}\text{e}^{\text{m}\sin^{-1}\text{x}}+\text{C}$
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Question 3035 Marks
Evaluate the following intregals: $\int\frac{1}{\sqrt{3}\sin\text{x}+\cos\text{x}}\ \text{dx}$
Answer
Let $\text{I}=\int\frac{1}{\sqrt{3}\sin\text{x}+\cos\text{x}}\ \text{dx}$
Let $\sqrt{3}=\text{r}\cos\theta,\text{and }1=\text{r}\sin\theta$
$\tan\theta=\frac{1}{\sqrt{3}}$
$\theta=\frac{\pi}{6}$
$\text{r}=\sqrt{3+1}=2$
$\text{I}=\int\frac{1}{\text{r}\cos\theta\sin\text{x}+\text{r}\sin\theta\cos\text{x}}\ \text{dx}$
$=\frac{1}{\text{r}}\int\frac{1}{\sin(\text{x}+\theta)}\text{dx}$
$=\frac{1}{2}\int\text{cosec}(\text{x}+\theta)\text{dx}$
$=\frac{1}{2}\log\Big|\tan\Big(\frac{\text{x}}{2}+\frac{\theta}{2}\Big)\Big|+\text{c}$
$\text{I}=\frac{1}{2}\log\Big|\tan\Big(\frac{\text{x}}{2}+\frac{\pi}{12}\Big)\Big|+\text{C}$
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Question 3045 Marks
Evaluate the following integrals:
$\int\text{e}^{2\text{x}}\Big(\frac{1-\sin2\text{x}}{1-\cos2\text{x}}\Big)\text{dx}$
Answer
WE have,
$\text{I}=\int\text{e}^{2\text{x}}\Big(\frac{1-\sin2\text{x}}{1-\cos2\text{x}}\Big)\text{dx}$
$=\int\text{e}^{2\text{x}}\Big(\frac{1-2\sin\text{x}\cos\text{x}}{2\sin^2\text{x}}\Big)\text{dx}$
Put $\text{t}=2\text{x}.$ Then $\text{dt}=2\text{dx}$
Therefore,
$\text{I}=\frac{1}{2}\int\text{e}^{\text{t}}\bigg(\frac{1-2\sin\frac{\text{t}}{2}\cos\frac{\text{t}}{2}}{2\sin^2\frac{\text{t}}{2}}\bigg)\text{dt}$
$=\frac{1}{4}\int\text{e}^{\text{t}}\bigg(\frac{1-2\sin\frac{\text{t}}{2}\cos\frac{\text{t}}{2}}{\sin^2\frac{\text{t}}{2}}\bigg)\text{dt}$
$=\frac{1}4{\int\text{e}^{\text{t}}}\bigg(\frac{1}{\sin^2\frac{\text{t}}{2}}-\frac{2\sin\frac{\text{t}}{2}\cos\frac{\text{t}}{2}}{\sin^2\frac{\text{t}}{2}}\bigg)\text{dt}$
$=\frac{1}{4}\int\text{e}^{\text{t}}\big(\text{cosec}^2\frac{\text{t}}{2}-2\cot\frac{\text{t}}{2}\big)\text{dt}$
$=-\frac{1}{4}\int\text{e}^{\text{t}}\big(2\cot\frac{\text{t}}{2}-\text{cosec}^2\frac{\text{t}}{2}\big)\text{dt}$
Consider, $\text{f(x)}=2\cot\frac{\text{t}}{2},$ then $\text{f}'\text{(x)}=-\text{cosec}^2\frac{\text{t}}{2}$
Thus, the given integrand is of the from $\text{e}^{\text{x}}\big[\text{f(x)}+\text{f'}\text{(x)}\big].$
Therefore, $\text{I}=-\frac{1}{4}\big(2\cot\frac{\text{t}}{2}\big)\text{e}^{\text{t}}+\text{C}$
$=-\frac{1}{4}\big(2\cot\frac{2\text{x}}{2}\big)\text{e}^{2\text{x}}+\text{C}$
Hence, $\int\text{e}^{2\text{x}}\Big(\frac{1-\sin2\text{x}}{1-\cos2\text{x}}\Big)\text{dx}=-\frac{1}{2}(\cot\text{x})\text{e}^{2\text{x}}+\text{C}$
