Question 2015 Marks
Evaluate the following intregals:
$\int\frac{1}{\text{x}(\text{x}^4+1)}\ \text{dx}$
AnswerLet $\frac{1}{\text{x}(\text{x}^4+1)}+\frac{\text{A}}{\text{x}}+\frac{\text{Bx}^3+\text{Cx}^2+\text{Dx}+\text{E}}{\text{x}^4+1}$$\Rightarrow1=\text{A}(\text{x}^4+1)+(\text{Bx}^3+\text{Cx}^2+\text{Dx}+\text{E})\text{x}$
$=(\text{A}+\text{B})\text{x}^4+\text{Cx}^3+\text{Dx}^2+\text{Ex}+\text{A}$
Equating similar terms, we get,
$\text{A}+\text{B}+0,\text{C}=0,\text{E}=0,\text{A}=1$
$\therefore\text{B}=-1$
Thus,
$\text{I}=\int\frac{\text{dx}}{\text{x}}+\int-\frac{\text{x}^3\text{dx}}{\text{x}^4+1}$
$=\log|\text{x}|-\frac{1}{4}\log|\text{x}^4+1|+\text{C}$
$\text{I}=\frac{1}{4}\log\Big|\frac{\text{x}^4}{\text{x}^4+1}\Big|+\text{C}$
View full question & answer→Question 2025 Marks
Evaluate the following integrals:$\int\text{e}^{\text{x}}.\frac{\sqrt{1-\text{x}^2}\sin^{-1}\text{x}+1}{\sqrt{1-\text{x}^2}}\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\bigg[\frac{\sqrt{1-\text{x}^2}\sin^{-1}\text{x}+1}{\sqrt{1-\text{x}^2}}\bigg]\text{dx}$
$=\int\text{e}^{\text{x}}\Big[\sin^{-1}\text{x}+\frac{1}{\sqrt{1-\text{x}^2}}\Big]\text{dx}$
Here, $\text{f(x)}=\sin^{-1}\text{x}$
$\Rightarrow\text{f}'\text{(x)}=\frac{1}{\sqrt{1-\text{x}^2}}$
Put $\text{e}^{\text{x}}\text{f(x)}=\text{t}$
$\Rightarrow\text{e}^{\text{x}}\sin^{-1}\text{x}=\text{t}$
Diff both sides w.r.t x
$\Big(\text{e}^{\text{x}}\sin^{-1}\text{x}+\text{e}^{\text{x}}\times\frac{1}{\sqrt{1-\text{x}^2}}\Big)\text{dx = dt}$
$\because\text{I}=\int\text{dt}$
$=\text{t + C}$
$=\text{e}^{\text{x}}\sin^{-1}\text{x + C}$
View full question & answer→Question 2035 Marks
Evaluate the following integrals:$\int\frac{\text{x}^2+\text{x}-1}{\text{x}^2+\text{x}-6}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2+\text{x}-1}{\text{x}^2+\text{x}-6}\text{ dx}$
$=\int\Big[1+\frac{5}{\text{x}^2+\text{x}-6}\Big]\text{dx}$
$=\text{x}+\int\frac{5}{\text{x}^2+\text{x}-6}\text{ dx}+\text{C}_1\ ....(1)$
$\text{I}_1=5\int\frac{1}{\text{x}^2+\text{x}-6}\text{ dx}$
$=5\int\frac{1}{\text{x}^2+2\text{x}\big(\frac{1}{2}\big)+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2-6}\text{ dx}$
$=5\int\frac{1}{\big(\text{x}+\frac{1}{2}\big)^2-\big(\frac{5}{2}\big)^2}\text{ dx}$
$5\times\frac{1}{2\big(\frac{5}{2}\big)}\log\bigg|\frac{\text{x}+\frac{1}{2}-\frac{5}{2}}{\text{x}+\frac{1}{2}+\frac{5}{2}}\bigg|+\text{C}_2$
$\Big[\text{since},\int\frac{1}{\text{x}^2-\text{a}^2}\text{ dx}=\frac{1}{2\text{a}}\log\Big|\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\Big|+\text{C}\Big]$
$\text{I}_1=\log\Big|\frac{\text{x}-2}{\text{x}+3}\Big|+\text{C}_2\ ....(2)$
Using equation (1) and (2)
$\text{I}=\text{x}+\log\Big|\frac{\text{x}-2}{\text{x}+3}\Big|+\text{C}$
View full question & answer→Question 2045 Marks
Evaluate the following integrals:$\int\sin\text{x}\log(\cos\text{x})\text{dx}$
AnswerLet $\text{I}=\int\sin\text{x}\cdot\log(\cos\text{x})\text{dx}$
Let $\cos\text{x = t}$
$\Rightarrow-\sin\text{x dx =}\text{ dt}$
$\Rightarrow\sin\text{x dx =}-\text{dt}$
$\therefore\text{I}=-\int\log\text{t dt}$
$=-\int1\cdot\log\text{t dt}$
Taking log t as the first function and 1 as the second function.
$=\log\text{t}\int1\text{dt}-\int\big\{\frac{\text{d}}{\text{dt}}(\log\text{t})\int1\text{dt}\big\}\text{dt}$
$=-[\log\text{t}\cdot\text{t}-\int\frac{1}{\text{t}}\times\text{t dt}]$
$=-[\log\text{t}\cdot\text{t}-\text{t}]+\text{C}$
$=-\text{t}(\log\text{t}-1)+\text{C} \dots(1)$
Substituting the value of t in eq (1)
$=-\cos\text{x}\{\log(\cos\text{x})-1\}+\text{C}$
$=\cos\text{x}\{1-\log(\cos\text{x})\}+\text{C}$
View full question & answer→Question 2055 Marks
Evaluate the following integrals:$\int\frac{1}{\sqrt{(\text{x}-\alpha)(\beta-\text{x})}}\text{ dx},(\beta>\alpha)$
AnswerLet $\text{I}=\int\frac{1}{\sqrt{(\text{x}-\alpha)(\beta-\text{x})}}\text{ dx},(\beta>\alpha)$
$=\int\frac{1}{-\text{x}^2-\text{x}(\alpha+\beta)-\alpha\beta}\text{ dx}$
$=\int\frac{1}{\sqrt{-\Big[\text{x}^2-2\text{x}\big(\frac{\alpha+\beta}{2}\big)+\big(\frac{\alpha+\beta}{2}\big)^2-\big(\frac{\alpha+\beta}{2}\big)^2+\alpha\beta\Big]}}\text{ dx}$
$=\int\frac{1}{\sqrt{-\Big[\big(\text{x}-\frac{\alpha+\beta}{2}\big)^2-\big(\frac{\alpha+\beta}{2}\big)^2\Big]}}\text{ dx}$
$=\int\frac{1}{\sqrt{\Big[\big(\frac{\beta-\alpha}{2}\big)^2-\big(\text{x}-\frac{\alpha+\beta}{2}\big)^2\Big]}}\text{ dx}$ $[\therefore\ \beta>\alpha]$
Let $\Big(\text{x}-\frac{\alpha+\beta}{2}\Big)=\text{t}$
$\Rightarrow\text{dx}=\text{dt}$
$\text{I}=\int\frac{1}{\sqrt{\big(\frac{\beta-\alpha}{2}\big)^2-\text{t}^2}}\text{ dt}$
$\text{I}=\sin^{-1}\bigg(\frac{\text{t}}{\frac{\beta-\alpha}{2}}\bigg)+\text{C}$ $\Big[\text{Since }\int\frac{1}{\sqrt{\text{a}^2-\text{x}^2}}\text{ dx}=\sin^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$
$\text{I}=\sin^{-1}\Bigg(\frac{2\big(\text{x}-\frac{\alpha+\beta}{2}\big)}{\beta-\alpha}\Bigg)+\text{C}$
$\text{I}=\sin^{-1}\Big(\frac{2\text{x}-\alpha-\beta}{\beta-\alpha}\Big)+\text{C}$
View full question & answer→Question 2065 Marks
Evaluate the following integrals:
$\int\text{x}\cos^3\text{x}^2\sin\text{x}^2\text{ dx}$
Answer$\int\text{x}\cos^3\text{x}^2\sin\text{x}^2\text{ dx}$
Let $\text{x}^2=\text{t}$
$2\text{x}\text{dx}=\text{dt}$
$\text{x}\text{ dx}=\frac{\text{dt}}{2}$
Now, $\int\text{x}\cos^3\text{x}^2\sin\text{x}^2\text{ dx}$
$=\frac{1}{2}\int\cos^3\text{t}\cdot\sin\text{t}\text{ dt}$
Again let $\cos\text{t}=\text{p}$
$-\sin\text{t}\text{ dt}=\text{dp}$
$\sin\text{t}\text{ dt}=-\text{dp}$
So, $\frac{1}{2}\int\cos^3\text{t}\cdot\sin\text{t}\text{ dt}$
$=-\frac{1}{2}\text{p}^3\text{ dp}$
$=-\frac{1}{2}\Big(\frac{\text{p}^4}{4}\Big)+\text{C}$
$=-\frac{\text{p}^4}{8}+\text{C}$
$=-\frac{\cos^4\text{t}}{8}+\text{C}$
$=-\frac{\cos^4\text{x}^2}{8}+\text{C}$
View full question & answer→Question 2075 Marks
