Question 1515 Marks
Evaluate the following integrals:$\int\frac{\text{x}}{\text{x}^4-\text{x}^2+1}\text{dx}$
Answer$\int\frac{\text{x dx}}{\text{x}^4-\text{x}^2+1}$
Let $\text{x}^2=\text{t}$
$\Rightarrow2\text{x dx = dt}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$
Now, $\int\frac{\text{x dx}}{\text{x}^4-\text{x}^2+1}$
$ =\frac{1}{2}\int\frac{\text{dt}}{\text{t}^2-\text{t}+1}$
$=\frac{1}{2}\int\frac{\text{dt}}{\text{t}^2-\text{t}+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2+1}$
$=\frac{1}{2}\int\frac{\text{dt}}{\big(\text{t}-\frac{1}{2}\big)^2+\frac{3}{4}}$
$=\frac{1}2{}\int\frac{\text{dt}}{\Big(\text{t}-\frac{1}{2}\Big)^2+\Big(\frac{\sqrt{3}}{2}\Big)^2}$
$=\frac{1}{2}\times\frac{2}{\sqrt{3}}\tan^{-1}\Bigg(\frac{\text{t}-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\Bigg)+\text{C}$
$=\frac{1}{\sqrt{3}}\tan^{-1}\Big(\frac{2\text{t}-1}{\sqrt{3}}\Big)+\text{C}$
$=\frac{1}{\sqrt{3}}\tan^{-1}\Big(\frac{2\text{x}^2-1}{\sqrt{3}}\Big)+\text{C}$
View full question & answer→Question 1525 Marks
Evalute the following integrals:
$\int\frac{-\sin\text{x}+2\cos\text{x}}{2\sin\text{x}+\cos\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{-\sin\text{x}+2\cos\text{x}}{2\sin\text{x}+\cos\text{x}}\text{dx}\ .....\text{(i)}$
Let $2\sin\text{x}+\cos\text{x}=\text{t}$ then,
$\text{d}(2\sin\text{x}+\cos\text{x})=\text{dt}$
$\Rightarrow(2\cos\text{x}-\sin\text{x})\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{-\sin\text{x}+2\cos\text{x}}$
Putting $2\sin\text{x}+\cos\text{x}=\text{t and dx}=\frac{\text{dt}}{-\sin\text{x}+2\cos\text{x}}$ in equation (i), we get,
$\text{I}=\int\frac{-\sin\text{x}+2\cos\text{x}}{\text{t}}\times\frac{\text{dt}}{-\sin\text{x}+2\cos\text{x}}$
$=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|2\sin\text{x}+\cos\text{x}|+\text{C}$
$\therefore\text{I}=\log|2\sin\text{x}+\cos\text{x}|+\text{C}$
View full question & answer→Question 1535 Marks
Evaluate the following integrals:$\int\frac{1}{2\text{x}^2-\text{x}-1}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{2\text{x}^2-\text{x}-1}\text{dx}$
$=\frac{1}{2}\int\frac{1}{\text{x}^2-\frac{\text{x}}{2}-\frac{1}{2}}\text{dx}$
$=\frac{1}{2}\int\frac{1}{\text{x}^2-2\text{x}\times\frac{1}{4}+\big(\frac{1}{4}\big)^2-\big(\frac{1}{4}\big)^2-\frac{1}{2}}\text{dx}$
$=\frac{1}{2}\int\frac{1}{\big(\text{x}-\frac{1}{4}\big)^2-\frac{9}{16}}\text{dx}$
Let $\text{x}-\frac{1}{4}=\text{t}$
$\Rightarrow\text{dx = dt}$
$\text{I}=\frac{1}{2}\int\frac{1}{\text{t}^2-\big(\frac{3}{4}\big)^2}\text{dt}$
$\text{I}=\frac{1}{2}\times\frac{1}{2\times\big(\frac{3}{4}\big)}\log\Bigg|\frac{\text{t}-\frac{3}{4}}{\text{t}+\frac{3}{4}}\Bigg|+\text{C} $ $\Big[\text{Since,}\int\frac{1}{\text{x}^2-\text{a}^2}\text{dx}=\frac{1}{2\text{a}}\log\bigg|\frac{\text{x}-\text{a}}{\text{x+a}}\bigg|+\text{C}\Big]$
$\text{I}=\frac{1}{3}\log\Bigg|\frac{\text{x}-\frac{1}{4}-\frac{3}{4}}{\text{x}-\frac{1}{4}+\frac{3}{4}}\Bigg|+\text{C}$
$\text{I}=\frac{1}{3}\log\bigg|\frac{\text{x}-1}{2\text{x}+1}\bigg|+\text{C}$
View full question & answer→Question 1545 Marks
Evalute the following integrals:
$\int\frac{1}{\cos(\text{x}+\text{a})\cos(\text{x}+\text{b})}\text{dx}$
Answer$\int\frac{1}{\cos(\text{x}+\text{a})\cos(\text{x}+\text{b})}\text{dx}$
Multiplying and Dividing by $\sin\big[(\text{x}+\text{b})-(\text{x}+\text{a})\big],$ we get
$=\int\frac{1}{\sin\big[(\text{x}+\text{b})-(\text{x}+\text{a})\big]}\times\frac{\sin\big[(\text{x}+\text{b})-(\text{x}+\text{a})\big]}{\cos(\text{x}+\text{a})\cos(\text{x}+\text{b})}\text{dx}$
$=\int\frac{1}{\sin(\text{b}-\text{a})}\times\frac{\sin\big[(\text{x}+\text{b})-(\text{x}+\text{a})\big]}{\cos(\text{x}+\text{a})\cos(\text{x}+\text{b})}\text{dx}$
$=\frac{1}{\sin(\text{b}-\text{a})}\int\frac{\sin(\text{x}+\text{b})\cos(\text{x}+\text{a})-\sin(\text{x}+\text{a})\cos(\text{x}+\text{b})}{\cos(\text{x}+\text{a})\cos(\text{x}+\text{b})}\text{dx}$
$=\frac{1}{\sin(\text{b}-\text{a})}\Big[\int\frac{\sin(\text{x}+\text{b})}{\cos(\text{x}+\text{b})}\text{dx}-\int\frac{\sin(\text{x}+\text{a})}{\cos(\text{x}+\text{a})}\text{dx}\Big]$
$=\frac{1}{\sin(\text{b}-\text{a})}\Big[\int\tan(\text{x}+\text{b})\text{dx}-\int\tan(\text{x}+\text{a})\text{dx}\Big]$
$=\frac{1}{\sin(\text{b}-\text{a})}\big[\log(\sec(\text{x}+\text{b}))-\log(\sec(\text{x}+\text{a}))\big]+\text{C}$
$=\frac{1}{\sin(\text{b}-\text{a})}\bigg[\log\Big(\frac{\sec(\text{x}+\text{b})}{\sec(\text{x}+\text{a})}\Big)\bigg]+\text{C}$
Hence, $\int\frac{1}{\cos(\text{x}+\text{a})\cos(\text{x}+\text{b})}\text{dx}=\frac{1}{\sin(\text{b}-\text{a})}\bigg[\log\Big(\frac{\sec(\text{x}+\text{b})}{\sec(\text{x}+\text{a})}\Big)\bigg]+\text{C}$
View full question & answer→Question 1555 Marks
$\int(2\text{x}^2+3)\sqrt{\text{x+2}}\text{dx}$
AnswerLet $\text{I}=\int(2\text{x}^2+3)\sqrt{\text{x+2}}\text{dx}$
Substituting x + 2 = t and dx = dt, we get
$\text{I}=\int\big[2(\text{t}-2)^2+3\big]\sqrt{\text{t}}\text{dt}$
$=\int\big[2(\text{t}^2+4-4\text{t})+3\big]\sqrt{\text{t}}\text{dt}$
$=\int\big[2\text{t}^2+8-8\text{t}+3\big]\sqrt{\text{t}}\text{dt}$
$=\int\Big(2\text{t}^{\frac{5}{2}}+11\text{t}^{-\frac{1}{2}}-8\text{t}^{\frac{3}{2}}\Big)\text{dt}$
$=\frac{4}{7}\text{t}^{\frac{7}{2}}+\frac{22}{3}\text{t}^{\frac{3}{2}}-\frac{16}{5}\text{t}^{\frac{5}{2}}+\text{C}$
$=\frac{4}{7}(\text{x}+2)^\frac{\text{7}}{2}-\frac{16}{5}(\text{x}+2)^\frac{\text{5}}{2}+\frac{22}{3}(\text{x}+2)^\frac{\text{3}}{2}+\text{C}$
$\therefore\ \text{I}=\frac{4}{7}(\text{x}+2)^\frac{\text{7}}{2}-\frac{16}{5}(\text{x}+2)^\frac{\text{5}}{2}+\frac{22}{3}(\text{x}+2)^\frac{\text{3}}{2}+\text{C}$
View full question & answer→Question 1565 Marks
Evaluate the following integrals:
$\int\sin^3\sqrt{\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\sin^3\sqrt{\text{x}}\text{dx}$
$\sqrt{\text{x}}=\text{t}$
$\text{x = t}^2$
$\text{dx}=2\text{t dt}$
$\text{I}=2\int\text{t}\sin^3\text{t dt}$
$=2\int\text{t}\Big(\frac{3\sin\text{t}-\sin3\text{t}}{4}\Big)\text{dt}$
$=\frac{1}{2}\int\text{t}(3\sin\text{t}-\sin3\text{t})\text{dt}$
Using integration by parts,
$\text{I}=\frac{1}2{}\Big[\text{t}\Big(-3\cos\text{t}+\frac{1}{3}\cos3\text{t}\Big)-\int\Big(-3\cos\text{t}+\frac{\cos3\text{t}}{3}\Big)\text{dt}\Big]$
$=\frac{1}{2}\Big[\frac{-9\text{t}\cos\text{t}+\text{t}\cos3\text{t}}{3}-\Big\{-3\sin\text{t}+\frac{\sin3\text{t}}{9}\Big\}\Big]+\text{C}$