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Question 3055 Marks
Evaluate the following integrals:
$\int\text{x}^3\tan^{-1}\text{x dx}$
Answer
$\int\text{x}^3\tan^{-1}\text{x dx}$
$\int\text{x}^3\tan^{-1}\text{x dx}=\tan^{-1}\text{x}\int\text{x}^3\text{dx}-\Big(\int\frac{\text{d}\tan^{-1}\text{x}}{\text{dx}}\big(\int\text{x}^3\text{dx}\big)\text{dx}\Big)$
$=\tan^{-1}\text{x}\frac{\text{x}^4}{4}-\Big(\int\frac{1}{1+\text{x}^2}\Big(\frac{\text{x}^4}{4}\Big)\text{dx}\Big)$
$=\tan^{-1}\text{x}\frac{\text{x}^4}{4}-\Big(\int\frac{1}{1+\text{x}^2}\Big(\frac{\text{x}^4}{4}\Big)\text{dx}\Big)$
$\int\frac{1}{1+\text{x}^2}\Big(\frac{\text{x}^4}{4}\Big)\text{dx}=\frac{1}{4}\Big[\int\frac{1}{1+\text{x}^2}\text{dx}+(\text{x}^2-1)\text{dx}\Big]$
$\int\frac{1}{1+\text{x}^2}\Big(\frac{\text{x}^4}{4}\Big)\text{dx}=\frac{1}{4}\Big[\tan^{-1}\text{x}+\frac{\text{x}^3}{3}-\text{x}\Big]$
$\int\text{x}^3\tan^{-1}\text{x dx}=\frac{\text{x}^4}{4}\tan^{-1}\text{x}-\frac{1}{4}\Big[\tan^{-1}\text{x}+\frac{\text{x}^3}{3}-\text{x}\Big]+\text{C}$
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Question 3065 Marks
Evaluate the following integrals:
$\int\frac{1}{4\text{x}^2+12\text{x}+5}\text{dx}$
Answer
Let $\text{I}=\int\frac{1}{4\text{x}^2+12\text{x}+5}\text{dx}$
$=\frac{1}{4}\int\frac{1}{\text{x}^2+3\text{x}+\frac{5}{4}}\text{dx}$
$=\frac{1}4{}\int\frac{1}{\text{x}^2+2\times\text{x}\times\big(\frac{3}{2}\big)+\big(\frac{3}{2}\big)^2-\big(\frac{3}{2}\big)^2+\frac{5}{4}}\text{dx}$
$\text{I}=\frac{1}{4}\int\frac{1}{\Big(\text{x}+\frac{3}{2}\Big)^2-1}\text{dx}$
Let $\Big(\text{x}+\frac{3}{2}\Big)=\text{t}\ \dots(1)$
$\Rightarrow\text{dx = dt}$
So,
$\text{I}=\frac{1}{4}\int\frac{1}{\text{t}^2-(1)^2}\text{dt}$
$\text{I}=\frac{1}{4}\times\frac{1}{2\times(1)}\log\bigg|\frac{\text{t}-1}{\text{t}+1}\bigg|+\text{C}$ $\Big[\text{Since,} \int\frac{1}{\text{x}^2-\text{a}^2}\text{dx}=\frac{1}{2\text{a}}\log\bigg|\frac{\text{x}-\text{a}}{\text{x+a}}\bigg|+\text{C}\Big]$
$\text{I}=\frac{1}{8}\log\Bigg|\frac{\text{x}+\frac{3}{2}-1}{\text{x}+\frac{3}{2}+1}\Bigg|+\text{C}$ [using (1)]
$\text{I}=\frac{1}{8}\log\bigg|\frac{2\text{x}+1}{2\text{x}+5}\bigg|+\text{C}$
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Question 3075 Marks
Evaluate the following integrals:
$\int{\frac{\text{e}^{\text{x}}}{\text{e}^{2\text{x}}+5\text{e}^{\text{x}}+6}}\text{dx}$
Answer
$\int\frac{\text{e}^\text{x}\text{dx}}{\text{e}^{2\text{x}}+5\text{e}^{\text{x}}+6}$