Evaluate the following intregals:
$\int\frac{2\text{x}+1}{(\text{x}+2)(\text{x}-3)^2}\text{ dx}$
AnswerLet $\frac{2\text{x}+1}{(\text{x}+2)(\text{x}-3)^2}=\frac{\text{A}}{\text{x}+2}+\frac{\text{B}}{\text{x}-3}+\frac{\text{C}}{(\text{x}-3)^2}$
$\Rightarrow 2x + 1 = A (x - 3)^2 + B (x + 2) (x - 3) + C (x + 2) = (A + B)x^2 + (-6A - B + C)x + (9A - 6B + 2C)$
Equating similar terms,
we get, $A + B = 0 $
$\Rightarrow A = -B -6A - B +C = 2 $
$\Rightarrow 5B + C = 2 9A - 6B + 2C = 1 $
$\Rightarrow -15B + 2C = 1$
Solving, we get, $\text{B}=\frac{3}{25},\text{C}=\frac{7}{5},\text{A}=-\frac{3}{25}$
thus,
$\text{I}=-\frac{3}{25}\int\frac{\text{dx}}{\text{x}+2}+\frac{3}{25}\int\frac{\text{dx}}{\text{x}-3}+\frac{7}{5}\int\frac{\text{dx}}{(\text{x}-3)^2}$
$\text{I}=-\frac{3}{25}\log|\text{x}+2|+\frac{3}{25}\log|\text{x}-3|-\frac{7}{5(\text{x}-3)}+\text{C}$
View full question & answer→Question 2085 Marks
Evalute the following integrals:
$\int\frac{\sin(\text{x}-\text{a})}{\sin(\text{x}-\text{b})}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sin(\text{x}-\text{a})}{\sin(\text{x}-\text{b})}\text{dx}$
Putting x - b = t
⇒ x = b + t
& dx = dt
$\therefore\text{I}=\int\frac{\sin(\text{b}+\text{t}-\text{a})}{\sin\text{t}}\text{dt}$
$=\int\frac{\sin\big\{(\text{b}-\text{a})+\text{t}\big\}}{\sin\text{t}}\text{dt}$
$=\int\frac{\sin(\text{b}-\text{a})\cos\text{t}}{\sin\text{t}}+\int\frac{\cos(\text{b}-\text{a})\sin\text{t}}{\sin\text{t}}\text{dt}$
$=\int\sin(\text{b}-\text{a})\cot\text{t dt}+\int\cos(\text{b}-\text{a})\text{dt}$
$=\sin(\text{b}-\text{a})\text{in}|\sin\text{t}\big|+\text{t}\cos(\text{b}-\text{a})+\text{C}$
$=\sin(\text{b}-\text{a})\text{in}|\sin(\text{x}-\text{b})|+(\text{x}-\text{b})\cos(\text{b}-\text{a})+\text{C}$
$\big[\because\text{t}=\text{x}-\text{b}\big]$
View full question & answer→Question 2095 Marks
Evaluate the following integrals:$\int\text{e}^{\text{x}}\frac{(1-\text{x})^2}{(1+\text{x}^2)^2}\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\bigg[\frac{(1-\text{x})^2}{(1+\text{x}^2)^2}\bigg]\text{dx}$
$=\int\text{e}^\text{x}\bigg[\frac{1+\text{x}^2-2\text{x}}{(1+\text{x}^2)^2}\bigg]\text{dx}$
$=\int\text{e}^{\text{x}}\bigg[\frac{1}{1+\text{x}^2}-\frac{2\text{x}}{(1+\text{x}^2)^2}\bigg]\text{dx}$
Here, $\text{f(x)}=\frac{1}{1+\text{x}^2}$
$\Rightarrow\text{f}'(\text{x})=\frac{-2\text{x}}{(1+\text{x}^2)^2}$
Put $\text{e}^{\text{x}}\text{f(x)}=\text{t}$
$\Rightarrow\text{e}^{\text{x}}\frac{1}{1+\text{x}^2}=\text{t}$
Diff. both sides w.r.t w
$\text{e}^{\text{x}}\frac{1}{1+\text{x}^2}+\text{e}^{\text{x}}\frac{-1}{(1+\text{x}^2)^2}2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^{\text{x}}\bigg[\frac{1}{1+\text{x}^2}-\frac{2\text{x}}{(1+\text{x}^2)^2}\bigg]\text{dx = dt}$
$\therefore\int\text{e}^{\text{x}}\bigg[\frac{1}{1+\text{x}^2}-\frac{2\text{x}}{(1+\text{x}^2)^2}\bigg]\text{dx}=\int\text{dt}$
$\Rightarrow\text{I}=\text{t}+\text{C}$
$=\frac{\text{e}^{\text{x}}}{1+\text{x}^2}+\text{C}$
View full question & answer→Question 2105 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^2}{(\text{a}-\text{x}^2)^{\frac{3}{2}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2}{(\text{a}-\text{x}^2)^{\frac{3}{2}}}\text{ dx}$
Let $\text{x}=\text{a}\cos\theta$
On differentiating both sides, we get
$\text{dx}=-\text{a}\sin\theta\text{ d}\theta$
$\therefore\ \text{I}=\int\frac{\text{a}^2\cos^2\theta}{(\text{a}^2-\text{a}^2\cos^2\theta)^\frac{3}{2}}\times-\text{a}\sin\theta\text{ d}\theta$
$=-\int\frac{\text{a}^3\cos^2\theta\sin\theta}{\text{a}^3(1-\cos^2\theta)^{\frac{3}{2}}}\text{ d}\theta$
$=-\int\frac{\cos^2\theta\sin\theta}{\sin^3\theta}\text{ d}\theta$
$=-\int\cot^2\theta\text{ d}\theta$
$=-\int(\text{cosec}^2\theta-1)\text{d}\theta$
$=-(-\cot\theta-\theta)+\text{C}$
$=\cot\theta+\theta+\text{C}$
$=\cot\Big(\cos^{-1}\frac{\text{x}}{\text{a}}\Big)+\cos^{-1}\frac{\text{x}}{\text{a}}+\text{C}$
$=\cot\Big(\cos^{-1}\frac{\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}\Big)+\cos^{-1}\frac{\text{x}}{\text{a}}+\text{C}$
$=\frac{\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}+\cos^{-1}\frac{\text{x}}{\text{a}}+\text{C}$
Hence, $\int\frac{\text{x}^2}{(\text{a}^2-\text{x}^2)^{\frac{3}{2}}}\text{ dx}=\frac{\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}+\cos^{-1}\frac{\text{x}}{\text{a}}+\text{C}$
View full question & answer→Question 2115 Marks
Evaluate the following integrals:
$\int\frac{1}{(1+\text{x}^2)\sqrt{1-\text{x}^2}}\text{ dx}$
AnswerWe have
$\text{I}=\int\frac{1}{(1+\text{x}^2)\sqrt{1-\text{x}^2}}\text{ dx}$
Putting $\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(1+\frac{1}{\text{t}^2}\big)\sqrt{1-\frac{1}{\text{t}^2}}}$
$=\int\frac{-\frac{1}{\text{t}^2}\text{ dt}}{\frac{(\text{t}^2+1)}{\text{t}^2}\frac{\sqrt{\text{t}^2-1}}{\text{t}}}$
$=-\int\frac{\text{t dt}}{(\text{t}^2+1)\sqrt{\text{t}^2-1}}$
Again Putting $\text{t}^2-1=\text{u}^2$
$2\text{tdt}=2\text{udu}$
$\text{tdt}=\text{udu}$
$\therefore\ \text{I}=-\int\frac{\text{u du}}{(\text{u}^2+2)\text{u}}$
$=-\int\frac{\text{du}}{\text{u}^2+(\sqrt{2})^2}$
$=-\frac{1}{\sqrt{2}}\tan^{-1}\Big(\frac{\text{u}}{\sqrt{2}}\Big)+\text{C}$
$=-\frac{1}{\sqrt{2}}\tan^{-1}\bigg(\frac{\sqrt{\text{t}^2-1}}{\sqrt{2}}\bigg)+\text{C}$
$=-\frac{1}{\sqrt{2}}\tan^{-1}\Bigg(\sqrt{\frac{\frac{1}{\text{x}^2}-1}{2}}\Bigg)+\text{C}$
$=-\frac{1}{\sqrt{2}}\tan^{-1}\bigg(\sqrt{\frac{1-\text{x}^2}{2\text{x}^2}}\bigg)+\text{C}$
View full question & answer→Question 2125 Marks
$\int\text{x}(1-\text{x})^{23}\text{dx}$
AnswerLet $\text{I}=\int\text{x}(1-\text{x})^{23}\text{dx}$
Substituting 1 - x = t and dx = -dt, we get
$\text{I}=\int(1-\text{t})^{23}\text{dt}$
$=-\int(\text{t}^{23}-\text{t}^{24})\text{dt}$
$=-\int\Big(\frac{\text{t}^{24}}{24}-\frac{\text{t}^{25}}{25}\Big)+\text{C}$