$=\frac{1}{2}\Big[\frac{-9\text{t}\cos\text{t}+\text{t}\cos3\text{t}}{3}+\frac{27\sin\text{t}-3\sin3\text{t}}{9}\Big]+\text{C}$
$=\frac{1}{18}\big[-27\text{t}\cos\text{t}+3\text{t}\cos3\text{t}+27\sin\text{t}-3\sin3\text{t}\big]+\text{C}$
$\text{I}=\frac{1}{18}\big[3\sqrt{\text{x}}\cos3\sqrt{\text{x}}+27\sin\sqrt{\text{x}}-27\sqrt{\text{x}}\cos\sqrt{\text{x}}-3\sin3\sqrt{\text{x}}\big]+\text{C}$
View full question & answer→Question 1575 Marks
Evalute the following integrals:
$\int\frac{1-\cot\text{x}}{1+\cot\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1-\cot\text{x}}{1+\cot\text{x}}\text{dx}$ then,
$\text{I}=\int\frac{1-\frac{\cos\text{x}}{\sin\text{x}}}{1+\frac{\cos\text{x}}{\sin\text{x}}}\text{dx}$
$=\int\frac{\frac{\sin\text{x}-\cos\text{x}}{\sin\text{x}}}{\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}}}\text{dx}$
$\Rightarrow\text{I}=\int\frac{\sin\text{x}-\cos\text{x}}{\sin\text{x}+\cos\text{x}}\text{dx}\ .....(\text{i})$
Let $\sin\text{x}+\cos\text{x}=\text{t},$ then,
$\text{d}(\sin\text{x}+\cos\text{x})=\text{dt}$
$\Rightarrow(\cos\text{x}-\sin\text{x})\text{dx}=\text{dt}$
$\Rightarrow-(\sin\text{x}-\cos\text{x})\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=-\frac{\text{dt}}{\sin\text{x}-\cos\text{x}}$
Putting $\sin\text{x}+\cos\text{x}=\text{t and dx}=-\frac{\text{dt}}{\sin\text{x}-\cos\text{x}}$ in equation (i), we het
$\text{I}=\int\frac{\sin\text{x}-\cos\text{x}}{\text{t}}\times\frac{-\text{dt}}{\sin\text{x}-\cos\text{x}}$
$=\int\frac{-\text{dt}}{\text{t}}$
$=-\log|\text{t}|+\text{C}$
$=-\log|\sin\text{x}+\cos\text{x}|+\text{C}$
View full question & answer→Question 1585 Marks
Evalute the following integrals:
$\int\tan2\text{x}\tan3\text{x}\tan5\text{x dx}$
AnswerLet $\text{I}=\int1+\tan\text{x}\tan(\text{x}+\theta)\text{dx}$
$=\int1+\tan\text{x}\Big(\frac{\tan\text{x}+\tan\theta}{1-\tan\text{x}\tan\theta}\Big)\text{dx}$
$=\int\frac{1+\tan^2\text{x}}{1-\tan\text{x}\tan\theta}\text{dx}$
$=\int\frac{\sec^2\text{x dx}}{1-\tan\text{x}\tan\theta}$
Putting $\tan\text{x}=\text{t}$
$\Rightarrow\sec^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{\sec^2\text{x}}$
$\therefore\text{I}\approx\int\frac{1}{1-\text{t}\tan\theta}\text{dt}$
$=\frac{-1}{\tan\theta}\text{ ln}|1-\text{t}\tan\theta|+\text{C}$
$\Big[\because\int\frac{1}{\text{ax}+\text{b}}\text{dx}=\frac{1}{\text{a}}\text{ ln}|\text{ax}+\text{b}|+\text{C}\Big]$
$=-\cot\theta\text{ ln}|1-\tan\text{ x }\tan\theta|+\text{C}$
$=\cot\theta\text{ ln}\Big|\frac{1}{1-\tan\text{ x }\tan\theta}\Big|+\text{C}$
$=\cot\theta\text{ ln}\Big|\frac{\cos\text{ x }\cos\theta}{\cos\text{x}\cos\theta-\sin\text{x}\sin\theta}\Big|+\text{C}$
$=\cot\theta\text{ ln}\Big|\frac{\cos\text{x}}{\cos(\text{x}+\theta)}\Big|+\text{C }\big[\text{Let C}'=\text{C}+\cot\theta\text{ ln}\cos\theta\big]$
View full question & answer→Question 1595 Marks
Evaluate the following integrals:$\int\frac{\text{x}^3\sin^{-1}\text{x}^2}{\sqrt{1-\text{x}^4}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^3\sin^{-1}\text{x}^2}{\sqrt{1-\text{x}^4}}\text{dx}$
Let $\sin^{-1}\text{x}^2=\text{t}$
$\frac{1}{\sqrt{1-\text{x}^4}}(2\text{x})\text{dx}=\text{dt}$
$\text{I}=\int\frac{\text{x}^2\sin^{-1}\text{x}^2}{\sqrt{1-\text{x}^2}}\text{x dx}$
$=\int(\sin\text{t})\text{t}\frac{\text{dt}}{2}$
$=\frac{1}{2}\int\text{t}\sin\text{t dt}$
$=\frac{1}{2}\big[\text{t}\int\sin\text{t dt}-\int(1\int\sin\text{t dt})\text{dt}\big]$
$=\frac{1}{2}\big[\text{t}(-\cos\text{t})-\int(-\cos\text{t})\text{dt}\big]$
$=\frac{1}{2}\big[-\text{t}\cos\text{t}+\sin\text{t}\big]+\text{C}$
$\text{I}=\frac{1}{2}\Big[\text{x}^2-\sqrt{1-\text{x}^4}\sin^{-1}\text{x}^2\Big]+\text{C}$
View full question & answer→Question 1605 Marks
Evaluate the following integrals:
$\int \frac{\sin^5\text{x}}{\cos^4\text{x}}\text{ dx}$
Answer$\int \frac{\sin^5\text{x}}{\cos^4\text{x}}\text{ dx}$ $=\int\Big(\frac{\sin^4\text{x}.\sin\text{x}}{\cos^4\text{x}}\Big)\text{dx}$ $=\int\frac{(\sin^2\text{x})^2\sin\text{x}}{\cos^4\text{x}}\text{ dx}$ $=\int\frac{(1-\cos^2\text{x})^2\sin\text{x}}{\cos^4\text{x}}\text{ dx}$ $=\int\Big(\frac{1+\cos^4\text{x}-2\cos^2\text{x}}{\cos^4\text{x}}\Big)\sin\text{x dx }$ $=\int\Big(\frac{1}{\cos^4\text{x}}+1-\frac{2}{\cos^2\text{x}}\Big)\sin\text{x dx}$ Let $\cos\text{x}=\text{t}$ $\Rightarrow-\sin\text{x}=\frac{\text{dt}}{\text{dx}}$ $$$\Rightarrow\sin\text{x dx}=-\text{dt}$Now, $=\int\Big(\frac{1}{\cos^4\text{x}}+1-\frac{2}{\cos^2\text{x}}\Big)\sin\text{x dx}$
$=-\int\big(\text{t}^{-4}+1-2\text{t}^{-2}\big)\text{ dt}$ $=-\Big[-\frac{\text{t}^{-4+1}}{-4+1}+\text{t}-\frac{2\text{t}^{-2+1}}{-2+1}\Big]+\text{C}$ $=-\Big[-\frac{1}{3\text{t}^3}+\text{t}+\frac{2}{\text{t}}\Big]+\text{C}$ $=\frac{1}{3\text{t}^3}-\text{t}-\frac{2}{\text{t}}+\text{C}$ $=\frac{1}{3\cos^3\text{x}}-\cos\text{x}-\frac{2}{\cos\text{x}}+\text{C}$
View full question & answer→Question 1615 Marks
$\int\frac{2-3\text{x}}{\sqrt{1+3\text{x}}}\text{dx}$
AnswerLet I $=\int\frac{2-3\text{x}}{\sqrt{1+3\text{x}}}\times\text{dx}.$ Then,
$\text{I}=\int\frac{2-3\text{x}-1+1}{\sqrt{1+3\text{x}}}\text{dx}$
$=\int\frac{-3\text{x}-1+3}{\sqrt{1+3\text{x}}}\text{dx}$
$=\int-\frac{(3\text{x}+1)}{\sqrt{1+3\text{x}}}\text{dx}+3\int\frac{1}{\sqrt{1+3\text{x}}}\text{dx}$
$=-1\int\frac{1+3\text{x}}{\sqrt{1+3\text{x}}}\text{dx}+3\int\frac{1}{\sqrt{1+3\text{x}}}\text{dx}$
$=-1\int(1+3\text{x})^\frac{1}{2}\text{dx}+3\int(1+3\text{x})^\frac{-1}{2}\text{dx}$
$=-1\times\frac{(1+3\text{x})^\frac{3}{2}}{\frac{3}{2}\times3}+3\times\frac{(1+3\text{x})^\frac{1}{2}}{\frac{1}{2}\times3}+\text{C}$
$=-\frac{2}{9}\times(1+3\text{x})^\frac{3}{2}+2(1+3\text{x})^\frac{1}{2}+\text{C}$
$=2(1+3\text{x})^\frac{1}{2}\Big[-\frac{1}{9}(1+3\text{x})^1+1\Big]+\text{C}$
$=2(1+3\text{x})^\frac{1}{2}\Big[\frac{-1-3\text{x}+9}{9}\Big]+\text{C}$
$=2(1+3\text{x})^\frac{1}{2}\Big[\frac{8-3\text{x}}{9}\Big]+\text{C}$
$=\frac{2}{9}\sqrt{1+3\text{x}}(8-3\text{x})+\text{C}$
$\therefore\text{I}=\frac{2}{9}(8-9\text{x})\sqrt{1+3\text{x}}+\text{C}$
View full question & answer→Question 1625 Marks
Evaluate the following integrals:
$\int\frac{1}{\sin^3\text{x}\cos\text{x}}\text{dx}$
Answer$\int\frac{1}{\sin^3\text{x}\cos\text{x}}\text{ dx}$
Dividing numerator & denominator by $\sin^4\text{x}$
$=\int\frac{\frac{1}{\sin^4\text{x}}\text{ dx}}{\frac{\sin^3\text{x}\cdot\cos\text{x}}{\sin^4\text{x}}}$