Let $\text{e}^{\text{x}}=\text{t}$
$\Rightarrow \text{e}^{\text{x} }\text{dx = dt}$
Now, $\int\frac{\text{e}^{\text{x}}\text{ dx}}{\text{e}^{2\text{x}}+5\text{e}^{\text{x}}+6}$
$=\int\frac{\text{dt}}{\text{t}^2+5\text{t}+6}$
$=\int\frac{\text{dt}}{\text{t}^2+5\text{t}+\big(\frac{5}{2}\big)^2-\big(\frac{5}{2}\big)^2+6}$
$=\int\frac{\text{dt}}{\big(\text{t}+\frac{5}{2}\big)^2-\frac{25}{4}+6}$
$=\int\frac{\text{dt}}{\big(\text{t}+\frac{5}{2}\big)^2-\frac{25+24}{4}}$
$=\int\frac{\text{dt}}{\big(\text{t}+\frac{5}{2}\big)^2-\big(\frac{1}{2}\big)^2}$
$=\frac{1}{2\times\frac{1}{2}}\log\Bigg|\frac{\text{t}+\frac{5}{2}-\frac{1}{2}}{\text{t}+\frac{5}{2}+\frac{1}{2}}\Bigg|+\text{C}$
$=\log\bigg|\frac{\text{t}+2}{\text{t}+3}\bigg|+\text{C}$
$=\log\bigg|\frac{\text{e}^\text{x}+2}{\text{e}^\text{x}+3}\bigg|+\text{C}$
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Question 3085 Marks
Evaluate the following integrals:$\int\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)\text{dx}$
Answer
Let $\text{I}=\int\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)\text{dx}$
Let $\text{x}=\tan\text{t}$
$\text{dx}=\sec^2\text{t dt}$
$\text{I}=\int\cos^{-1}\Big(\frac{1-\tan^2\text{t}}{1+\tan^2\text{t}}\Big)\sec^2\text{t dt}$
$=\int\cos^{-1}(\cos2\text{t})\sec^2\text{t dt}$
$=\int2\text{t}\sec^2\text{x dx}$
$=2\Big[\text{t}\int\sec^2\text{t dt}-\int(1\int\sec^2\text{t dt})\text{dt}\Big]$
$=2[\text{t}\tan^2\text{t}-\int\tan\text{t dt}]$
$=2[\text{t}\tan^2\text{t}-\log\sec\text{t}]+\text{C}$
$=2\Big[\text{x}\tan^{-1}\text{x}-\log\sqrt{1+\text{x}^2}\Big]+\text{C}$
$\text{I}=2\text{x}\tan^{-1}\text{x}-\log|1+\text{x}^2|+\text{C}$
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Question 3095 Marks
Evaluate the following integrals:
$\int(\tan^{-1}\text{x}^2)\text{x dx}$
Answer
Let $\text{I}=\int(\tan^{-1}\text{x}^2)\text{x dx}$
Let $\text{x}^2=\text{t}$
$2\text{x dx = dt}$
$\text{I}=\frac{1}{2}\int\tan^{-1}\text{t dt}$
$=\frac{1}{2}\int1\tan^{-1}\text{t dt}$
$=\frac{1}{2}\Big[\tan^{-1}\text{t}\int\text{dt}-\Big(\int\frac{1}{1+\text{t}^2}\int\text{dt}\Big)\text{dt}\Big]$
$=\frac{1}{2}\Big[\text{t}\tan^{-1}\text{t}-\int\frac{\text{t}}{1+\text{t}^2}\text{dt}\Big]$
$=\frac{1}{2}\text{t}\tan^{-1}\text{t}-\frac{1}{4}\int\frac{2\text{t}}{1+\text{t}^2}\text{dt}$
$=\frac{1}{2}\text{t}\tan^{-1}\text{t}-\frac{1}{4}\log\big|1+\text{t}^2\big|+\text{C}$
$\text{I}=\frac{1}{2}\text{x}^2\tan^{-1}\text{x}^2-\frac{1}{4}\log\big|1+\text{x}^4\big|+\text{C}$
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Question 3105 Marks