$=\frac{\text{t}^{25}}{24}-\frac{\text{t}^{24}}{25}+\text{C}$
$=\frac{(1-\text{x})^{25}}{25}-\frac{(1-\text{x})^{24}}{24}+\text{C}$
$\therefore\ \text{I}=\frac{(1-\text{x})^{25}}{25}-\frac{(1-\text{x})^{24}}{24}+\text{C}$
$=\frac{1}{600}(1-\text{x})^{24}\big[24(1-\text{x})-25\big]+\text{C}$
$=\frac{1}{600}(1-\text{x})^{24}\big[24-24\text{x}-25\big]+\text{C}$
$=\frac{1}{600}(1-\text{x})^{24}\big[-1-24\text{x}\big]+\text{C}$
$=\frac{1}{600}(1-\text{x})^{24}\times-\big[1+24\text{x}\big]+\text{C}$
$=\frac{1}{600}(1-\text{x})^{24}(1+24\text{x})+\text{C}$
View full question & answer→Question 2135 Marks
Evaluate the following integrals:
$\int\cot^5\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\cot^5\text{x}\text{ dx}$ Then
$\text{I}=\int\cot^3\text{x}\times\big(\text{cosec}^2-1\big)\text{dx}$
$=\int\cot^3\text{x}\times\big(\text{cosec}^2\text{x}-1\big)\text{dx}$
$=\int\cot^3\text{x}\text{ cosec}^2\text{x}\text{ dx}-\int\cot^3\text{x}\text{dx}$
$=\int\cot^3\text{x}\text{ cosec}^2\text{x}\text{ dx}-\int\big(\text{cosec}^2\text{x}-1\big)\cot\text{dx}$
$=\int\cot^3\text{x}\text{cosec}^2\text{x}\text{ dx}-\int\text{cosec}^2\text{x}\cot\text{x}\text{ dx}+\int\cot\text{x}\text{ dx}$
$\text{I}=\int\cot^3\text{x}\text{cosec}^2\text{x}\text{ dx}-\int\text{cosec}^2\text{x}\cot\text{x}\text{ dx}+\int\cot\text{x}\text{ dx}$
Substituting $\cot\text{x}=\text{t}$ and $-\text{cosec}^2\text{x}\text{ dx}=\text{dt}$ in first two integral, we get
$\text{I}=\int\text{t}^3(-\text{dt})-\int\text{t}\times(-\text{dt})+\int\cot\text{x}\text{ dx}$
$=-\frac{\text{t}^4}{4}+\frac{\text{t}^2}{2}+\log|\sin\text{x}|+\text{C}$
$=-\frac{1}{4}\cot^4\text{x}+\frac{1}{2}\cot^2\text{x}+\log|\sin\text{x}|+\text{C}$
$\therefore\ \text{I}=-\frac{1}{4}\cot^4\text{x}+\frac{1}{2}\cot^2\text{x}+\log|\sin\text{x}|+\text{C}$
View full question & answer→Question 2145 Marks
Evaluate the following intregals:
$\int\frac{1}{4\cos\text{x}-1}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{4\cos\text{x}-1}\ \text{dx}$
Putting $\cos\text{x}=\frac{1-\tan^2\frac{\text{x}}{2}}{1-\tan^2\frac{\text{x}}{2}}$
$\Rightarrow\text{I}=\int\frac{1}{4\Bigg(\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)-1}\ \text{dx}$
$=\int\frac{1}{\frac{4\Big(1-\tan^2\frac{\text{x}}{2}\Big)-\Big(1+\tan^2\frac{\text{x}}{2}\Big)}{\Big(1+\tan^2\frac{\text{x}}{2}\Big)}}$
$=\int\frac{\Big(1+\tan^2\frac{\text{x}}{2}\Big)\text{dx}}{4-4\tan^2\big(\frac{\text{x}}{2}\big)-1-\tan^2\big(\frac{\text{x}}{2}\big)}$
$=\int\frac{\sec^2\big(\frac{\text{x}}{2}\big)\text{dx}}{3-5\tan^2\big(\frac{\text{x}}{2}\big)}$
Let $\tan\Big(\frac{\text{x}}{2}\Big)=\text{t}$
$\Rightarrow\frac{1}{2}\sec^2\Big(\frac{\text{x}}{2}\Big)\text{dx}=\text{dt}$
$\Rightarrow\sec^2\Big(\frac{\text{x}}{2}\Big)\text{dx}=2\text{dt}$
$\therefore\ \text{I}=2\int\frac{\text{x}}{3-5\text{t}^2}$
$=\frac{2}{5}\int\frac{\text{dt}}{\frac{3}{5}-\text{t}^2}$
$=\frac{2}{5}\int\frac{\text{dt}}{\Big(\frac{\sqrt{3}}{\sqrt{5}}\Big)^2-\text{t}^2}$
$=\frac{2}{5}\times\frac{\sqrt{5}}{2\sqrt{3 }}\ln\begin{vmatrix}\frac{\frac{\sqrt{3}}{\sqrt{5}}+\text{t}}{\frac{\sqrt{3}}{\sqrt{5}}-\text{t}}\end{vmatrix}+\text{C}$
$=\frac{1}{\sqrt{15}}\ln\bigg|\frac{\sqrt{3}+\sqrt{5}\text{t}}{\sqrt{3}-\sqrt{5}\text{t}}\bigg|+\text{C}$
$=\frac{1}{\sqrt{15}}\ln\begin{vmatrix}\frac{\sqrt{3}+\sqrt{5}\tan\big(\frac{\text{x}}{2}\big)}{\sqrt{3}-\sqrt{5}\tan\big(\frac{\text{x}}{2}\big)}\end{vmatrix}+\text{C}$
View full question & answer→Question 2155 Marks
Evaluate the following integrals:
$\int\sqrt{2\text{x}^2+3\text{x}+4}\text{dx}$
Answer$\text{I}=\int\sqrt{2\text{x}^2+3\text{x}+4}\text{dx}$
$=\sqrt2\int\sqrt{\text{x}^2+\frac{3}{2}\text{x}+2}\text{dx}$
$=\sqrt2\int\sqrt{\text{x}^2+\frac{3}{2}\text{x}+\frac{9}{16}+\frac{23}{16}}\text{dx}$
$=\sqrt2\int\sqrt{\Big(\text{x}+\frac{3}{4}\Big)^2+\Big(\frac{\sqrt{23}}{4}\Big)^2}\text{dx}$
$=\sqrt2\begin{Bmatrix}\frac{\big(\text{x}+\frac{3}{4}\big)}{2}\sqrt{\text{x}^2+\frac{3}{2}\text{x}+2}+\frac{23}{32}\\\times\log\bigg|\Big(\text{x}+\frac{3}{4}\Big)+\sqrt{\text{x}^2+\frac{3}{2}\text{x}+2}\bigg|+\text{C}\end{Bmatrix}$
$\therefore\ \text{I}=\frac{4\text{x}+3}{8}\sqrt{2\text{x}^2+3\text{x}+4}+\frac{23\sqrt2}{32}\\\times\log\bigg|\Big(\text{x}+\frac{3}{4}\Big)+\sqrt{\text{x}^2+\frac{3}{2}\text{x}+2}\bigg|+\text{C}$
View full question & answer→Question 2165 Marks
Evaluate the following integrals:
$\int\frac{\text{e}^{\text{x}}(\text{x}-4)}{(\text{x}-2)^3}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{e}^{\text{x}}(\text{x}-4)}{(\text{x}-2)^3}\text{dx}$
$=\int\text{e}^{\text{x}}\bigg\{\frac{(\text{x}-2)-2}{(\text{x}-2)^3}\bigg\}\text{dx}$
$=\int\text{e}^{\text{x}}\bigg\{\frac{1}{(\text{x}-2)^2}-\frac{2}{(\text{x}-2)^3}\bigg\}\text{dx}$
Here, $\text{f(x)}=\frac{1}{(\text{x}-2)^2}$ and $\text{f}'\text{(x)}=\frac{-2}{(\text{x}-2)^3}$
And we know that,
$\int\text{e}^{\text{ax}}(\text{af(x)}+\text{f}'(\text{x}))\text{dx}=\text{e}^{\text{ax}}\text{f(x)}+\text{C}$
$\therefore\int\text{e}^{\text{x}}\bigg\{\frac{1}{(\text{x}-2)^2}-\frac{2}{(\text{x}-2)^3}\bigg\}\text{dx}=\frac{\text{e}^{\text{x}}}{(\text{x}-2)^2}+\text{C}$
$\therefore\text{I}=\frac{\text{e}^{\text{x}}}{(\text{x}-2)^2}+\text{C}$
View full question & answer→Question 2175 Marks
Evaluate the following intregals:
$\int\frac{1}{5+4\cos\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{5+4\cos\text{x}}\ \text{dx}$
Put $\cos\text{x}=\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}$
$\text{I}=\int\frac{1}{5+4\Bigg(\frac{1-\tan^2\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(1-\tan^2\frac{\text{x}}{2}\Big)}\ \text{dx}$
$=\int\frac{\sec^2\frac{\text{x}}{2}}{9+\tan^2\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}}{(3)^2+\text{t}^2}$
$=2\times\frac{1}{3}\tan^{-1}(\text{t})+\text{C}$
$\text{I}=\frac{2}{3}\tan^{-1}\Big(\frac{\tan\frac{\text{x}}{2}}{3}\Big)+\text{C}$