$=\int\frac{\text{cosec}^4\text{x}\text{ dx}}{\cot\text{x}}$
$=\int\frac{\text{cosec}^2\text{x}\cdot\text{cosec}^2\text{x}\text{ dx}}{\cot\text{x}}$
$=\int\frac{(1+\cot^2\text{x})\cdot\text{cosec}^2\text{x}\text{ dx}}{\cot\text{x}}$
Let $\cot\text{x}=\text{t}$
$\text{cosec}^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\text{cosec}^2\text{x}\text{dx}=-\text{dt}$
Now, $\int\frac{(1+\cot^2\text{x})\cdot\text{cosec}^2\text{x}}{\cot\text{x}}\text{ dx}$
$=\int\frac{(1+\text{t}^2)\cdot(-\text{dt})}{\text{t}}$
$=-\int\Big(\frac{1}{\text{t}}+\text{t}\Big)\text{dt}$
$=-\log|\text{t}|-\frac{\text{t}^2}{2}+\text{C}$
$=-\log|\cot\text{x}|-\frac{\cot^2\text{x}}{1}+\text{C}$
$=\log|\cot\text{x}|^{-1}-\frac{(\text{cosec}^2\text{x}-1)}{2}+\text{C}$
$=\log\Big|\frac{1}{\cot\text{x}}\Big|-\frac{\text{cosec}^2\text{x}}{2}+\frac{1}{2}+\text{C}$
$=\log|\tan\text{x}|-\frac{1}{2\sin^2\text{x}}+\text{C}'$ $\Big[\therefore\text{C}'=\text{C}+\frac{1}{2}\Big]$
View full question & answer→Question 1635 Marks
Evaluate the following integrals:
$\int\sec^4\text{x}\tan\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\sec^4\text{x}\tan\text{x}\text{ dx}\ ....(1)$
Let $\tan\text{x}=\text{t}$ then,
$\Rightarrow\text{d}(\tan\text{x})=\text{dt}$
$\Rightarrow\sec^2\text{x}\text{ dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{\sec^2\text{x}}$
Putting $\tan\text{x}=\text{t}$ and $\text{dx}=\frac{\text{dt}}{\sec^2\text{x}}$ in equation (1), we get,
$\text{I}=\int\sec^4\text{x}\tan\text{x}\frac{\text{dt}}{\sec^2\text{x}}$
$=\int\sec^2\text{x}\text{ t dt}$
$=\int\big(1+\tan^2\text{x}\big)\text{t dt}$
$=\int\big(1+\text{t}^2\big)\text{t dt}$
$=\int\big(\text{t}+\text{t}^3\big)\text{dt}$
$=\frac{\text{t}^2}{2}+\frac{\text{t}^4}{4}+\text{C}$
$=\frac{\tan^2\text{x}}{2}+\frac{\tan^4\text{x}}{4}+\text{C}$
$\text{I}=\frac{1}{2}\tan^2\text{x}+\frac{1}{4}\tan^4\text{x}+\text{C}$
View full question & answer→Question 1645 Marks
Evaluate the following intregals: $\int\frac{1}{1-2\sin\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{1-2\sin\text{x}}\ \text{dx}$
Put $\sin\text{x}=\frac{2\tan^\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}$
$\text{I}=\int\frac{1}{1-2\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)}\ \text{dx}$
$=\int\frac{1+\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}-4\tan\frac{\text{x}}{2}}\ \text{dx}$
$=\int\frac{\sec^2\frac{\text{x}}{2}}{\tan^2\frac{\text{x}}{2}-4\tan\frac{\text{x}}{2}+1}\ \text{dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$
$\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{dx}=\text{dt}$
$=\int\frac{2\text{dt}}{\text{t}^2-4\text{t}+1}$
$=\int\frac{2\text{dt}}{\text{t}^2-2\text{t}(2)+(2)^2-(2)^2+1}$
$\text{I}=2\int\frac{\text{dt}}{(\text{t}-2)^2+(\sqrt{3})^2}$
$=2\times\frac{1}{2\sqrt{3}}\log\Big|\frac{\text{t}-2-\sqrt{3}}{\text{t}-2+\sqrt{3}}\Big|+\text{C}$
$\text{I}=\frac{1}{\sqrt{3}}\log\Bigg|\frac{\tan\frac{\text{x}}{2}-2-\sqrt{3}}{\tan\frac{\text{x}}{2}-2+\sqrt{3}}\Bigg|+\text{C}$
View full question & answer→Question 1655 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^2}{\text{x}^4+\text{x}^2-2}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2}{\text{x}^4+\text{x}^2-2}\ \text{dx}$
we express
$\frac{\text{x}^2}{\text{x}^4+\text{x}^2-2}=\frac{\text{x}^2}{\text{x}^4+2\text{x}^2-\text{x}^2-2}$
$=\frac{\text{x}^2}{(\text{x}^2+2)(\text{x}^2-1)}$
$=\frac{\text{A}}{\text{x}^2+2}+\frac{\text{B}}{\text{x}^2-1}$
$\Rightarrow\text{x}^2=\text{A}(\text{x}^2-1)+\text{B}(\text{x}^2+2)$
Equating the coefficient of x and constants, we get
1 = A + B and 0 = -A + 2B or
$\text{A}=\frac{2}{3}$ and $\text{B}=\frac{1}{3}$
$\therefore\text{I}=\int\Big(\frac{\frac{2}{3}}{\text{x}^2+2}+\frac{\frac{1}{3}}{\text{x}^2-1}\Big)\text{dx}$
$=\frac{2}{3}\int\frac{1}{\text{x}^2+2}\ \text{dx}+\frac{1}{3}\int\frac{1}{\text{x}^2-1}\ \text{dx}$
$=\frac{\sqrt{2}}{3}\tan^{-1}\frac{\text{x}}{\sqrt{2}}+\frac{1}{6}\log\Big|\frac{\text{x}-1}{\text{x}+1}\Big|+\text{C}$
Hence, $\int\frac{\text{x}^2}{\text{x}^4+\text{x}^2-2}\ \text{dx}=\frac{\sqrt{2}}{3}\tan^{-1}\frac{\text{x}}{\sqrt{2}}+\frac{1}{6}\log\Big|\frac{\text{x}-1}{\text{x}+1}\Big|+\text{C}$
View full question & answer→Question 1665 Marks
Evaluate the following integrals:
$\int\frac{3\text{x}+5}{\text{x}^3-\text{x}^2-\text{x}+1}\ \text{dx}$
AnswerTo evaluate the integrals follow the steps:
$\int\frac{3\text{x}+5}{\text{x}^3-\text{x}^2-\text{x}+1}\ \text{dx}$
Let $\frac{3\text{x}-5}{(\text{x}-1)^2(\text{x}+1)}=\frac{\text{A}}{\text{x}-1}+\frac{\text{B}}{(\text{x}-1)^2}+\frac{\text{C}}{\text{x}+1}$
$3\text{x}+5=\text{A}(\text{x}-1)(\text{x}+1)+\text{B}(\text{x}+1)+\text{C}(\text{x}-1)^2$
For x = 1 B = 4
$\text{For x} = -1\ \text{C}=\frac{1}{2}$
$\text{For x} = 0\ \text{A}=-\frac{1}{2}$
Therefore
$\int\frac{3\text{x}+5}{(\text{x}-1)^2(\text{x}+1)}\ \text{dx}=-\frac{1}{2}\int\frac{\text{dx}}{\text{x}-1}+\frac{1}{2}\int\frac{\text{dx}}{\text{x}+1}$
$=-\frac{1}{2}\ln|(\text{x}-1)|-\frac{4}{(\text{x}-1)}+\frac{1}{2}\ln|(\text{x}+1)|+\text{C}$
$=\frac{1}{2}\ln\big|\frac{\text{x}+1}{\text{x}-1}\big|-\frac{4}{(\text{x}-1)}+\text{C}$
View full question & answer→Question 1675 Marks
Evaluate the following intregals:
$\int\frac{\text{x}^2+6\text{x}-8}{\text{x}^3-4\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2+6\text{x}-8}{\text{x}^3-4\text{x}}\ \text{dx}$ $\Rightarrow\text{I}=\int\frac{\text{x}^2+6\text{x}-8}{\text{x}(\text{x}+2)(\text{x}-2)}\text{ dx}$Now,
Let $\frac{\text{x}^2+6\text{x}-8}{\text{x}(\text{x}+2)(\text{x}-2)}=\frac{\text{A}}{\text{x}}+\frac{\text{B}}{\text{x}+2}+\frac{\text{C}}{\text{x}-2}$
$\Rightarrow\text{x}^2+6\text{x}-8=\text{A}(\text{x}^2-4)+\text{B}(\text{x}-2)\text{x}+\text{C}(\text{x}+2)\text{x}$ Put x = 0 ⇒ -8 = -4A ⇒ A = 2 Put x = -2 ⇒ -16 = 8B ⇒ B = -2 Put x = 2 ⇒ 8 = 8C ⇒ C = 1 Thus,$\text{I}=\int\frac{2\text{dx}}{\text{x}}-\int\frac{2\text{dx}}{\text{x}+2}+\int\frac{\text{dx}}{\text{x}-2}$
$=2\log|\text{x}|-2\log|\text{x}+2|+\log|\text{x}-2|+\text{C}$
$\therefore\text{I}=\log\Big|\frac{\text{x}^2(\text{x}-2)}{(\text{x}+2)^2}\Big|+\text{C}$
View full question & answer→Question 1685 Marks
Evaluate the following integrals:$\int\text{x}^2\sin^{-1}\text{x dx}$
AnswerLet $\text{I}=\int\text{x}^2\sin^{-1}\text{x dx}$
$\text{I}=\sin^{-1}\text{x}\int\text{x}^2\text{dx}-\int\Big(\frac{1}{\sqrt{1-\text{x}^2}}\int\text{x}^2\text{dx}\Big)\text{dx}$