Evaluvate the following intregals:
$\int\frac{1}{\text{x}(\text{x}-2)(\text{x}-4)}\ \text{dx}$
Answer
Let $\int\frac{1}{\text{x}(\text{x}-2)(\text{x}-4)}\ \text{dx}=\frac{\text{A}}{\text{x}}+\frac{\text{B}}{\text{x}-2}+\frac{\text{C}}{\text{x}-4}$ $\Rightarrow1=\text{A}(\text{x}-2)(\text{x}-4)+\text{B}(\text{x})(\text{x}-4)+\text{Cx}(\text{x}-2)$Put x = 0
$\Rightarrow1=8\text{A}\Rightarrow\text{A}=\frac{1}{8}$ Put x = 2 $\Rightarrow1=-4\text{B}\Rightarrow\text{B}=-\frac{1}{4}$ Put x = 4 $\Rightarrow1=8\text{C}\Rightarrow\text{C}=\frac{1}{8}$ So, $\int\frac{1}{\text{x}(\text{x}-2)(\text{x}-4)}\ \text{dx}=\frac{1}{8}\int\frac{\text{dx}}{\text{x}}+\Big(-\frac{1}{4}\Big)\int\frac{\text{dx}}{\text{x}-2}+\frac{1}{8}\int\frac{\text{dx}}{\text{x}-4}$ $=\frac{1}{8}\log|\text{x}|-\frac{1}{4}\log|\text{x}-2|+\frac{1}{8}\log|\text{x}-4|+\text{C}$ $=\frac{1}{8}\log\Big|\frac{\text{x}(\text{x}-4)}{(\text{x}-2)^2}\Big|+\text{C}$ $\text{I}=\frac{1}{8}\log\Big|\frac{\text{x}(\text{x}-4)}{(\text{x}-2)^2}\Big|+\text{C}$
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Question 3115 Marks
$\int\frac{\text{x}-1}{\sqrt{\text{x}+4}}\ \text{dx}$
Answer
$\text{Let I} =\int\Big(\frac{\text{x}-1}{\sqrt{\text{x}+4}}\Big)\text{dx}$
Putting x + 4 = t
Then, x = t - 4
Difference both sides
dx = dt
Now integral becomes,
$\text{I}=\int\Big(\frac{\text{t}-4-1}{\sqrt{\text{t}}}\Big)\text{dt}$
$=\int\Big(\frac{\text{t}}{\sqrt{\text{t}}}-\frac{5}{\sqrt{\text{t}}}\Big)\text{dt}$
$=\int\Big(\text{t}^{\frac{1}{2}}-5\text{t}^{-\frac{1}{2}}\Big)\text{dt}$
$=\frac{\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}-5\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\text{C}$
$=\frac{2}{3}\text{t}^\frac{3}{2}-10\sqrt{\text{t}}+\text{C}$
$=\frac{2}{3}(\text{x}+4)^\frac{3}{2}-10(\text{x}+4)^\frac{1}{2}+\text{C}$
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Question 3125 Marks
Evaluate the following integrals:
$\int\text{x}\sqrt{\text{x}^4+1}\text{dx}$
Answer
$\text{I}=\int\text{x}\sqrt{\text{x}^4+1}\text{dx}$
$=\int\text{x}\sqrt{(\text{x}^2)^2+1}\text{dx}$
Putting $\text{x}^2=\text{t}$
$\Rightarrow2\text{x dx}=\text{dt}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$
$\therefore\ \text{I}=\frac{1}{2}\int\sqrt{\text{t}^2+1}\text{dt}$
$=\frac{1}{2}\int\sqrt{\text{t}^2+1}\text{dt}$
$=\frac{1}{2}\Big[\frac{\text{t}}{2}\sqrt{\text{t}^2+1}+\frac{1^2}{2}\log\big|\text{t}+\sqrt{\text{t}^2+1}\big|\Big]+\text{C}$