View full question & answer→Question 2185 Marks
$\int\frac{\text{x}^2}{\sqrt{1-\text{x}}}\text{dx}$
Answer$\int\frac{\text{x}^2}{\sqrt{1-\text{x}}}\text{dx}$
Let $1-\text{x}=\text{t}$
$\Rightarrow\text{x}=1-\text{t}$
$\Rightarrow1=-\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{dx}=-\text{dt}$
Now, $\int\frac{\text{x}^2}{\sqrt{1-\text{x}}}\text{dx}$
$=\int\frac{(1-\text{t})^2}{\sqrt{\text{t}}}\text{dt}$
$=\int\Big(\frac{1-\text{t}^2-2\text{t}}{\sqrt{\text{t}}}\Big)\text{dt}$
$=\int\Big(\frac{1}{\sqrt{\text{t}}}+\frac{\text{t}^2}{\sqrt{\text{t}}}-\frac{2\text{t}}{\sqrt{\text{t}}}\Big)\text{dt}$
$=\int\Big(\text{t}^{-\frac{1}{2}}+\text{t}^{\frac{3}{2}}-2\text{t}^{\frac{1}{2}}\Big)\text{dt}$
$=\Bigg[\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\frac{\text{t}^{\frac{3}{2}+1}}{\frac{3}{2}+1}-\frac{2\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\Bigg]+\text{C}$
$=2\text{t}^{\frac{1}{2}}+\frac{2}{5}\text{t}^{\frac{5}{2}}-\frac{4}{3}\text{t}^{\frac{3}{2}}+\text{C}$
$=2\text{t}^{\frac{1}{2}}\Big[1+\frac{\text{t}^2}{5}-\frac{2}{3}\text{t}\Big]+\text{C}$
$=2\text{t}^{\frac{1}{2}}\Big[\frac{15+3\text{t}^2-10\text{t}}{15}\Big]+\text{C}$
$=2\sqrt{1-\text{x}}\Big[\frac{15+3(1-\text{x})^2-10(1-\text{x})}{15}\Big]+\text{C}$
$=\frac{2}{15}\sqrt{1-\text{x}}\big[15+3(1^2+\text{x}^2-2\text{x})-10+10\text{x}\big]+\text{C}$
$=\frac{2}{15}\sqrt{1-\text{x}}\big[15+3+3\text{x}^2-6\text{x}+10+10\text{x}\big]+\text{C}$
$=\frac{2}{15}\sqrt{1-\text{x}}\big[3\text{x}^2+4\text{x}+8\big]+\text{C}$
View full question & answer→Question 2195 Marks
Evaluate the following integrals:
$\int\frac{\text{x}}{(\text{x}^2+1)\sqrt{\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{x}}{(\text{x}^2+1)\sqrt{\text{x}}}\text{ dx}$
Let $\text{x}=\text{t}^2$
$\text{dx}=2\text{t dt}$
$\therefore\ 2\int\frac{\text{t dt}}{(\text{t}^2+1)\text{t}}$
$=2\Big|\frac{\text{dt}}{\text{t}^4+1}\Big|$
Dividing numerator and denominator by $t^2$
$\text{I}=2\int\frac{\frac{\text{t}}{\text{t}^2}}{\big(\text{t}^2+\frac{1}{\text{t}^2}\big)}\text{ dt}$
$=\int\frac{\Big(1+\frac{1}{\text{t}^2}\big)-\Big(1-\frac{1}{\text{t}^2}\Big)}{\Big(\text{t}^2+\frac{1}{\text{t}^2}\Big)}\text{ dt}$
Let $\text{t}-\frac{1}{\text{t}}=\text{z}$
$\Big(1+\frac{1}{\text{t}^2}\Big)\text{ dt}=\text{dz}$ [For $I^{st}$ part]
and, $\text{t}+\frac{1}{\text{t}}=\text{y}$
$\Big(1+\frac{1}{\text{t}^2}\Big)\text{ dt}=\text{dy}$ [For $II^{nd}$ part]
$\therefore\ \text{I}=\int\frac{\text{dz}}{\text{z}^2+2}-\int\frac{\text{dy}}{\text{y}^2-2}$
$=\frac{1}{\sqrt{2}}\tan^{-1}\Big(\frac{\text{z}}{\sqrt{2}}\Big)-\frac{1}{2\sqrt{2}}\log\bigg|\frac{\text{y}-\sqrt{2}}{\text{y}+\sqrt{2}}\bigg|+\text{C}$
$=\frac{1}{\sqrt{2}}\tan^{1}\Big(\frac{\text{t}^2-1}{\sqrt{2}\text{t}}\Big)-\frac{1}{2\sqrt{2}}\log\bigg|\frac{\text{x}+1-\sqrt{2\text{x}}}{\text{x}+1+\sqrt{2\text{x}}}\bigg|+\text{C}$
View full question & answer→Question 2205 Marks
Evaluate the following integrals:
$\int(\text{x}+3)\sqrt{3-4\text{x}-\text{x}^2}\text{dx}$
AnswerLet $\text{I}=\int(\text{x}+3)\sqrt{3-4\text{x}-\text{x}^2}\text{dx}$We express $\text{x}+3=\text{A}\frac{\text{d}}{\text{dx}}(3-4\text{x}-\text{x}^2)+\text{B}$
$\text{x}+3=\text{A}(-4-2\text{x})+\text{B}$
Equating the co-efficient of x and constants, we get
$1=-2\text{A}$ and $3=-4\text{A + B}$
or $\text{A}=-\frac{1}{2}$ and $\text{B}=1$
$\therefore\ \text{I}=\int\Big(-\frac{1}{2}(-4-2\text{x})+1\Big)\sqrt{3-4\text{x}-\text{x}^2}\text{dx}$
$=-\frac{1}{2}\int(-4-2\text{x})\sqrt{3-4\text{x}-\text{x}^2}\text{dx}+\int\sqrt{3-4\text{x}-\text{x}^2}\text{dx}$
$=-\frac{1}{2}\text{I}_1+\text{I}_2\ \dots(1)$
Now, $\text{I}_1=\int(-4-2\text{x})\sqrt{3-4\text{x}-\text{x}^2}\text{dx}$
Let $3-4\text{x}-\text{x}^2=\text{u}$
On differentiating both sides, we get
$(-4-2\text{x})\text{dx = du}$
$\therefore\ \text{I}_1=\int\sqrt{\text{u}}\text{du}$
$=\frac{2}{3}\text{u}^{\frac{3}{2}}+\text{c}_1$
$=\frac{2}{3}(3-4\text{x}-\text{x}^2)^{\frac{3}{2}}+\text{c}_1\ \dots(2)$
And, $\text{I}_2=\int\sqrt{3-4\text{x}-\text{x}^2}\text{dx}$
$=\int\sqrt{3+4-4-4\text{x}-\text{x}^2}\text{dx}$
$=\int\sqrt{(\sqrt7)^2-(\text{x}+2)^2}\text{dx}$
Let $(\text{x}+2)=\text{u}$
On differentiating both sides, we get
$\text{dx = du}$
$\therefore\ \text{I}_2=\int\sqrt{(\sqrt7)^2-(\text{u})^2}\text{du}$
$=\frac{\text{u}}{2}\sqrt{(\sqrt7)^2-(\text{u})^2}+\frac{1}{2}(\sqrt7)^2\sin^{-1}\Big(\frac{\text{u}}{\sqrt7}\Big)+\text{c}_2$
$=\frac{\text{x}+2}{2}\sqrt{7-(\text{x}+2)^2}+\frac{7}{2}\sin^{-1}\Big(\frac{\text{x}+2}{\sqrt7}\Big)+\text{c}_2$
$=\frac{\text{x}+2}{2}\sqrt{3-4\text{x}-\text{x}^2}+\frac{7}{2}\sin^{-1}\Big(\frac{\text{x}+2}{\sqrt7}\Big)+\text{c}_2\ \dots(3)$
From (1), (2) and (3), we get
$\therefore\ \text{I}=-\frac{1}{2}\Big(\frac{2}{3}(3-4\text{x}-\text{x}^2)^{\frac{3}{2}}+\text{c}_1\Big)\\+\Big(\frac{\text{x}+2}{2}\sqrt{3-4\text{x}-\text{x}^2}+\frac{7}{2}\sin^{-1}\Big(\frac{\text{x}+2}{\sqrt7}\Big)+\text{c}_2\Big)$
$=-\frac{1}{3}(3-4\text{x}-\text{x}^2)^{\frac{3}{2}}+\frac{\text{x}+2}{2}\sqrt{3-4\text{x}-\text{x}^2}\\+\frac{7}{2}\sin^{-1}\Big(\frac{\text{x}+2}{\sqrt7}\Big)+\text{C}$
Hence,
$\int(\text{x}+3)\sqrt{3-4\text{x}-\text{x}^2}\text{dx}=-\frac{1}{3}(3-4\text{x}-\text{x}^2)^{\frac{3}{2}}\\+\frac{\text{x}+2}{2}\sqrt{3-4\text{x}-\text{x}^2}+\frac{7}{2}\sin^{-1}\Big(\frac{\text{x}+2}{\sqrt7}\Big)+\text{C}$
View full question & answer→Question 2215 Marks
$\int\frac{\text{x}^2+\text{x}+5}{3\text{x}+2}\text{dx}$
Answer$\int\frac{(\text{x}^2+\text{x}+5)}{(3\text{x}+2)}\text{dx}$
$=\frac{1}{9}\int\frac{9\text{x}^2+9\text{x}+45}{(3\text{x}+2)}\text{dx}$
$=\frac{1}{9}\Big[\int\frac{9\text{x}^2-4}{3\text{x}+2}\text{dx}+\int\frac{9\text{x}+6}{3\text{x}+2}\text{dx}+\int\frac{43}{3\text{x}+2}\text{dx}\Big]$