$=\frac{\text{x}^3}{3}\sin^{-1}\text{x}-\int\frac{\text{x}^3}{3\sqrt{1-\text{x}^2}}\text{dx}$
$\text{I}=\frac{\text{x}^3}{3}\sin^{-1}\text{x}-\frac{1}{3}\text{I}_1+\text{C}_1\dots(1)$
$\text{I}_1=\int\frac{\text{x}^3}{\sqrt{1-\text{x}^2}}\text{dx}$
Let $1-\text{x}^2=\text{t}^2$
$-2\text{x dx}=2\text{t dt}$
$-\text{x dx}=\text{t dt}$
$\text{I}_1=-\int\frac{(1-\text{t}^2)\text{tdt}}{\text{t}}$
$=\int(\text{t}^2-1)\text{dt}$
$=\frac{\text{t}^3}{3}-\text{t}+\text{C}_2$
$=\frac{(1-\text{x}^2)^{\frac{3}{2}}}{3}-(1-\text{x}^2)^{\frac{1}{2}}+\text{C}_2$
Now,
$\text{I}=\frac{\text{x}^3}{3}\sin^{-1}\text{x}-\frac{1}{9}(1-\text{x}^2)^{\frac{3}{2}}+\frac{1}{3}(1-\text{x}^2)^{\frac{1}{2}}+\text{C}$
View full question & answer→Question 1695 Marks
Evaluate the following integrals:
$\int\text{e}^{2\text{x}}\cos(3\text{x}+4)\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{2\text{x}}\cos(3\text{x}+4)\text{dx}$
Integrating by parts
$\text{I}=\text{e}^{2\text{x}}\frac{\sin(3\text{x}+4)}{3}-\int2\text{e}^{2\text{x}}\frac{\sin(3\text{x}+4)}{3}\text{dx}$
$=\frac{1}{3}\text{e}^{2\text{x}}\sin(3\text{x}+4)-\frac{2}{3}\int\text{e}^{2\text{x}}\sin(3\text{x}+4)\text{dx}$
$=\frac{1}{3}\text{e}^{2\text{x}}\sin(3\text{x}+4)-\frac{2}{3}\Big[-\text{e}^{2\text{x}}\frac{\cos(3\text{x}+4)}{3}+\int2\text{e}^{2\text{x}}\frac{\cos(3\text{x}+4)}{3}\text{dx}\Big]+\text{C}$
$\text{I}=\frac{\text{e}^{2\text{x}}}{9}\{2\cos(3\text{x}+4)+3\sin(3\text{x}+4)\}+\text{C}$
Hence,
$\text{I}=\frac{\text{e}^{2\text{x}}}{13}\{2\cos(3\text{x}+4)+3\sin(3\text{x}+4)\}+\text{C}$
View full question & answer→Question 1705 Marks
Evaluate the following integrals:
$\int\frac{1}{\text{x}^2(\text{x}^4+1)^{\frac{3}{4}}}\text{ dx}$
Answer$\int\frac{1}{\text{x}^2(\text{x}^4+1)^{\frac{3}{4}}}\text{ dx}$
Multiplying and dividing by $x^{-3},$ we obtain
$\frac{\text{x}^{-3}}{\text{x}^2.\text{x}^{-3}(\text{x}^4+1)^{\frac{3}{4}}}=\frac{\text{x}^{-3}(\text{x}^4+1)^{\frac{-3}{4}}}{\text{x}^2.\text{x}^{-3}}$
$=\frac{(\text{x}^4+1)^{\frac{-3}{4}}}{\text{x}^5.(\text{x}^{4})^{-\frac{3}{4}}}$
$=\frac{1}{\text{x}^5}\Big(\frac{\text{x}^4+1}{\text{x}^4}\Big)^{-\frac{3}{4}}$
$=\frac{1}{\text{x}^5}\Big(1+\frac{1}{\text{x}^4}\Big)^{-\frac{3}{4}}$
Let, $\frac{1}{\text{x}^4}=\text{t}$
$\Rightarrow-\frac{4}{\text{x}^5}\text{ dx}=\text{dt}$
$\Rightarrow\frac{1}{\text{x}^5}\text{ dx}=-\frac{\text{dt}}{4}$
View full question & answer→Question 1715 Marks
Evaluate the following intregals:
$\int\frac{\text{x}+1}{\sqrt{\text{x}^2+1}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}+1}{\sqrt{\text{x}^2+1}}\text{dx}$
let $\text{x}+1=\lambda\frac{\text{d}}{\text{dx}}(\text{x}^2+1)+\mu$
$\text{x}+1=\lambda(2\text{x})+\mu$
Compairing the coefficient of like powewrs of x,
$2\lambda=1\ \Rightarrow\lambda=\frac{1}{2}$
$\Rightarrow\mu=1$
So, $\text{I}=\int\frac{\frac{1}{2}(2\text{x})+1}{\sqrt{\text{x}^2+1}}\text{dx}$
$=\frac{1}{2}\int\frac{(2\text{x})}{\sqrt{\text{x}^2+1}}\text{dx}+\int\frac{1}{\sqrt{\text{x}^2+1}}\text{dx}$
$\text{I}=\frac{1}{2}\times2\sqrt{\text{x}^2+1}+\log\big|\text{x}+\sqrt{\text{x}^2+1}\big|+\text{C}$ $\big[\text{since}, \int\frac{1}{\sqrt{\text{x}}}\text{dx}=2\sqrt{\text{x}}+\text{c},\int\frac{1}{\sqrt{\text{x}^2+1}}\text{dx}=\log\big|\text{x}+\sqrt{\text{x}^2-\text{x}^2}\big|+\text{C}\big]$
$\text{I}=\sqrt{\text{x}^2+1}+\log\big|\text{x}+\sqrt{\text{x}^2+1}\big|+\text{C}$
View full question & answer→Question 1725 Marks
Evaluate the following integrals:$\int\text{x}\tan^2\text{x dx}$
AnswerLet $\text{I}=\int\text{x}\tan^2\text{x dx}$
$=\int\text{x}(\sec^2\text{x}-1)\text{dx}$
$=\int\text{x}\sec^2\text{x dx}-\int\text{x dx}$
$=[\text{x}\int\sec^2\text{x dx}-\int(1\int\sec^2\text{x dx})\text{dx}]-\frac{\text{x}^2}{2}$
$=\text{x}\tan\text{x}-\int\tan\text{x dx}-\frac{\text{x}^2}{2}$
$\text{I}=\text{x}\tan\text{x}-\log\sec\text{x}-\frac{\text{x}^2}{2}+\text{C}$
View full question & answer→Question 1735 Marks
Evalute the following integrals:
$\int\frac{1+\cot\text{x}}{\text{x}+\log\sin\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1+\cot\text{x}}{\text{x}+\log\sin\text{x}}\text{dx}\ .....(\text{i})$
Let $\text{x}+\log\sin\text{x}=\text{t}$ then,
$\text{d}(\text{x}+\log\sin\text{x})=\text{dt}$
$\Rightarrow(1+\cot\text{x})\text{dx}=\text{dt}\Big[\because\frac{\text{d}}{\text{dx}}(\log\sin\text{x})=\cot\text{x}\Big]$
$\Rightarrow\text{dx}=\frac{\text{dt}}{1+\cot\text{x}}$
Putting $\text{x}+\log\sin\text{x}=\text{t}$ and $\text{dx}=\frac{\text{dt}}{1+\cot\text{x}}$ In equation (i), we get,
$\text{I}=\int\frac{1+\cot\text{x}}{\text{t}}\times\frac{\text{dt}}{1+\cot\text{x}}$
$=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|\text{x}+\log\sin\text{x}|+\text{C}$
$\therefore\text{I}=\log|\text{x}+\log\sin\text{x}|+\text{C}$
View full question & answer→Question 1745 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^2}{1-\text{x}^4}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2}{1-\text{x}^4}\ \text{dx}$
we express
$\frac{\text{x}^2}{1-\text{x}^4}=\frac{\text{x}^2}{(1-\text{x})^2(1+\text{x})^2}$
$=\frac{\text{A}}{1-\text{x}^2}+\frac{\text{B}}{1+\text{x}^2}$
$\Rightarrow\text{x}^2=\text{A}(1+\text{x}^2)+\text{B}(1-\text{x})^2$
Equating the coefficient of x and constants, we get
1 = A - B and 0 = A + B or
$\text{A}=\frac{1}{2}$ and $\text{B}=-\frac{1}{2}$
$\therefore\text{I}=\int\Big(\frac{\frac{1}{2}}{1-\text{x}^2}+\frac{-\frac{1}{2}}{1+\text{x}^2}\Big)\text{dx}$
$=\frac{1}{2}\int\frac{1}{1-\text{x}^2}\ \text{dx}-\frac{1}{2}\int\frac{1}{1+\text{x}^2}\ \text{dx}$
$=\frac{1}{2}\times\frac{1}{2}\log\Big|\frac{1+\text{x}}{1-\text{x}}\Big|-\frac{1}{2}\tan^{-1}\text{x}+\text{C}$
$=\frac{1}{4}\log\Big|\frac{1+\text{x}}{1-\text{x}}\Big|-\frac{1}{2}\tan^{-1}\text{t}+\text{C}$
Hence, $\int\frac{\text{x}^2}{1-\text{x}^4}\ \text{dx}=\frac{1}{4}\log\Big|\frac{1+\text{x}}{1-\text{x}}\Big|-\frac{1}{2}\tan^{-1}\text{x}+\text{C}$
View full question & answer→Question 1755 Marks
Integrate the following integrals:
$\int\sin\text{x}\cos2\text{x}\sin3\text{x dx}$
Answer$\int\sin\text{x}\cos2\text{x}\sin3\text{x dx}$
$=\frac{1}{2}\int(2\sin\text{x}\cos2\text{x})\sin3\text{x dx}$
$=\frac{1}{2}\int\big[\sin(\text{x}+2\text{x})+\sin(\text{x}-2\text{x})\big]\sin(3\text{x) dx}$
$=\frac{1}{2}\int\big[\sin(3\text{x})-\sin(\text{x})\big]\sin(3\text{x) dx}$