$=\frac{1}{2}\Big[\frac{\text{x}^2}{2}\sqrt{\text{x}^4+1}+\frac{1}{2}\log\big|\text{x}^2+\sqrt{\text{x}^4+1}\big|\Big]+\text{C}$
$=\frac{\text{x}^2}{4}\sqrt{\text{x}^4+1}+\frac{1}{4}\log\big|\text{x}^2+\sqrt{\text{x}^4+1}+\text{C}$
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Question 3135 Marks
Evaluate the following integrals:
$\int\frac{\cot\text{x}}{\sqrt{\sin\text{x}}}\text{dx}$
Answer
Let I $=\int\frac{\cot\text{x}}{\sqrt{\sin\text{x}}}\text{dx}\ .....(1)$
Let $\sin\text{x}=\text{t}$ then,
$\text{d}(\sin\text{x})=\text{dt}$
$\Rightarrow\cos\text{x}\text{ dx}=\text{dt}$
$\text{Now,}\text{I}=\int\frac{\cot\text{x}}{\sqrt{\sin\text{x}}}\text{dx}$
$=\int\frac{\cot\text{x}}{\sin\text{x}\sqrt{\sin\text{x}}}\text{dx}$
$\int\frac{\cos\text{x}}{(\sin\text{x})^{\frac{3}{2}}}\text{dx}$
$\Rightarrow\ =\int\frac{\cos\text{x}}{(\sin\text{x})^\frac{3}{2}}\text{dx}\ ...(2)$
Putting $\sin\text{x}=\text{t}$ and $\cos\text{x}\text{ dx}=\text{dt}$ in equation (2), we get
$\text{I}=\int\frac{\text{dt}}{\text{t}^\frac{3}{2}}$
$=\int\text{t}^{-\frac{3}{2}}\text{dt}$
$=-2\text{t}^{-\frac{1}{2}}+\text{C}$
$=\frac{-2}{\sqrt{\text{t}}}+\text{C}$
$=\frac{-2}{\sqrt{\sin\text{x}}}+\text{C}$
$\therefore\text{I}=\frac{-2}{\sqrt{\sin\text{x}}}+\text{C}$
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Question 3145 Marks
Evaluate the following integrals:$\int\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\text{dx}$
Answer
Let $\text{x}=\tan\theta\Rightarrow\text{dx}=\sec^2\theta\text{d}\theta$
$\therefore\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)=\sin^{-1}\Big(\frac{2\tan\theta}{1+\tan^2\theta}\Big)=\sin^{-1}(\sin2\theta)=\theta$
$\Rightarrow\int\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\text{dx}=\int2\theta\cdot\sec^2\theta\text{d}\theta=2\int\theta\cdot\sec^2\theta\text{d}\theta$
Integrating by parts, we obtain
$2\Big[\theta\cdot\int\sec^2\theta\text{d}\theta-\int\Big\{\Big(\frac{\text{d}}{\text{d}\theta}\theta\Big)\int\sec^2\theta\text{d}\theta\Big\}\text{d}\theta\Big]$
$=2\big[\theta\cdot\tan\theta-\int\tan\theta\text{d}\theta\big]$
$=2\big[\theta\tan\theta+\log|\cos\theta|\big]+\text{C}$
$=2\Big[\text{x}\tan^{-1}\text{x}+\log\Big|\frac{1}{\sqrt{1+\text{x}^2}}\Big|\Big]+\text{C}$
$=2\text{x}\tan^{-1}\text{x}+2\log(1+\text{x}^2)^{-\frac{1}{2}}+\text{C}$
$=2\text{x}\tan^{-1}\text{x}+2\Big[-\frac{1}{2}\log(1+\text{x}^2)\Big]+\text{C}$
$=2\text{x}\tan^{-1}\text{x}-\log(1+\text{x}^2)+\text{C}$
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Question 3155 Marks