$=\frac{1}{9}\Big[\int\frac{(3\text{x}-2)(3\text{x}+2)}{(3\text{x}+2)}\text{dx}+\int\frac{3(3\text{x}+2)}{3\text{x}+2}\text{dx}+43\int\frac{\text{dx}}{3\text{x}+2}\Big]$
$=\frac{1}{9}\Big[\int(3\text{x}-2)\text{dx}+3\int1\text{dx}+43\int\frac{\text{dx}}{3\text{x}+2}\Big]$
$=\frac{1}{9}\Big[\Big(3\frac{\text{x}^2}{2}-2\text{x}\Big)+3\text{x}+\frac{43}{3}\text{ln}|3\text{x}+2|+\text{c}\Big]$
$=\frac{1}{9}\Big[\frac{3}{2}\text{x}^2+\text{x}-\frac{43}{3}\text{ln}|3\text{x}+2|+\text{c}\Big]$
$=\frac{1}{6}\text{x}^2+\frac{1}{9}\text{x}-\frac{43}{27}\text{ln}|3\text{x}+2|+\text{c}$
View full question & answer→Question 2225 Marks
Evaluate the following integral:
$\int\frac{\text{x}^4+1}{\text{x}^2+1}\text{ dx}$
Answer$\int\Big(\frac{\text{x}^4+1}{\text{x}^2+1}\Big)\text{ dx}$
$=\int\Big(\frac{\text{x}^4-1+1+1}{\text{x}^2+1}\Big)\text{ dx}$
$=\int\Big[\frac{(\text{x}^4-1)}{\text{x}^2+1}+\frac{2}{\text{x}^2+1}\Big]\text{ dx}$
$=\int\Big[\frac{(\text{x}^2-1)(\text{x}^2+1)}{(\text{x}^2+1)}+\frac{2}{\text{x}^2+1}\Big]\text{ dx}$
$=\int\Big[(\text{x}^2-1)+\frac{2}{\text{x}^2+1}\Big]\text{ dx}$
$=\frac{\text{x}^3}{3}-\text{x}+2\tan^{-1}(\text{x})+\text{C}$
View full question & answer→Question 2235 Marks
Evalute the following integrals:
$\int\frac{\sin(\text{x}-\alpha)}{\sin(\text{x}+\alpha)}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sin(\text{x}-\alpha)}{\sin(\text{x}+\alpha)}\text{dx}$ then,
$\text{I}=\int\frac{\sin(\text{x}-\alpha+\alpha-\alpha)}{\sin(\text{x}+\alpha)}\text{dx}$
$=\int\frac{\sin(\text{x}+\alpha-2\alpha)}{\sin(\text{x}+\alpha)}\text{dx}$
$=\int\frac{\sin(\text{x}+\alpha)\cos2\alpha-\cos(\text{x}+\alpha)\sin2\alpha}{\sin(\text{x}+\alpha)}\text{dx}$
$=\int\Big[\frac{\sin(\text{x}+\alpha)\cos2\alpha}{\sin(\text{x}+\alpha)}-\frac{\cos(\text{x}+\alpha)\sin2\alpha}{\sin(\text{x}+\alpha)}\Big]\text{dx}$
$=\int\big(\cos2\alpha-\cot(\text{x}+\alpha)\sin2\alpha\big)\text{dx}$
$=\cos2\alpha\int\text{dx}-\sin2\alpha\int\cot(\text{x}+\alpha)\text{dx}$
$=\text{x}\cos2\alpha-\sin2\alpha\log|\sin(\text{x}+\alpha)|+\text{C}$
$\therefore\text{I}=\text{x}\cos2\alpha-\sin2\alpha\log|\sin(\text{x}+\alpha)|+\text{C}$
View full question & answer→Question 2245 Marks
Evaluate the following intregals:
$\int\frac{\sin2\text{x}}{(1+\sin\text{x})(2+\sin\text{x})}\text{ dx}$
AnswerLet $\int\frac{\sin2\text{x}}{(1+\sin\text{x})(2+\sin\text{x})}\text{ dx}=\frac{\text{A}}{1+\sin\text{x}}+\frac{\text{B}}{2+\sin\text{x}}$
$\Rightarrow\sin2\text{x}=\text{A}(2+\sin\text{B})+\text{B}(1+\sin\text{B})$
$\Rightarrow2\sin\text{x}\cos\text{x}=(2\text{A}+\text{B})+(\text{A}+\text{B})\sin\text{x}$
Equating similar terms, we get,
$2\text{A}+\text{B}=0\Rightarrow\text{B}=-2\text{A}\text{ and}$
$\text{A}+\text{B}=2\cos\Rightarrow-\text{A}=2\cos\text{x}$
$\Rightarrow\text{A}=-2\cos\text{x}$
Thus,
$\text{I}=\int-\frac{2\cos\text{x}}{1+\sin\text{x}}\text{ dx}+\int\frac{4\cos\text{x}}{1+\sin\text{x}}\text{ dx}$
$=-2\log|1+\sin\text{x}|+4\log|2+\sin\text{x}|+\text{C}$
$\text{I}=\log\Big|\frac{(2+\sin\text{x})^4}{(1+\sin\text{x})^2}\Big|+\text{C}$
View full question & answer→Question 2255 Marks
Evaluate the following integrals:
$\int\sqrt{2\text{x}-\text{x}^2}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{2\text{x}-\text{x}^2}\text{dx}$
$=\int\sqrt{\text{x}(2-\text{x})}\text{dx}$
Let $\text{x}=1+\sin\text{u}$
or, $\text{dx}=\cos\text{u du}$
$\Rightarrow\text{I}=\int\sqrt{(1+\sin\text{u})(1-\sin\text{u})}\cos\text{u du}$
$\Rightarrow\text{I}=\int\cos^2\text{u du}$
$\Rightarrow\text{I}=\frac{1}{2}\int(\cos2\text{u}+1)\text{du}$
$\Rightarrow\text{I}=\frac{1}{2}\Big(\frac{1}{2}\sin2\text{u}+\text{u}\Big)+\text{C}$
$\Rightarrow\text{I}=\frac{1}{2}(\sin\text{u}\cos\text{u}+\text{u})+\text{C}$
$\Rightarrow\text{I}=\frac{1}{2}\big(\sin\text{u}\sqrt{1-\sin^2\text{u}}+\text{u}\big)+\text{C}$
$\therefore\ \text{I}=\frac{1}{2}(\text{x}-1)\sqrt{2\text{x}-\text{x}^2}+\frac{1}{2}\sin^{-1}(\text{x}-1)+\text{C}$ $\big[\because\text{u}=\sin^{-1}(\text{x}-1)\big]$
View full question & answer→Question 2265 Marks
Integrate the following integrals:
$\int\sin2\text{x}\sin4\text{x}\sin6\text{x dx}$
Answer$\int\sin2\text{x}\sin4\text{x}\sin6\text{x dx}$
$=\frac{1}{2}\int(2\sin2\text{x}\sin4\text{x})\sin6\text{x dx}$
$=\frac{1}{2}\int\big[\cos(2\text{x}-4\text{x})-\cos(2\text{x}+4\text{x})\big]\sin6\text{x dx}$
$=\frac{1}{2}\int\big[\cos(2\text{x})-\cos(6\text{x})\big]\sin6\text{x dx}$
$=\frac{1}{2}\big[\int\cos(2\text{x})\sin(6\text{x})\text{dx}-\int\cos(6\text{x})\sin(6\text{x})\text{dx}\big]$
$=\frac{1}{4}\big[\int2\cos(2\text{x})\sin(6\text{x})\text{dx}-\int2\cos(6\text{x})\sin(6\text{x})\text{dx}\big]$
$=\frac{1}{4}\Big\{\int\big[\sin(2\text{x}+6\text{x})-\sin(2\text{x}-6\text{x})\big]\text{dx}-\int\sin(12\text{x})\text{dx}\Big\}$
$=\frac{1}{4}\Big[\int\sin(8\text{x})\text{dx}+\int\sin(4\text{x})\text{dx}-\int\sin(12\text{x})\text{dx}\Big]$
$=\frac{1}{4}\Big[\frac{-\cos(8\text{x})}{8}+\frac{-\cos(4\text{x})}{4}+\frac{\cos(12\text{x})}{12}\Big]+\text{C}$
$=-\frac{\cos(8\text{x})}{32}-\frac{\cos(4\text{x})}{16}+\frac{\cos(12\text{x})}{48}+\text{C}$
View full question & answer→Question 2275 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^2\sin^{-1}\text{x}}{(1-\text{x}^2)^{\frac{3}{2}}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2\sin^{-1}\text{x dx}}{(1-\text{x}^2)^{\frac{3}{2}}}$
Putting $\text{x}=\sin\theta$
$\Rightarrow\text{dx}=\cos\theta\text{d}\theta$
$\&\ \theta=\sin^{-1}\text{x}$
$\therefore\text{I}=\int\frac{\sin^2\theta.\theta.\cos\theta\text{d}\theta}{(1-\sin^2\theta)^{\frac{3}{2}}}$
$=\int\frac{\sin^2\theta.\theta.\cos\theta\text{d}\theta}{(\cos^2\theta)^{\frac{3}{2}}}$
$=\int\frac{\sin^2\theta.\theta.\cos\theta\text{d}\theta}{\cos^3\theta}$
$=\int\tan^2\theta.\theta.\text{d}\theta$
$=\int(\sec^2\theta-1)\theta.\text{d}\theta$.