$=\frac{1}{2}\big[\int\sin^2(3\text{x})\text{dx}-\int\sin(\text{x})\sin(3\text{x})\text{dx}\big]$
$=\frac{1}{4}\big[\int2\sin^2(3\text{x})\text{dx}-\int2\sin(\text{x})\sin(3\text{x})\text{dx}\big]$
$=\frac{1}{4}\Big\{\int\big[1-\cos(6\text{x})\big]\text{dx}-\int\big[\cos(\text{x}-3\text{x})-\cos(\text{x}+3\text{x})\big]\text{dx}\Big\}$
$=\frac{1}{4}\big[\int1\text{dx}-\int\cos(6\text{x})\text{dx}-\int\cos(2\text{x})\text{dx}+\int\cos(4\text{x})\text{dx}\big]$
$=\frac{1}{4}\Big[\text{x}-\frac{\sin(6\text{x})}{6}-\frac{\sin(2\text{x})}{2}+\frac{\sin(4\text{x})}{4}\Big]+\text{C}$
$=\frac{\text{x}}{4}-\frac{\sin(6\text{x})}{24}-\frac{\sin(2\text{x})}{8}+\frac{\sin(4\text{x})}{16}+\text{C}$
Hence, $\int\sin\text{x}\cos2\text{x}\sin3\text{x}\text{ dx}$ $=\frac{\text{x}}{4}-\frac{\sin(6\text{x})}{24}-\frac{\sin(2\text{x})}{8}+\frac{\sin(4\text{x})}{16}+\text{C}$
View full question & answer→Question 1765 Marks
Evaluate the following integrals:
$\int(4\text{x}+2)\sqrt{\text{x}^2+\text{x}+1}\text{ dx}$
Answer$\int(4\text{x}+2)\sqrt{\text{x}^2+\text{x}+1}\text{ dx}$
$=2\int(2\text{x}+1)\sqrt{\text{x}^2+\text{x}+1}\text{ dx}$
$\text{Let }\text{x}^2+\text{x}+1=\text{t}$
$\Rightarrow(2\text{x}+1)=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(2\text{x}+1)\text{dx}=\text{dt}$
$\text{Now, }2\int(2\text{x}+1)\sqrt{\text{x}^2+\text{x}+1}\text{ dx}$
$=2\int\sqrt{\text{t}}\text{ dt}$
$=2\int\text{t}^\frac{1}{2}\text{dt}$
$=2\bigg[\frac{\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\bigg]+\text{C}$
$=2\times\frac{2}{3}\text{t}^\frac{3}{2}+\text{C}$
$=\frac{4}{3}\text{t}^\frac{3}{2}+\text{C}$
$=\frac{4}{3}(\text{x}^2+\text{x}+1)^\frac{3}{2}+\text{C}$
View full question & answer→Question 1775 Marks
Evaluate the following integrals:
$\int\text{e}^{\text{x}}\sin^2\text{x }\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\sin^2\text{x }\text{dx}$
$=\frac{1}{2}\int\text{e}^\text{x}\ 2\sin^2\text{x dx}$
$=\frac{1}{2}\int\text{e}^\text{x}(1-\cos2\text{x})\text{dx}$
$=\frac{1}{2}\int\text{e}^\text{x}\text{dx}-\frac{1}{2}\int\text{e}^\text{x}\cos2\text{ x dx}$
$\because\ \int\text{e}^{2\text{x}}\cos\text{bx dx}=\frac{\text{e}^{2\text{x}}}{\text{a}^2+\text{b}^2}\{\text{a}\cos\text{bx}-\text{b}\sin\text{bx}\}+\text{C}$
$\therefore\ \text{I}=\frac{1}{2}\big[\text{e}^\text{x}-\frac{\text{e}^\text{x}}{5}\{\cos2\text{x}+2\sin2\text{x}\}\big]+\text{C}$
$\therefore\ \text{I}=\frac{\text{e}^\text{x}}{2}-\frac{\text{e}^\text{x}}{10}\{\cos2\text{x}+2\sin2\text{x}\}+\text{C}$
View full question & answer→Question 1785 Marks
Evaluate the following intregals:
$\int\frac{5\text{x}}{(\text{x}+1)(\text{x}^2-4)}\text{dx}$
Answer$\int\frac{5\text{x}}{(\text{x}+1)(\text{x}^2-4)}=\frac{5\text{x}}{(\text{x}+1)(\text{x}+2)(\text{x}-2)}$
Let $\frac{5\text{x}}{(\text{x}+1)(\text{x}+2)(\text{x}-2)}=\frac{\text{A}}{(\text{x}+1)}+\frac{\text{B}}{(\text{x}+2)}+\frac{\text{C}}{(\text{x}-2)}$
$5\text{x}=\text{A}(\text{x}+2)(\text{x}-2)+\text{B}(\text{x}+1)(\text{x}-2)\\+\text{C}(\text{x}+1)(\text{x}+2)\ \dots(1)$
Substituting x = -1, -2 and 2 respectively in equation (1), we obtain
$\text{A}=\frac{5}{3},\text{B}=\frac{5}{2},\text{and }\text{C}=\frac{5}{6}$
$\therefore\frac{5\text{x}}{(\text{x}+1)(\text{x}+2)(\text{x}-2)}=\frac{5}{3(\text{x}+1)}-\frac{5}{2(\text{x}+2)}+\frac{5}{6(\text{x}-2)}$
$\Rightarrow\frac{5\text{x}}{(\text{x}+1)(\text{x}^2-4)}\ \text{dx}=\frac{5}{3}\int\frac{1}{(\text{x}+1)}\ \text{dx}-\frac{5}{2}\int\frac{1}{(\text{x}+2)}\\\ \text{dx}+\frac{5}{6}\int\frac{1}{(\text{x}-2)}\ \text{dx}$
$=\frac{5}{3}\log|\text{x}+1|-\frac{5}{2}\log|\text{x}+2|+\frac{5}{6}\log|\text{x}-2|+\text{C}$
View full question & answer→Question 1795 Marks
Evaluate the following integrals:
$\int\cos\Big\{2\cot^{-1}\sqrt{\frac{1+\text{x}}{1-\text{x}}}\Big\}\text{dx}$
AnswerLet $\text{I}=\int\cos\Big\{2\cot^{-1}\sqrt{\frac{1+\text{x}}{1-\text{x}}}\Big\}\text{dx}$
Let $\text{x}=\cos2\theta$
On differentiating both sides, we get
$\text{dx}=-2\sin2\theta\text{ d}\theta$
$\therefore\ \text{I}=\int\cos\Big\{2\cot^{-1}\sqrt{\frac{1+\cos2\theta}{1-\cos2\theta}}\Big\}2\sin2\theta\text{ d}\theta$
$=-2\int\cos\Big\{2\cot^{-1}\sqrt{\frac{2\cos^2\theta}{2\sin^2\theta}}\Big\}\sin2\theta\text{ d}\theta$
$=-2\int\cos\{2\cot^{-1}(\cot\theta)\}\sin2\theta\text{ d}\theta$
$=-2\int\cos2\theta\sin2\theta\text{ d}\theta$
$=\frac{\cos4\theta}{4}+\text{C}_1$
$=-\int\sin4\theta\text{ d}\theta$
$=\frac{1}{4}(2\cos^2\theta-1)+\text{C}_1$
$=\frac{1}{2}\text{x}^2-\frac{1}{4}+\text{C}_1$
$=\frac{1}{2}\text{x}^2+\text{C},$ where $\text{C}=-\frac{1}{4}+\text{C}_1$
Hence, $\int\cos\Big\{2\cot^{-1}\sqrt{\frac{1+\text{x}}{1-\text{x}}}\Big\}\text{dx}=\frac{1}{2}\text{x}^2+\text{C}$
View full question & answer→Question 1805 Marks
Evaluate the following intregals:
$\int\frac{\text{x}^2}{(\text{x}-1)(\text{x}-2)(\text{x}-3)}\ \text{dx}$
Answer$\int\frac{\text{x}^2}{(\text{x}-1)(\text{x}-2)(\text{x}-3)}\ \text{dx}$
Let $\int\frac{\text{x}^2}{(\text{x}-1)(\text{x}-2)(\text{x}-3)}\ \text{dx}=\frac{\text{A}}{\text{x}-1}+\frac{\text{B}}{\text{x}-2}+\frac{\text{C}}{\text{x}-3}$
$\Rightarrow\frac{\text{x}^2}{(\text{x}^2-1)(\text{x}-2)(\text{x}-3)}=\frac{\text{A}(\text{x}-2)(\text{x}-3)+\text{B}(\text{x}-1)(\text{x}-3)+\text{C}(\text{x}-1)(\text{x}-2)}{(\text{x}-1)(\text{x}-2)(\text{x}-3)}$
$\Rightarrow\text{x}^2=\text{A}(\text{x}-2) (\text{x}-3)+\text{B}(\text{x}-1)(\text{x}-3)\\+\text{C}(\text{x}-1)(\text{x}-2)\ \dots(1)$
Putting x - 1 = 0 or x = 1 in eq (1)
⇒ 1 = A (1 - 2) (1 - 3)
⇒ 1 = A (-1) (-2)
$\text{A}=\frac{1}{2}$
Putting x - 2 = 0 or x = 2 in eq (1)
⇒ 4 = B (2 - 1)(2 - 3)
⇒ B = -4
Putting x - 3 = 0 or x = 3 in eq (1)
⇒ 9 = C (3 - 1) (3 - 2)
$\Rightarrow\text{C}=\frac{9}{2}$
$\therefore\frac{\text{x}^2}{(\text{x}-1)(\text{x}-2)(\text{x}-3)}=\frac{1}{2(\text{x}-1)}-\frac{4}{\text{x}-2}+\frac{9}{2(\text{x}-3)}$
$\int\frac{\text{x}^2}{(\text{x}-1)(\text{x}-2)(\text{x}-3)}\ \text{dx}=\frac{1}{2}\int\frac{1}{\text{x}-1}\ \text{dx}-4\int\frac{1}{\text{x}-2}+\frac{9}{2(\text{x}-3)}\ \text{dx}$
$=\frac{1}{2}\ln|\text{x}-1|-4\ln|\text{x}-2|+\frac{9}{2}\ln|\text{x}-3|+\text{C}$
View full question & answer→Question 1815 Marks
Evaluate the following integrals:$\int\frac{1}{\sqrt{7-3\text{x}-2\text{x}^2}}\text{ dx}$