Evaluate the following intregals:
$\int\frac{1}{4\sin^2\text{x}+5\cos^2\text{x}}\text{ dx}$
Answer
Let $\text{I}=\int\frac{1}{4\sin^2\text{x}+5\cos^2\text{x}}\text{ dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{\frac{1}{\cos^2\text{x}}}{4\tan^2\text{x}+5}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{4\tan^2\text{x}+5}\ \text{dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x}\ \text{dx}=\text{dt}$
$\text{I}=\int\frac{\text{dt}}{4+9(\text{t})^2}$
$=\int\frac{\text{dt}}{4\text{t}^2+5}$
Let $2\text{t}=\text{u}$
$2\text{dt}=\text{du}$
$\text{I}=\frac{1}{2}\int\frac{\text{du}}{(4)^2+(\sqrt{5})^2}$
$=\frac{1}{2}\times\frac{1}{\sqrt{5}}\times\tan^{-1}\Big(\frac{\text{u}}{\sqrt{5}}\Big)+\text{C}$
$=\frac{1}{2\sqrt{5}}\tan^{-1}\Big(\frac{2\text{t}}{\sqrt{5}}\Big)+\text{C}$
$\text{I}=\frac{1}{2\sqrt{5}}\tan^{-1}\Big(\frac{2\tan\text{x}}{\sqrt{5}}\Big)+\text{C}$
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Question 3165 Marks
Evaluate the following integrals:
$\int\tan^{-1}(\sqrt{\text{x}})\text{dx}$
Answer
Let $\text{I}=\int\tan^{-1}(\sqrt{\text{x}})\text{dx}$Let $\text{x}=\text{t}^2$
$\text{dx}=2\text{t dt}$
$\text{I}=\int2\text{t}\tan^{-1}\text{t dt}$
$=2\Big[\tan^{-1}\text{t}\int\text{t dt}-\int\Big(\frac{1}{1+\text{t}^2}\int\text{t dt}\Big)\text{dt}\Big]$
$=2\Big[\frac{\text{t}^2}{2}\tan^{-1}\text{t}-\int\frac{\text{t}^2}{2(1+\text{t}^2)}\text{dt}\Big]$
$=\text{t}^2\tan^{-1}\text{t}-\int\frac{\text{t}^2+1-1}{1+\text{t}^2}\text{dt}$
$=\text{t}^2\tan^{-1}\text{t}-\int\Big(1-\frac{1}{1+\text{t}^2}\Big)\text{dt}$
$=\text{t}^2\tan^{-1}\text{t}-\text{t}+\tan^{-1}\text{t + C}$
$=(\text{t}^2+1)\tan^{-1}\text{t}-\text{t + C}$
$\text{I}=(\text{x}+1)\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\text{C}$
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Question 3175 Marks
Evaluate the following integrals:
$\int\frac{1}{(\text{x}^2-1)\sqrt{\text{x}^2+1}}\text{ dx}$
Answer
Let $\text{I}=\int\frac{1}{(\text{x}^2-1)\sqrt{\text{x}^2+1}}\text{ dx}$
Let $\text{x}=\frac{1}{\text{t}}$
$\text{dx}=-\frac{1}{\text{t}^2}\text{ dt}$
$\therefore\ \text{I}=-\int\frac{\frac{1}{\text{t}^2}\text{ dt}}{\Big(\frac{1}{\text{t}^2}-1\Big)\sqrt{\Big(\frac{1}{\text{t}^2}+1\Big)}}$
$=-\int\frac{\text{t dt}}{(1-\text{t}^2)\sqrt{1+\text{t}^2}}$
Let $1+\text{t}^2=\text{u}^2$
$2\text{tdt}=2\text{udt}$
$\text{I}=\int\frac{\text{udu}}{(\text{u}^2-2)\text{u}}$
$=\int\frac{\text{du}}{\text{u}^2-2}$
$\therefore\ \text{I}=\frac{1}{2\sqrt{2}}\log\bigg|\frac{\text{u}-\sqrt{2}}{\text{u}+\sqrt{2}}\bigg|+\text{C}$