$=\int\theta.\sec^2\theta\text{d}\theta-\int\theta.\text{d}\theta$
$=\theta\int\sec^2\theta\text{d}\theta-\int\Big\{\frac{\text{d}}{\text{d}\theta}(\theta)\int\sec^2\theta\text{d}\theta\Big\}\text{d}\theta-\int\theta.\text{d}\theta$
$=\theta\tan\theta-\int1.\tan\theta\text{d}\theta-\frac{\theta^{\ 2}}{2}$
$=\theta.\tan\theta-\ln\big|\sec\theta\big|-\frac{\theta^{\ 2}}{2}+\text{C}$
$=\theta.\frac{\sin\theta}{\cos\theta}+\ln\big|\cos\theta\big|-\frac{\theta^{\ 2}}{2}+\text{C}$
$=\theta.\frac{\sin\theta}{\cos\theta}+\ln\Big|\sqrt{1-\sin^2\theta}\Big|-\frac{\theta^{\ 2}}{2}+\text{C}$
$=\frac{\theta.\sin\theta}{\sqrt{1-\sin^2\theta}}+\frac{1}2\ln\Big|1-\sin^2\theta\Big|-\frac{\theta^{\ 2}}{2}+\text{C}$
$=\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}+\frac{1}{2}\ln\big(1-\text{x}^2\big)-\frac{1}{2}\big(\sin^{-1}\text{x}\big)^2+\text{C}$$\Big[\because\theta=\sin^{-1}\text{x}\Big]$
View full question & answer→Question 2285 Marks
Evaluate the following integrals:
$\int\text{x}\sin^3\text{x dx}$
AnswerLet $\text{I}=\int\text{x}\sin^3\text{x dx}$
$\sin(3\text{A})=3\sin\text{A}-4\sin^3\text{A}$
$\sin^3\text{A}=\frac{1}4{}\big[3\sin\text{A}-\sin3\text{A}\big]$
$\therefore\text{I}=\frac{1}{4}\int\text{x}(3\sin\text{x}-\sin3\text{x})\text{dx}$
$=\frac{3}{4}\int\text{x}\sin\text{x dx}-\frac{1}{4}\int\text{x}\sin(3\text{x})\text{dx}$
$=\frac{3}{4}\big[\text{x}(-\cos\text{x})-\int1.(-\cos\text{x})\text{dx}\big]\\-\frac{1}{4}\Big[\text{x}\Big(-\frac{\cos3\text{x}}{3}\Big)-\int1.\Big(-\frac{\cos3\text{x}}{3}\Big)\text{dx}\Big]$
$=-\frac{3\text{x}\cos\text{x}}{4}+\frac{3}{4}\sin\text{x}+\frac{\text{x}\cos3\text{x}}{12}-\frac{1}{36}\sin3\text{x}+\text{C}$
View full question & answer→Question 2295 Marks
Evaluate the following integrals:
$\int\text{e}^{2\text{x}}\cos^2\text{x }\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{2\text{x}}\cos^2\text{x }\text{dx}$
$=\frac{1}{2}\int\text{e}^{2\text{x}}2\cos^2\text{x dx}$
$=\frac{1}{2}\int\text{e}^{2\text{x}}(1+\cos2\text{x})\text{dx}$
$=\frac{1}{2}\int\text{e}^{2\text{x}}\text{dx}+\frac{1}{2}\int\text{e}^{2\text{x}}\cos2\text{x }\text{dx}$
$\because\ \int\text{e}^{2\text{x}}\cos\text{bx dx}=\frac{\text{e}^{2\text{x}}}{\text{a}^2+\text{b}^2}\{\text{a}\cos\text{bx}-\text{b}\sin\text{bx}\}+\text{C}$
$\therefore\ \text{I}=\frac{1}{4}\text{e}^{2\text{x}}+\frac{1}{2}\frac{\text{e}^{2\text{x}}}{8}\{2\cos2\text{x}+2\sin2\text{x}\}+\text{C}$
Hence,
$\text{I}=\frac{\text{e}^{2\text{x}}}{4}+\frac{\text{e}^{2\text{x}}}{16}\{2\cos2\text{x}+2\sin2\text{x}\}+\text{C}$
or
$\text{I}=\frac{\text{e}^{2\text{x}}}{4}+\frac{\text{e}^{2\text{x}}}{8}\{\cos2\text{x}+\sin2\text{x}\}+\text{C}$
View full question & answer→Question 2305 Marks
Evaluate the following integrals:
$\int\sqrt{3-2\text{x}-2\text{x}^2}\text{dx}$
Answer$\text{I}=\int\sqrt{3-2\text{x}-2\text{x}^2}\text{dx}$
$=\sqrt2\int\sqrt{\frac{3}{2}-\text{x}-\text{x}^2}\text{dx}$
$=\sqrt2\int\sqrt{\frac{7}{4}-\Big(\frac{1}{4}+\text{x}+\text{x}^2\Big)}\text{dx}$ $\Big[\text{Adding and subtracting }\frac{1}{4}\Big]$
$=\sqrt2\int\sqrt{\Big(\frac{\sqrt7}{2}\Big)^2-\Big(\text{x}+\frac{1}{2}\Big)^2}\text{dx}$
$=\sqrt2\begin{Bmatrix}\frac{\text{x}+\frac{1}{2}}{2}\sqrt{\frac{3}{2}-\text{x}-\text{x}^2}+\frac{7}{8}\sin^{-1}\bigg(\frac{\text{x}+\frac{1}{2}}{\frac{\sqrt7}{2}}\bigg)+\text{C}\end{Bmatrix}$
$\therefore\ \text{I}=\frac{2\text{x}+1}{4}\sqrt{3-2\text{x}-2\text{x}^2}+\frac{7\sqrt2}{8}\sin^{-1}\Big(\frac{2\text{x}+1}{\sqrt7}\Big)+\text{C}$
View full question & answer→Question 2315 Marks
Evaluate the following integrals:
$\int\text{cosec x}\log({\text{cosec x}-\cot\text{x})}\text{dx}$
AnswerLet $\text{I}=\int\text{cosec x}\log(\text{cosec x}-\cot\text{x})\text{dx}\ ....(1)$
Let $\log(\text{cosec x}-\cot\text{x})=\text{t}$ then,
$\text{dx}[\log(\text{cosec x}-\cot\text{x})]=\text{dt}$
$\Rightarrow\text{cosec x}\text{ dx}=\text{dt}$ $\Big[\because\ \frac{\text{d}}{\text{dx}}(\log(\text{cosec x}-\cot\text{x}))=\text{cosec x}\Big]$
Putting $\log(\text{cosec x}-\cot\text{x})=\text{t}$ and $\text{cosec x}\text{ dx}=\text{dt}$ in equation (1), we get
$\text{I}=\int\text{t dt}$
$=\frac{\text{t}^2}{2}+\text{C}$
$\text{I}=\frac{1}{2}[\log(\text{cosec x}-\cot\text{x})]^2+\text{C}$
View full question & answer→Question 2325 Marks
$\int\frac{1}{\sqrt{\text{x}}+\sqrt[4]{\text{x}}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\sqrt{\text{x}}+\sqrt[4]{\text{x}}}\text{dx}$Let $\text{x}=\text{t}^{4}$
On differentiating both sides, we get $\text{dx}=4\text{t}^{3}\text{dt}$ $\therefore\ \text{I}=\int\frac{4\text{t}^{3}}{\sqrt{\text{t}^{4}}+\sqrt[4]{\text{t}^{4}}}\text{dt}$ $=\int\frac{4\text{t}^3}{\text{t}^{2}+\text{t}}\text{dt}$ $=4\int\frac{\text{t}^{2}}{\text{t}+1}\text{dt}$ $4\int\frac{(\text{t}-1)(\text{t}+1)+1}{\text{t}+1}\text{dt}$ $=4\int\Big[(\text{t}-1)+\frac{1}{\text{t}+1}\Big]\text{dt}$ $=4\Big[\frac{\text{t}^2}{2}-\text{t}+\log(\text{t}+1)\big]+\text{C}$ $=2\sqrt{x}-4\sqrt[4]{\text{x}}+4\log\big(\sqrt[4]{\text{x}+1}\big)+\text{C}$ Hence, $\int\frac{1}{\sqrt{\text{x}}+\sqrt[4]{\text{x}}}\text{dx}=2\sqrt{x}-4\sqrt[4]{\text{x}}+4\log\big(\sqrt[4]{\text{x}+1}\big)+\text{C}$
View full question & answer→Question 2335 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^2+9}{\text{x}^4+81}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2+9}{\text{x}^4+81}\ \text{dx}$Dividing numerator and denominator by $x^2$
$\text{I}=\int\frac{1+\frac{9}{\text{x}^2}}{\text{x}^2+\frac{81}{\text{x}^2}}\ \text{dx}$