Answer$\int\frac{\text{dx}}{\sqrt{7-3\text{x}-2\text{x}^2}}$ $=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{7}{2}-\frac{3}{2}\text{x}-\text{x}^2}}$ $=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{7}{2}-\big(\text{x}^2-\frac{3}{2}\text{x}\big)}}$ $=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\Big(\frac{\sqrt7}{\sqrt2}\Big)^2-\Big(\text{x}^2+\frac{3}{2}\text{x}+\big(\frac{3}{4}\big)^2-\big(\frac{3}{4}\big)^2\Big)}}$ $=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\Big(\frac{\sqrt7}{\sqrt2}\Big)^2-\big(\text{x}+\frac{3}{4}\big)^2+\frac{9}{16}}}$ $=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{7}{2}+\frac{9}{16}-\big(\text{x}+\frac{3}{4}\big)^2}}$ $=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{56+9}{16}-\big(\text{x}+\frac{3}{4}\big)^2}}$$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\Big(\frac{\sqrt{65}}{4}\Big)^2-\big(\text{x}+\frac{3}{4}\big)^2}}$
$=\frac{1}{\sqrt2}\sin^{-1}\Bigg[\frac{\text{x}+\frac{3}{4}}{\frac{\sqrt{65}}{4}}{}\Bigg]+\text{C}$ $=\frac{1}{\sqrt2}\sin^{-1}\Big[\frac{4\text{x}+3}{\sqrt{65}}\Big]+\text{C}$
View full question & answer→Question 1825 Marks
Evaluate the following integrals:
$\int\sqrt{1+\text{x}-2\text{x}^2}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{1+\text{x}-2\text{x}^2}\text{dx}$
$=\sqrt{2}\int\sqrt{\frac{1}{2}+\frac{\text{x}}{2}-\text{x}^2}\text{dx}$
$=\sqrt{2}\int\sqrt{\frac{9}{16}-\Big(\frac{1}{16}-\frac{\text{x}}{2}+\text{x}^2\Big)}\text{dx}$
$=\sqrt{2}\int\sqrt{\Big(\frac{3}{4}\Big)^2-\Big(\text{x}-\frac{1}{4}\Big)^2}\text{dx}$
$=\sqrt{2}\begin{Bmatrix}\frac{\Big(\text{x}-\frac{1}{4}\Big)}{2}\sqrt{\frac{1}{2}+\frac{\text{x}}{2}-\text{x}^2}+\frac{9}{32}\sin^{-1}\bigg(\frac{\text{x}-\frac{1}{4}}{\frac{3}{4}}\bigg)\end{Bmatrix}+\text{C}$
$\text{I}=\frac{1}{8}(4\text{x}-1)\sqrt{1+\text{x}-2\text{x}^2}+\frac{9\sqrt{2}}{32}\sin^{-1}\Big(\frac{4\text{x}-1}{3}\Big)+\text{C}$
View full question & answer→Question 1835 Marks
Evaluate the following intregals:
$\int\frac{5\text{x}^2-1}{\text{x}(\text{x}-1)(\text{x}+1)}\ \text{dx}$
AnswerLet $\int\frac{5\text{x}^2-1}{\text{x}(\text{x}-1)(\text{x}+1)}\ \text{dx}=\frac{\text{A}}{\text{x}}+\frac{\text{B}}{\text{x}-1}+\frac{\text{C}}{\text{x}+1}$
$\Rightarrow5\text{x}^2-1=\text{A}(\text{x}^2-1)+\text{B}(\text{x}+1)\text{x}+\text{C}(\text{x}-1)\text{x}$
Put x = 0
⇒ -1 = -A ⇒ A = 1
Put x = +1
⇒ 4 = 2B ⇒ B = 2
Put x = -1
⇒ 4 = 2C ⇒ C = 2
So,
$\text{I}=\int\frac{\text{dx}}{\text{x}}+\int\frac{2\text{dx}}{\text{x}-1}+\int\frac{2\text{dx}}{\text{x}+1}$
$=\log|\text{x}|+2\log|\text{x}-1|+2\log|\text{x}+1|+\text{C}$
$\text{I}=\log|\text{x}(\text{x}^2-1)^2|$
View full question & answer→Question 1845 Marks
Evaluate the following integrals:
$\int\tan^5\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\tan^5\text{x}\text{ dx}$ Then
$\text{I}=\int\tan^2\text{x}\tan^3\text{x}\text{ dx}$
$=\int(\sec^2\text{x}-1)\tan^3\text{x}\text{ dx}$
$=\int\sec^2\text{x}\tan^3\text{x}\text{ dx}-\int\tan^3\text{x}\text{ dx}$
$=\int\sec^2\tan^3\text{x}\text{ dx}-\int(\sec^2\text{x}-1)\tan\text{x}\text{ dx}$
Substituting $\tan\text{x}=\text{t}$ and $\sec^2\text{x}\text{ dx}=\text{dt}$ in first two integral, we get
$\text{I}=\int\text{t}^3\text{dt}-\int\text{tdt}+\int\tan\text{x}\text{ dx}$
$=\frac{\text{t}^4}{4}-\frac{\text{t}^2}{2}+\log|\sec\text{x}|+\text{C}$
$=\frac{\tan^4\text{x}}{4}-\frac{\tan^2\text{x}}{2}+\log|\sec\text{x}|+\text{C}$
$\therefore\ \text{I}=\frac{\tan^4\text{x}}{4}-\frac{\tan^2\text{x}}{2}+\log|\sec\text{x}|+\text{C}$
View full question & answer→Question 1855 Marks
Evalute the following integrals:
$\int\frac{1}{\text{e}^\text{x}+1}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\text{e}^\text{x}+1}\text{dx}$ then,
$\text{I}=\int\frac{1}{\text{e}^\text{x}\Big[1+\frac{1}{\text{e}^\text{x}}\Big]}\text{dx}$
$\Rightarrow\text{I}=\int\frac{1}{\text{e}^\text{x}\big[1+\text{e}^{-\text{x}}\big]}\text{dx}\ .....\text{(i)}$
Let $1+\text{e}^{-\text{x}}=\text{t}$ then,
$\text{d}(1+\text{e}^{-\text{x}})=\text{dt}$
$\Rightarrow-\text{e}^{-\text{x}}\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{-\text{dt}}{\text{e}^{-\text{x}}}$
$\text{dx}=-\text{dt}\times\text{e}^\text{x}$
Putting $1 + e^{-x} = t $and $dx = -e^x $dt in equation (i), we get,
$\text{I}=\int\frac{1}{\text{e}^\text{x}\times\text{t}}\times-\text{e}^\text{x}\text{dt}$
$=-\int\frac{\text{dt}}{\text{t}}$
$=-\log|\text{t}|+\text{C}$
$=-\log|1+\text{e}^{-\text{x}}|+\text{C}$
$\therefore -\log|1+\text{e}^{-\text{x}}|+\text{C}$
View full question & answer→Question 1865 Marks
Evaluate the following integrals:$\int\frac{\sin2\text{x}}{\sqrt{\sin^4\text{x}+4\sin^2\text{x}-2}}\text{ dx}$
Answer$\int\frac{\sin(2\text{x})\text{dx}}{\sqrt{\sin^4\text{x}+4\sin^2\text{x}-2}}$
Let $\sin^2\text{x}=\text{t}$
$\Rightarrow2\sin\text{x}\cos\text{x}\text{ dx}=\text{dt}$
$\Rightarrow\sin(2\text{x})\text{ dx}=\text{dt}$
Now, $\int\frac{\sin(2\text{x})\text{dx}}{\sqrt{\sin^4\text{x}+4\sin^2\text{x}-2}}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}^2+4\text{t}-2}}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}^2+4\text{t}+4-4-2}}$
$=\int\frac{\text{dt}}{\sqrt{(\text{t}+2)^2-\big(\sqrt6\big)^2}}$
$=\log\Big|\text{t}+2+\sqrt{(\text{t}+2)^2-6}\Big|+\text{C}$
$=\log\Big|\text{t}+2+\sqrt{\text{t}^2+4\text{t}-2}\Big|+\text{C}$
$=\log\Big|\sin^2\text{x}+2\sqrt{\sin^4\text{x}+4\sin^2\text{x}-2}\Big|+\text{C}$
View full question & answer→Question 1875 Marks
$\int\frac{\text{x}^2+5\text{x}+2}{\text{x}+2}\text{dx}$
Answer$\int\frac{(\text{x}^2+5\text{x}+2)}{(\text{x}+2)}\text{dx}$
$=\int\frac{\text{x}^2}{\text{x}+2}\text{dx}+5\int\frac{\text{x dx}}{\text{x}+2}+2\int\frac{\text{dx}}{\text{x}+2}$
$=\int\Big(\frac{\text{x}^2-4+4}{\text{x}+2}\Big)\text{dx}+5\int\Big(\frac{\text{x}+2-2}{\text{x}+2}\Big)\text{dx}+2\int\frac{\text{dx}}{\text{x}+2}$
$=\int\frac{(\text{x}-2)(\text{x}+2)}{(\text{x}+2)}\text{dx}+\int\frac{4}{\text{x}+2}\text{dx}+5\int\Big(1-\frac{2}{\text{x}+2}\Big)\text{dx}+2\int\frac{\text{dx}}{\text{x}+2}$
$=\int(\text{x}-2)\text{dx}+4\int\frac{\text{dx}}{\text{x}+2}+5\int\text{dx}-10\int\frac{\text{dx}}{\text{x}+2}+2\int\frac{\text{dx}}{\text{x}+2}$
$=\int(\text{x}-2)\text{dx}-4\int\frac{\text{dx}}{\text{x}+2}+5\int\text{dx}$
$=\Big(\frac{\text{x}^2}{2}-2\text{x}\Big)-4\text{ln|}\text{x}+2|+5\text{x}+\text{C}$
$=\frac{\text{x}^2}{2}+3\text{x}-4\text{ln}|\text{x}+2|+\text{C}$
View full question & answer→Question 1885 Marks
Evalute the following integrals:
$\int\frac{\cos4\text{x}-\cos2\text{x}}{\sin4\text{x}-\sin2\text{x}}\text{dx}$
Answer$\int\Big(\frac{\cos4\text{x}-\cos2\text{x}}{\sin4\text{x}-\sin2\text{x}}\Big)\text{dx}$
$=\int\frac{-2\sin\Big(\frac{4\text{x}+2\text{x}}{2}\Big)\sin\Big(\frac{4\text{x}-2\text{x}}{2}\Big)}{2\cos\Big(\frac{4\text{x}+2\text{x}}{2}\Big)\sin\Big(\frac{4\text{x}-2\text{x}}{2}\Big)}\text{dx}$
$\bigg[\because\cos\text{A}-\cos\text{B}=-2\sin\Big(\frac{\text{A}+\text{B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\ \& \\ \sin\text{A}-\sin\text{B}=2\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\bigg]$
$=-\int\frac{\sin3\text{x}}{\cos3\text{x}}\text{dx}$
$=-\int\tan3\text{x dx}$
$=\frac{-\text{In}|\sec3\text{x}|}{3}+\text{C}$
$=\frac{1}{3}\text{ln}\big(|\sec3\text{x}|\big)^{-1}+\text{C}$
$=\frac{1}{3}\text{ln}|\cos3\text{x}|+\text{C}$
View full question & answer→Question 1895 Marks
Evaluate the following intregals:
$\int\frac{\cos\text{x}}{(1-\sin\text{x})(2-\sin\text{x})}\ \text{dx}$
AnswerWe have,
$\text{I}=\int\frac{\cos\text{x}\text{ dx}}{(1-\sin\text{x})(2-\sin\text{x})}$
putting $\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{(1-\text{t})(2-\text{t})}$
$\therefore\text{I}=\int\frac{\text{dt}}{(1-\text{t})(2-\text{t})}$
Let $\frac{1}{(\text{t}-1)(\text{t}-2)}=\frac{\text{A}}{\text{t}-1}+\frac{ \text{B}}{\text{t}-2}$
$\Rightarrow\frac{1}{(\text{t}-1)(\text{t}-2)}=\frac{\text{A}(\text{t}-2)+\text{B}(\text{t}-1)}{(\text{t}-1)(\text{t}-2)}$
$\Rightarrow1=\text{A}(\text{t}-2)+\text{B}(\text{t}-1)$
Putting t - 1 = 0
⇒ t = 1
$\therefore$ 1 = A (1 - 2) + B × 0
⇒ A = -1
Putting t - 2 = 0
⇒ t = 2
$\therefore$ 1 = A × 0 + B(2 - 1)
⇒ B = 1
$\therefore\text{I}=\int\frac{-\text{dt}}{\text{t}-1}+\int\frac{\text{dt}}{\text{t}-2}$
$=-\log|\text{t}-1|+\log|\text{t}-2|+\text{C}$
$=\log\Big|\frac{\text{t}-2}{\text{t}-1}\Big|+\text{C}$
$=\log\Big|\frac{\sin\text{x}-2}{\sin\text{x}-1}\Big|+\text{C}$
$=\log\Big|\frac{2-\sin\text{x}}{1-\sin\text{x}}\Big|+\text{C}$
View full question & answer→Question 1905 Marks
Evaluate the following intregals:
$\int\frac{5\text{x}^2+20\text{x}+6}{\text{x}^2+2\text{x}^2+\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{5\text{x}^2+20\text{x}+6}{\text{x}^2+2\text{x}^2+\text{x}}\ \text{dx}$$=\int\frac{5\text{x}^2+20\text{x}+6}{\text{x}(\text{x}+1)^2}\ \text{dx}$
Now,
Let $\frac{5\text{x}^2+20\text{x}+6}{\text{x}(\text{x}+1)^2}=\frac{\text{A}}{\text{x}}+\frac{\text{B}}{\text{x}+1}+\frac{\text{C}}{(\text{x}+1)^2}$
$\Rightarrow5\text{x}^2+20\text{x}+6=\text{A}(\text{x}+1)^2+\text{Bx}(\text{x}+1)+\text{Cx}$
Equating similar terms, we get,
A + B = 5, 2A + B + C = 20, A = 6
Solving, we get, B = -1, C = 9
Thus,
$\text{I}=\int\frac{6\text{dx}}{\text{x}}-1\int\frac{\text{dx}}{\text{x}+1}+9\int\frac{\text{dx}}{(\text{x}+1)^2}$
$\therefore\text{I}=6\log|\text{x}|-\log|\text{x}+1|-\frac{9}{\text{x}+1}+\text{C}$
View full question & answer→Question 1915 Marks
Evaluate the following intregals:
$\int\frac{1}{4\cos^2\text{x}+3\sin^2\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{4\cos^2\text{x}+3\sin^2\text{x}}\ \text{dx}$
Dividing numerator and demnominator by $\cos^2\text{x}$
$=\int\frac{\frac{1}{\cos^2\text{x}}}{4+9\tan^2\text{x}} \text {dx}$
$\text{I}=\int\frac{\sec^2\text{x}}{4+9\tan^2\text{x}}\ \text{dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x}\text{ dx}=\text{dt}$
$\text{I}=\int\frac{\text{dt}}{4+9(\text{t})^2}$
$=\int\frac{\text{dt}}{4+(3\text{t})^2}$
Let $3\text{t}=\text{u}$
$3\text{dt}=\text{du}$
$\text{I}=\frac{1}{3}\int\frac{\text{du}}{(2)^2+(\text{u})^2}$
$=\frac{1}{3}\times\frac{1}{2}\times\tan^{-1}\Big(\frac{\text{u}}{2}\Big)+\text{C}$
$\text{I}=\frac{1}{6}\tan^{-1}\Big(\frac{3\text{t}}{2}\Big)+\text{C}$
$\text{I}=\frac{1}{6}\tan^{-1}\Big(\frac{3\tan\text{x}}{2}\Big)+\text{C}$
View full question & answer→Question 1925 Marks
Evaluate the following intergrals:
$\int\text{e}^\text{ax}\cos\text{bx dx}$
AnswerLet $\text{I}=\int\text{e}^\text{ax}\cos\text{bx dx}$
Intergrating by parts,
$\text{I}=\text{e}^\text{ax}\frac{\sin\text{bx}}{\text{b}}-\text{a}\int\text{e}^\text{ax}\frac{\sin\text{bx}}{\text{x}}\text{dx}$
$=\frac{1}{\text{b}}\text{e}^\text{ax}\sin\text{bx}-\frac{\text{a}}{\text{b}}\int\text{e}^\text{ax}\sin\text{bx dx}$
$=\frac{1}{\text{a}}\text{e}^\text{ax}\sin\text{bx}-\frac{\text{a}}{\text{b}}\Big[-\text{e}^\text{ax}\frac{\cos\text{bx}}{\text{b}}+\int\text{ae}^\text{ax}\frac{\cos\text{bx}}{\text{b}}\text{dx}\Big]$
$=\frac{1}{\text{a}}\text{e}^\text{ax}\sin\text{bx}-\frac{\text{a}}{\text{b}^2}\text{e}^\text{ax}\cos\text{bx}-\frac{\text{a}^2}{\text{b}^2}\int\text{e}^\text{ax}\cos\text{bx dx}$
$\Rightarrow\text{I}=\frac{\text{e}^\text{ax}}{\text{b}^2}\big[\text{b}\sin\text{bx}+\text{a}\cos\text{bx}\big]-\frac{\text{a}^2}{\text{b}^2}\text{I}+\text{C}$
$\Rightarrow\text{I}.\Big\{\frac{\text{a}^2+\text{b}^2}{\text{b}^2}\Big\}=\frac{\text{e}^\text{ax}}{\text{b}^2}\big[\text{b}\cos\text{bx}+\text{a}\cos\text{bx}\big]+\text{C}$
Thus,
$\text{I}=\frac{\text{e}^\text{ax}}{\text{a}^2+\text{b}^2}\big[\text{a}\cos\text{bx}+\text{a}\cos\text{bx}\big]+\text{C}$
View full question & answer→Question 1935 Marks
Evalute the following integrals:
$\int\frac{\sec\text{x}}{\log(\sec\text{x}+\tan\text{x})}\text{dx}$
AnswerNote: Here, we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$
Let $\text{I}=\int\frac{\sec\text{x}}{\log(\sec\text{x}+\tan\text{x})}\text{dx}$
Putting $\log(\sec\text{x}+\tan\text{x})=\text{t}$
$\Rightarrow\frac{\sec\text{x}\tan\text{x}+\sec^2\text{x}}{\sec\text{x}+\tan\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\sec\text{x}\frac{(\sec\text{x}+\tan\text{x})}{\sec\text{x}+\tan\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\sec\text{x }\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|\log(\sec\text{x}+\tan\text{x})|+\text{C}$
View full question & answer→Question 1945 Marks
Evalute the following integrals:
$\int\frac{\cos\text{x}}{2+3\sin\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\cos\text{x}}{2+3\sin\text{x}}\text{dx}\ .....\text{(i)}$
Let $2+3\sin\text{x}=\text{t}$ then,
$\text{d}(2+3\sin\text{x})=\text{dt}$
$\Rightarrow3\cos\text{x dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{3\cos\text{x}}$
Putting $2+3\sin\text{x}=\text{t and dx}=\frac{\text{dt}}{3\cos\text{x}}$ in equation (i), we get,