$=\frac{1}{2\sqrt{2}}\log\bigg|\frac{\sqrt{1+\text{t}^2}-\sqrt{2}}{\sqrt{1+\text{t}^2}+\sqrt{2}}\bigg|+\text{C}$
Hence,
$\text{I}=-\frac{1}{2\sqrt{2}}\log\bigg|\frac{\sqrt{2}\text{x}+\sqrt{\text{x}^2+1}}{\sqrt{2}\text{x}-\sqrt{\text{x}^2+1}}\bigg|+\text{C}$
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Question 3185 Marks
Evaluate the following integral:
$\int\frac{1}{\sqrt{(2-\text{x})^2+1}}\text{ dx}$
Answer
Let $\text{I}=\int\frac{1}{\sqrt{(2-\text{x})^2+1}}\text{ dx}$
Let $2-\text{x}=\text{t}$
$-\text{dx}=\text{dt}$
$\text{dx}=-\text{dt}$
So, $\text{I}=-\int\frac{1}{\sqrt{\text{t}^2+(1)^2}}\text{ dt}$
$\text{I}=-\log\big|\text{t}+\sqrt{\text{t}^2+1}\big|+\text{C}$ $\Big[$Since $\int\frac{1}{\sqrt{\text{x}^2+\text{a}^2}}\text{ dx}=\log\big|\text{x}+\sqrt{\text{x}^2+\text{a}^2}\big|+\text{C}\Big]$
$\text{I}=-\log\big|(2-\text{x})+\sqrt{(2-\text{x})^2+1}\big|+\text{C}$ $[$Since $\text{t}=(2-\text{x})]$
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Question 3195 Marks
Evaluate the following intergrals:
$\int\text{e}^\text{ax}\sin(\text{bx}+\text{c})\text{dx}$
Answer
Let $\text{I}=\int\text{e}^\text{ax}\sin(\text{bx}+\text{c})\text{dx}$
$\Rightarrow-\text{e}^\text{ax}\frac{\cos(\text{bx}+\text{x})}{\text{b}}+\int\text{ae}^\text{ax}\frac{\cos(\text{bx}+\text{c})}{\text{b}}\text{dx}$
$=-\frac{1}{\text{b}}\text{e}^\text{ax}\cos(\text{bx}+\text{c})+\frac{\text{a}}{\text{b}}\int\text{e}^\text{ax}\cos(\text{bx}+\text{c})\text{dx}$
$=-\frac{1}{\text{b}}\text{e}^\text{ax}\cos(\text{bx}+\text{c})+\frac{\text{a}}{\text{b}}\Big[\int\text{e}^\text{ax}\frac{\sin(\text{bx}+\text{c})}{\text{b}}-\int\text{ae}^{\text{ax}}\frac{\sin(\text{bx}+\text{c})}{\text{b}}\text{dx}\Big]+\text{C}_1$
$=\frac{\text{e}^\text{ax}}{\text{b}^2}\big\{\text{a}\sin(\text{bx}+\text{c})-\text{b}\cos(\text{bx}+\text{c})\big\}\\-\frac{\text{a}^2}{\text{b}^2}\int\text{e}^\text{ax}\sin(\text{bx}+\text{c})\text{dx}+\text{C}_1$
$\Rightarrow\text{I}=\frac{\text{e}^\text{ax}}{\text{b}^2}\big\{\text{a}\sin(\text{bx}+\text{c})-\text{b}\cos(\text{bx}+\text{c})\big\}\\-\frac{\text{a}^2}{\text{b}^2}\text{I}+\text{C}_1$
$\Rightarrow\text{I}=\Big\{\frac{\text{a}^2+\text{b}^2}{\text{b}^2}\Big\}-\frac{\text{e}^\text{ax}}{\text{b}^2}\big\{\text{a}\sin(\text{bx}+\text{c})-\text{b}\cos(\text{bx}+\text{c})\big\}+\text{C}_1$
$\Rightarrow\text{I}=\frac{\text{e}^\text{ax}}{\text{a}^2+\text{b}^2}\big\{\text{a}\sin(\text{bx}+\text{c})-\text{b}\cos(\text{bx}+\text{c})\big\}+\text{C}_1$