$=\int\frac{1+\frac{9}{\text{x}^2}}{\Big(\text{x}-\frac{9}{\text{x}}\Big)^2+18}$
Let $\Big(\text{x}-\frac{9}{\text{x}}\Big)=\text{t}\Rightarrow\Big(1+\frac{9}{\text{x}^2}\Big)\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}^2+18}$
$\text{I}=\frac{1}{3\sqrt{2}}\tan^{-1}\Big(\frac{\text{t}}{3\sqrt{2}}\Big)+\text{C}$
Thus,
$\text{I}=\frac{1}{3\sqrt{2}}\tan^{-1}\Big(\frac{\text{x}^2-9}{3\sqrt{2}}\Big)+\text{C}$
View full question & answer→Question 2345 Marks
Evaluate the following integrals:
$\int\frac{1}{(2\text{x}^2+3)\sqrt{\text{x}^2-4}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1}{(2\text{x}^2+3)\sqrt{\text{x}^2-4}}\text{ dx}$
Let $\text{x}=\frac{1}{\text{t}}$
$\text{dx}=-\frac{1}{\text{t}^2}\text{ dt}$
$\therefore\ \text{I}=\int\frac{-\frac{1}{\text{t}^2}\text{ dt}}{\Big(\frac{2}{\text{t}^2+3}\Big)\sqrt{\big(\frac{1}{\text{t}^2}-4\big)}}$
$=-\int\frac{\text{t dt}}{(2+3\text{t}^2)\sqrt{1-4\text{t}^2}}$
Let $1-4\text{t}^2=\text{u}^2$
$-8\text{tdt}=2\text{udu}$
$\therefore\ \text{I}=\frac{1}{4}\int\frac{\text{u du}}{\frac{(11-3\text{u})^2}{4}\text{u}}$
$=\frac{1}{3}\int\frac{\text{du}}{\frac{11}{3}-\text{u}^2}$
$=\frac{1}{2\sqrt{33}}\log\begin{vmatrix}\frac{\text{u}-\sqrt{\frac{11}{3}}}{\text{u}+\sqrt{\frac{11}{3}}}+\text{C}\end{vmatrix}$
$=\frac{1}{2\sqrt{33}}\log\begin{vmatrix}\frac{\sqrt{1-4\text{t}^2}-\sqrt{\frac{11}{3}}}{\sqrt{1-4\text{t}^2}+\frac{11}{3}}+\text{C}\end{vmatrix}$
Hence,
$\text{I}=\frac{1}{2\sqrt{33}}\log\bigg|\frac{\sqrt{11}\text{x}+\sqrt{3\text{x}^2-12}}{\sqrt{11}\text{x}-\sqrt{3\text{x}^2-12}}\bigg|+\text{C}$
View full question & answer→Question 2355 Marks
Evaluate the following integrals:
$\int\text{e}^{2\text{x}}(-\sin\text{x}+2\cos\text{x})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{2\text{x}}(-\sin\text{x}+2\cos\text{x})\text{dx}$
$=-\int\text{e}^{2\text{x}}\sin\text{x dx}+2\int\text{e}^{2\text{x}}\cos\text{x dx}$
Applying by parts in the $2^{nd}$ integrand
$\therefore\text{I}=-\int\text{e}^{2\text{x}}\sin\text{x dx}+2\Big\{\frac{1}{2}\text{e}^{2\text{x}}\cos\text{x}+\int\frac{1}{2}\text{e}^{2\text{x}}\sin\text{x dx}\Big\}$
$=-\int\text{e}^{2\text{x}}\sin\text{x dx}+\text{e}^{2\text{x}}\cos\text{x}+\int\text{e}^{2\text{x}}\sin\text{x dx}+\text{C}$
$=\text{e}^{2\text{x}}\cos\text{x}+\text{C}$
Thus,
$\text{I}=\text{e}^{2\text{x}}\cos\text{x + C}$
View full question & answer→Question 2365 Marks
Evaluate the following integrals:
$\int\frac{1}{\text{x}^2-10\text{x}+34}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\text{x}^2-10\text{x}+34}\text{dx}$
$=\int\frac{1}{\text{x}^2-2\text{x}\times5+(5)^2-(5)^2+34}\text{dx}$
$=\int\frac{1}{(\text{x}-5)^2+9}\text{dx}$
Let $(\text{x}-1)=\text{t} \dots(1)$
$\Rightarrow\text{dx = dt}$
so,
$\text{I}=\int\frac{1}{\text{t}^2+(3)^2}\text{dt}$
$\text{I}=\frac{1}{3}\tan^{-1}\big(\frac{\text{t}}{3}\big)+\text{C}$ $\Big[\text{since,}\int\frac{1}{\text{x}^2+\text{a}^2}\text{dx}=\frac{1}{\text{a}}\tan^{-1}\big(\frac{\text{x}}{2}\big)+\text{C}\Big]$
$\text{I}=\frac{1}{3}\tan^{-1}\Big(\frac{\text{x}-5}{3}\Big)+\text{C}$ [using (1)]
View full question & answer→Question 2375 Marks
If f'(x) = x + b, f'(1) = 5, f'(2) = 13, find f'(x).
Answer$\text{f}'\text{(x)}=\text{x + b},\text{f}'(1)=5,\text{f}'(2)=13$
$\text{f}'\text{(x)}=\text{x + b}$
$\int\text{f}'\text{(x) dx}=\int(\text{x + b})\text{dx}$
$\text{f}'\text{(x)}=\frac{\text{x}^2}{2}+\text{bx}+\text{C}\ \dots(1)$
$\text{f}'(1)=5,\text{f}'(2)=13$ (Given)
putting x = 1 in (1)
$\text{f}'(1)=\frac{1^2}{2}+\text{b}_1+\text{C}$
$5=\frac{1}{2}+\text{b + C}\ \dots(2)$
Putting x = 2 in (1)
$\text{f}'(2)=\frac{2^2}{2}+\text{b}_2+\text{C}$
$13=\frac{4}{2}+2\text{b + C}$
$13=2+2\text{b + C}\ \dots(3)$
Solving (2) and (3) we get,
$\text{b}=\frac{13}{2}\text{ and }\text{C}=-2$
Thus, $\text{f}'\text{(x)}=\frac{\text{x}^2}{2}+\frac{13}{2}\text{x}-2$
View full question & answer→Question 2385 Marks
Evaluate the following integrals:
$\int\cos(\log\text{x})\text{dx}$
AnswerLet $\text{I}=\int\cos(\log\text{x})\text{dx}$
Let $\log\text{x}=\text{t}$
$\Rightarrow\text{x}=\text{e}^\text{t}$
$\Rightarrow\text{dx}=\text{e}^\text{t}\text{dt}$
$\text{I}=\int\text{e}^\text{t}\cos\text{(t)dt}$
Considering cos (t) as first function and $e^t$ as second function
$\text{I}=\cos\text{t}\text{e}^\text{t}-\int(-\sin\text{t})\text{e}^\text{t}\text{dt}$
$\Rightarrow\text{I}=\cos\text{t}\text{e}^\text{t}+\int\sin\text{t}\text{e}^\text{t}\text{dt}$
$\Rightarrow\text{I}=\cos\text{te}^\text{t}+\text{I}_1\ \dots(1)$
where $\text{I}_1=\int\text{e}^\text{t}\sin\text{t dt}$
$\text{I}_1=\int\text{e}^\text{t}\sin\text{t dt}$
Considering sin t as first function and $e^t$ as second function
$\text{I}_1=\sin\text{te}^\text{t}-\int\cos\text{te}^\text{t}\text{dt}$
$\Rightarrow\text{I}_1=\sin\text{te}^\text{t}-\text{I}\ \dots(2)$
From (1) & (2)
$\text{I}=\cos\text{te}^\text{t}+\sin\text{te}^\text{t}-\text{I}$
$\Rightarrow2\text{I}=\text{e}^\text{t}(\sin\text{t}+\cos\text{t})$
$\Rightarrow\text{I}=\frac{\text{e}^\text{t}(\sin\text{t}+\cos\text{t})}{2}+\text{C}$
$\Rightarrow\text{I}=\frac{\text{e}^{\log\text{x}}\big[\sin(\log\text{x})+\cos(\log\text{x})\big]}{2}+\text{C}$
$\Rightarrow\text{I}=\frac{\text{x}}{2}\big[\sin(\log\text{x})+\cos(\log\text{x})\big]+\text{C}$
View full question & answer→Question 2395 Marks
$\int(\text{x}+2)\sqrt{3\text{x}+5}\text{ dx}$
Answer$\text{Let I}=\int(\text{x}+2)\sqrt{3\text{x}+5}\text{ dx}$
$\text{Putting}\ 3\text{x}+5=\text{t}$