$\text{I}=\int\frac{\cos\text{x}}{\text{t}}\times\frac{\text{dt}}{3\cos\text{x}}$
$=\frac{1}{3}\int\frac{\text{dt}}{\text{t}}$
$=\frac{1}{3}\log|\text{t}|+\text{C}$
$=\frac{1}{3}\log|2+3\sin\text{x}|+\text{C}$
View full question & answer→Question 1955 Marks
Evaluate the following integrals:
$\int \frac{\text{x}+1}{(\text{x}-1)\sqrt{\text{x}+2}}\text{ dx}$
Answer Let $\text{I}=\int \frac{\text{x}+1}{(\text{x}-1)\sqrt{\text{x}+2}}\text{ dx}$
$\text{I}=\int\frac{(\text{x}-1)+2}{(\text{x}-1)\sqrt{\text{x}+2}}\text{ dx}$
$\text{I}=\int\frac{\text{dx}}{\sqrt{\text{x}+2}}+2\int\frac{\text{dx}}{(\text{x}+1)\sqrt{\text{x}+2}}\ ...(\text{i})$
Now, $\int\frac{\text{dx}}{\sqrt{\text{x}+2}}+2\sqrt{\text{x}+2}+\text{C}_1$
and, $\int \frac{1}{(\text{x}-1)\sqrt{\text{x}+2}}\text{ dx}$
Let $\text{x}+2=\text{t}^2$
$\text{dx}=2\text{t dt}$
$\therefore\ \int \frac{1}{(\text{x}-1)\sqrt{\text{x}+2}}=2\int\frac{\text{t dt}}{(\text{t}^2-3)\text{t}}=2\int\frac{\text{dt}}{\text{t}^2-3}$
$=\frac{2\times1}{2\sqrt{3}}\log\bigg|\frac{\text{t}-\sqrt{3}}{\text{t}+\sqrt{3}}\bigg|+\text{C}_2$
$=\frac{1}{\sqrt{3}}\log\bigg|\frac{\sqrt{\text{x}+2}-\sqrt{3}}{\sqrt{\text{x}+2}+\sqrt{3}}\bigg|+\text{C}_2$
Thus, from (i)
$\text{I}=2\sqrt{\text{x}+2}+\text{C}_1+\frac{2}{\sqrt{3}}\log\bigg|\frac{\sqrt{\text{x}+2}-\sqrt{3}}{\sqrt{\text{x}+2}+\sqrt{3}}\bigg|+\text{C}_2$
Hence, $\text{I}=2\sqrt{\text{x}+2}+\frac{2}{\sqrt{3}}\log\bigg|\frac{\sqrt{\text{x}+2}-\sqrt{3}}{\sqrt{\text{x}+2}+\sqrt{3}}\bigg|+\text{C}$
View full question & answer→Question 1965 Marks
$\int\frac{\text{x}^2+3\text{x}+1}{(\text{x}+1)^2}\text{dx}$
Answer$\int\Big(\frac{\text{x}^2+3\text{x}+1}{(\text{x}+1)^2}\Big)\text{dx}$
Let $\text{x}+1=\text{t}$
$\Rightarrow\text{x}=\text{t}-1$
$\Rightarrow1=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{dx}=\text{dt}$
Now, $\int\Big(\frac{\text{x}^2+3\text{x}+1}{(\text{x}+1)^2}\Big)\text{dx}$
$=\int\Big[\frac{(\text{t}-1)^2+3(\text{t}-1)+1}{\text{t}^2}\Big]\text{dt}$
$=\int\Big(\frac{\text{t}^2-2\text{t}+1+3\text{t}-3+1}{\text{t}^2}\Big)\text{dt}$
$=\int\Big(\frac{\text{t}^2+\text{t}-1}{\text{t}^2}\Big)\text{dt}$
$=\int\Big(1+\frac{1}{\text{t}}-\text{t}^{-2}\Big)\text{dt}$
$=\text{t}+\log|\text{t}|-\frac{\text{t}^{-2+1}}{-2+1}+\text{C}$
$=\text{t}+\log|\text{t}|-\frac{\text{1}}{\text{t}}+\text{C}$
$=\text{x}+1+\log|\text{x+1}|+\frac{1}{\text{x}+1}+\text{C}$
Let $1+\text{C}=\text{C}'$
$=\text{x}+\log|\text{x+1}|+\frac{1}{\text{x}+1}+\text{C}'$
View full question & answer→Question 1975 Marks
Evaluate the following intregals:
$\int\frac{2\text{x}+1}{(\text{x}-2)(\text{x}-3)}\ \text{dx}$
AnswerLet $\frac{2\text{x}+1}{(\text{x}-2)(\text{x}-3)}=\frac{\text{A}}{(\text{x}-2)}+\frac{\text{B}}{(\text{x}-3)}$
$\Rightarrow2\text{x}+1=\text{A}(\text{x}-3)+\text{B}(\text{x}-2)$
$=(\text{A}+\text{B})\text{x}+(-3\text{A}-2\text{B})$
equating similar terms, we get
A + B = 2, and -3A - 2B = 1
Thus,
$\text{I}=-5\int\frac{\text{dx}}{\text{x}-2}+7\int\frac{\text{dx}}{\text{x}-3}$
$=-5\log|\text{x}-2|+7\log|\text{x}-3|+\text{C}$
$\text{I}=\log\Big|\frac{(\text{x}-3)^7}{(\text{x}-2)^5}\Big|+\text{C}$
View full question & answer→Question 1985 Marks
Evaluate the following integrals:$\int\frac{1}{\sqrt{3\text{x}^2+5\text{x}+7}}\text{ dx}$
Answer$\int\frac{\text{dx}}{\sqrt{3\text{x}^2+5\text{x}+7}}$
$=\int\frac{\text{dx}}{\sqrt{3\big(\text{x}^2+\frac{5}{3}\text{x}+\frac{7}{3}}\big)}$
$=\frac{1}{\sqrt3}\int\frac{\text{dx}}{\sqrt{\text{x}^2+\frac{5}{3}\text{x}+\big(\frac{5}{6}\big)^2-\big(\frac{5}{6}\big)^2+\frac{7}{3}}}$
$=\frac{1}{\sqrt3}\int\frac{\text{dx}}{\sqrt{\big(\text{x}+\frac{5}{6}\big)^2-\frac{25}{36}+\frac{7}{3}}}$
$=\frac{1}{\sqrt3}\int\frac{\text{dx}}{\sqrt{\big(\text{x}+\frac{5}{6}\big)^2+\frac{-25+84}{36}}}$
$=\frac{1}{\sqrt3}\int\frac{\text{dx}}{\sqrt{\big(\text{x}+\frac{5}{6}\big)^2+\frac{59}{36}}}$
$=\frac{1}{\sqrt3}\int\frac{\text{dx}}{\sqrt{\big(\text{x}+\frac{5}{6}\big)^2+\Big(\sqrt{\frac{59}{36}}\Big)^2}}$
$=\frac{1}{\sqrt3}\log\Bigg|\text{x}+\frac{5}{6}+\sqrt{\big(\text{x}+\frac{5}{6}\big)^2+\frac{59}{36}}\Bigg|+\text{C}$
$=\frac{1}{\sqrt3}\log\Bigg|\text{x}+\frac{5}{6}+\sqrt{\text{x}^2+\frac{5}{3}\text{x}+\frac{7}{3}}\Bigg|+\text{C}$
View full question & answer→Question 1995 Marks
Evaluate the following intregals:
$\int\frac{1}{5-4\sin\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{5-4\sin\text{x}}\ \text{dx}$
Put $\sin\text{x}=\frac{2\tan^\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}$
$\text{I}=\int\frac{1}{5-4\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)}\ \text{dx}$
$=\int\frac{1+\tan^2\frac{\text{x}}{2}}{5\Big(1+\tan^2\frac{\text{x}}{2}\Big)-4\Big(2-\tan\frac{\text{x}}{2}\Big)}\ \text{dx}$
$=\int\frac{\sec^2\frac{\text{x}}{2}}{5+5\tan^2\frac{\text{x}}{2}-8\tan\frac{\text{x}}{2}}\ \text{dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$
$\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{dx}=\text{dt}$
$=\int\frac{2\text{dt}}{5\text{t}^2-8\text{t}+5}$
$=\frac{2}{5}\int\frac{\text{dt}}{\text{t}^2-2\text{t}\Big(\frac{4}{5}\Big)+\Big(\frac{4}{5}\Big)^2-\Big(\frac{4}{5}\Big)^2+1}$
$\text{I}=\frac{2}{5}\int\frac{\text{dt}}{\Big(\text{t}-\frac{4}{5}\Big)^2+\Big(\frac{3}{5}\Big)^2}$
$=\frac{2}{5}\times\frac{1}{\frac{3}{5}}\Bigg(\frac{\text{t}-\frac{4}{5}}{\frac{3}{5}}\Bigg)+\text{C}$
$=\frac{2}{3}\tan^{-1}\Big(\frac{5\text{t}-4}{3}\Big)+\text{C}$
$\text{I}=\frac{2}{3}\tan^{-1}\Big(\frac{5\tan\frac{\text{x}}{2}-4}{3}\Big)+\text{C}$
View full question & answer→Question 2005 Marks
Evaluate the following integrals:
$\int\text{e}^{\text{x}}\frac{\text{x}-1}{(\text{x}+1)^3}\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\frac{\text{x}+1-2}{(\text{x}+1)^3}\text{dx}$
$=\int\text{e}^{\text{x}}\Big\{\frac{1}{(\text{x}+1)^2}+\frac{-2}{({\text{x}+1})^3}\Big\}\text{dx}$
$=\int\text{e}^{\text{x}}\frac{1}{(\text{x}+1)^2}\text{dx}+\int\text{e}^{\text{x}}\frac{(-2)}{(\text{x}+1)^3}\text{dx}$
Integrating by parts
$=\text{e}^{\text{x}}\frac{1}{(\text{x}+1)^2}-\int\text{e}^{\text{x}}\Big(\frac{\text{d}}{\text{dx}}(\text{x}+1)^{-2}\Big)\text{dx}+\int\text{e}^{\text{x}}\frac{(-2)}{(\text{x}+1)^3}\text{dx}$
$=\text{e}^{\text{x}}\frac{1}{(\text{x}+1)^2}-\int\text{e}^{\text{x}}\frac{(-2)}{\text{(x+1)}^3}\text{dx}+\int\text{e}^{\text{x}}\frac{(-2)}{(\text{x}+1)^3}\text{dx}$
$=\text{e}^{\text{x}}\frac{1}{(\text{x}+1)}+\text{C}$
View full question & answer→