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Question 3205 Marks
Evaluate the follwing intregals:
$\int\frac{1}{\text{x}^4-1}\text{ dx}$
Answer
The Evaluate the integral follow the steps,
$\int\frac{1}{(\text{x}^4-1)}\text{ dx}$
Let $\frac{1}{(\text{x}^4-1)}=\frac{\text{A}}{\text{x}-1}+\frac{\text{B}}{\text{x}-1}+\frac{\text{C}}{\text{x}^2+1}$
$1={\text{A}}{(\text{x}-1})(\text{x}^2+1)+\text{B}(\text{x}+1)(\text{x}^2+1)+\text{C}(\text{x}+1)(\text{x}-1)$
$\text{For x}=1,\text{ B}=\frac{1}{4}$
$\text{For x}=-1,\text{ A}=-\frac{1}{4}$
$\text{For x}=0,\text{ C}=-\frac{1}{2}$
Therefore,
$\int\frac{1}{(\text{x}^4-1)}\text{ dx}=-\frac{1}{4}\int\frac{\text{dx}}{\text{x}+1}+\frac{1}{4}\frac{\text{dx}}{\text{x}-1}-\frac{1}{2}\int\frac{\text{dx}}{\text{x}^2+1}$
$=-\frac{1}{4}\ln\big|(\text{x}+1)\big|+\frac{1}{4}\ln\big|(\text{x}-1)\big|-\frac{1}{2}\tan^{-1}\text{x}+\text{C}$
$=\frac{1}{4}\ln\Big|\frac{\text{x}-1}{\text{x}+1}\Big|-\frac{1}{2}\tan^{-1}\text{x}+\text{C}$
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Question 3215 Marks
Evaluate the following integrals:$\int\frac{\text{x}^3}{\text{x}^4+\text{x}^2+1}\text{ dx}$
Answer
$\text{I}=\int\frac{\text{x}^3}{\text{x}^4+\text{x}^2+1}\text{ dx}$ $=\int\frac{\text{x}^2\cdot\text{x}}{(\text{x}^2)^2+\text{x}^2+1}\text{ dx}$ Let $\text{x}^2=\text{t}$ or $2\text{x}\text{ dx}=\text{dt}$ $\text{I}=\frac{1}{2}\int\frac{\text{t}}{\text{t}^2+\text{t}+1}\text{ dt}$ $=\frac{1}{4}\int\frac{2\text{t}}{\text{t}^2+\text{t}+1}\text{ dt}$$=\frac{1}{4}\int\frac{2\text{t}+1-1}{\text{t}^2+\text{t}+1}\text{ dt}$
$=\frac{1}{4}\int\Big[\frac{(2\text{t}+1)}{(\text{t}^2+\text{t}+1)}-\frac{1}{(\text{t}^2+\text{t}+1)}\Big]\text{dt}$
$=\frac{1}{4}\Big[\log\big|\text{t}^2+\text{t}+1\big|-\int\frac{1}{\big(\text{t}^2+\text{t}+\frac{1}{4}+\frac{3}{4}\big)}\text{ dt}\Big]$ $=\frac{1}{4}\begin{bmatrix}\log\big|\text{t}^2+\text{t}+1\big|-\int\frac{1}{\big(\text{t}+\frac{1}{2}\big)^2+\Big(\frac{\sqrt{3}}{2}\Big)^2}\text{ dt}\end{bmatrix}$ $=\frac{1}{4}\begin{bmatrix}\log\big|\text{t}^2+\text{t}+1\big|-\frac{2}{\sqrt3}\tan\frac{\big(\text{t}+\frac{1}{2}\big)}{\Big(\frac{\sqrt3}{2}\Big)}\end{bmatrix}+\text{C}$ $=\frac{1}{4}\Big[\log\big|\text{t}^2+\text{t}+1\big|-\frac{2}{\sqrt3}\tan\Big(\frac{2\text{t}+1}{\sqrt3}\Big)\Big]+\text{C}$ $=\frac{1}{4}\Big[\log\big|\text{x}^4+\text{x}^2+1\big|-\frac{2}{\sqrt3}\tan\Big(\frac{2\text{x}^2+1}{\sqrt3}\Big)\Big]+\text{C}$
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