$\Rightarrow\text{x}=\frac{\text{t}-5}{3}$
$\Rightarrow3\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{3}$
$\therefore\text{I}=\int\Big(\frac{\text{t}-5}{3}+2\Big)\sqrt{\text{t}}\frac{\text{dt}}{3}$
$=\frac{1}{3}\int\Big(\frac{\text{t}-5+6}{3}\Big)\sqrt{t}\text{ dt}$
$=\frac{1}{9}\int(\text{t}+1)\sqrt{\text{t}}\text{ dt}$
$=\frac{1}{9}\int\Big(\text{t}^\frac{3}{2}+\text{t}^\frac{1}{2}\Big)\text{dt}$
$=\frac{1}{9}\bigg[\frac{2}{5}(3\text{x}+5)^\frac{5}{2}+\frac{2}{3}(3\text{x}+5)^\frac{3}{2}\bigg]+\text{C}$ $[\because\text{t}=3\text{x}+5]$
$=\frac{2}{9}\bigg[(3\text{x}+5)^\frac{3}{2}\Big\{\frac{3\text{x}+5}{5}+\frac{1}{3}\Big\}\bigg]+\text{C}$
$=\frac{2}{9}\bigg[(3\text{x}+5)^\frac{3}{2}\Big\{\frac{9\text{x}+15+5}{15}\Big\}\bigg]+\text{C}$
$=\frac{2}{9}\bigg[(3\text{x}+5)^\frac{3}{2}\Big\{\frac{9\text{x}+20}{15}\Big\}\bigg]+\text{C}$
$=\frac{2}{135}(3\text{x}+5)^\frac{3}{2}(9\text{x}+20)+\text{C}$
View full question & answer→Question 2405 Marks
Evaluvate the following intregals:
$\int\frac{2\sin\text{x}+3\cos\text{x}}{3\sin\text{x}+4\cos\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{2\sin\text{x}+3\cos\text{x}}{3\sin\text{x}+4\cos\text{x}}\ \text{dx}$
Let $(2\sin\text{x}+3\cos\text{x})=\lambda\frac{\text{d}}{\text{dx}}(3\sin\text{x}+4\cos\text{x})+\mu(3\sin\text{x}+4\cos\text{x})+\text{v}$
$(2\sin\text{x}+3\cos\text{x})=\lambda(3\cos\text{x}-4\sin\text{x})+\mu(3\sin\text{x}+4\cos\text{x})+\text{v}$
$(2\sin\text{x}+3\cos\text{x})=(3\lambda+4\mu)\cos\text{x}+(-4\lambda+3\mu)\sin\text{x}+\text{v}$
Compairing the coefficient of $\sin\text{x},\cos\text{x}$ on both the sides,
$3\lambda+4\mu=3\dots\dots(1)$
$-4\lambda+3\mu=2\dots\dots(2)$
$\text{v}=0\dots\dots(3)$
Solving the equation (1), (2) and (3)
$\lambda=\frac{1}{25}$
$\mu=\frac{18}{25}$
$\text{v}=0$
$\text{I}=\frac{1}{25}\int\frac{(3\cos\text{x}-4\sin\text{x})}{(3\sin\text{x}+4\cos\text{x})}\text{dx}+\frac{18}{25}\int\text{dx}$
$\text{I}=\frac{1}{25}\log|3\sin\text{x}+4\cos\text{x}|+\frac{18}{25}\text{x}+\text{C}$
View full question & answer→Question 2415 Marks
Evaluate the following integrals:
$\int\frac{\text{x}}{\text{x}^4+2\text{x}^2+3}\text{dx}$
Answer$\int\frac{\text{x dx}}{\text{x}^4+2\text{x}^2+3}$
Let $\text{x}^2=\text{t}$
$\Rightarrow2\text{x dx = dt}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$
Now, $\int\frac{\text{x dx}}{\text{x}^4+2\text{x}^2+3}$
$=\frac{1}{2}\int\frac{\text{dt}}{\text{t}^2+2\text{t}+3}$
$=\frac{1}{2}\int\frac{\text{dt}}{\text{t}^2+2\text{t}+1+2}$
$=\frac{1}{2}\int\frac{\text{dt}}{(\text{t}+1)^2+(\sqrt2)^2}$
$=\frac{1}{2}\times\frac{1}{\sqrt{2}}\tan^{-1}\Big(\frac{\text{t}+1}{\sqrt{2}}\Big)+\text{C}$ $\Big[\because\int\frac{\text{dx}}{\text{x}^2+\text{a}^2}=\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$
$=\frac{1}{2\sqrt{2}}\tan^{}-1\Big(\frac{\text{x}^2+1}{\sqrt{2}}\Big)+\text{C}$
View full question & answer→Question 2425 Marks
Evaluate the following intregals:
$\int\frac{\text{x}^2}{(\text{x}-1)(\text{x}+1)^2}\ \text{dx}$
AnswerLet $\frac{\text{x}}{(\text{x}-1)^2(\text{x}+2)}=\frac{\text{A}}{\text{x}-1}+\frac{\text{B}}{(\text{x}-1)}+\frac{\text{C}}{\text{(x}+1)^2}$
$\Rightarrow x^2 = A(x + 1)^2+ B(x - 1) (x + 1) + C(x - 1)$
$= (A + B) x^2 + (2A + C) x + (A - B - C)$
Equating similar terms,
A + B = , 2A + C = 0, A - B - C = 0
Solving, we get, $\text{A}=\frac{1}{4},\text{B}=\frac{3}{4},\text{C}=-\frac{1}{2}$
Thus,
$\text{I}=\frac{1}{4}\int\frac{\text{dx}}{\text{x}-1}+\frac{3}{4}\int\frac{\text{dx}}{\text{x}+1}-\frac{1}{2}\int\frac{\text{dx}}{(\text{x}+1)}^2$
$=\frac{1}{4}\log|\text{x}-1|+\frac{3}{4}\log|\text{x}+1|+\frac{1}{2(\text{x}+1)}+\text{C}$
$\text{I}=\frac{1}{4}\log|\text{x}-1|+\frac{3}{4}\log|\text{x}+1|+\frac{1}{2(\text{x}+1)}\text{C}$
View full question & answer→Question 2435 Marks
Evaluate the following integrals:
$\int\text{cosec}^43\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\text{cosec}^43\text{x}\text{ dx}$ Then
$\text{I}=\int\text{cosec}^23\text{x }\text{cosec}^23\text{x}\text{ dx}$
$=\int\big(1+\cot^23\text{x}\big)\text{cosec}^23\text{x}\text{ dx}$
$=\int\big(\text{cosec}^23\text{x}+\cot^23\text{x }\text{cosec}^23\text{x}\big)\text{dx}$
$\text{I}=\int\text{cosec}^23\text{x}\text{ dx}+\int\cot^23\text{x }\text{cosec}^23\text{x}\text{ dx}$
Sunbstituting $\cot3\text{x}=\text{t}$ and $\text{cosec}^23\text{x}\text{ dx}=-\text{dt}$ in $2^{nd}$ integral, we get
$\text{I}=\int\text{cosec}^23\text{x}-\int\text{t}^2\frac{\text{dt}}{3}$
$=\frac{-1}{3}\cot3\text{x}-\frac{\text{t}^3}{9}+\text{C}$
$=\frac{-1}{3}\cot3\text{x}-\frac{\cot^33\text{x}}{9}+\text{C}$
$\therefore\ \text{I}=\frac{-1}{3}\cot3\text{x}-\frac{1}{9}\infty\text{t}^3\text{3x}+\text{C}$
View full question & answer→Question 2445 Marks
Evaluate the following integrals:
$\int\frac{\tan\text{x}}{\sqrt{\cos\text{x}}}\text{dx}$
Answer$\int\frac{\tan\text{x}}{\sqrt{\cos\text{x}}}\text{dx}$
$\Rightarrow\int\frac{\sin\text{x}}{\cos\text{x}\sqrt{\cos\text{x}}}\text{dx}$
$\Rightarrow\int\frac{\sin\text{x}}{\cos^\frac{3}{2}\text{x}}\text{dx}$
$\text{Let }\cos\text{x}=\text{t}$
$\Rightarrow-\sin\text{x}\text{ dx}=\text{dt}$
$\Rightarrow\sin\text{x}=-\frac{\text{dt}}{\text{dx}}$
$\text{Now,}\int\frac{\sin\text{x}}{\cos^\frac{3}{2}\text{x}}\text{dx}$
$=\int-\frac{1}{\text{t}^\frac{3}{2}}\text{dt}$
$=-\int\text{t}^{-\frac{3}{2}}\text{dt}$
$=-\bigg[\frac{\text{t}^{-\frac{3}{2}+1}}{\frac{-3}{2}+1}\bigg]+\text{C}$
$=\frac{2}{\sqrt{\text{t}}}+\text{C}$
$=\frac{2}{\sqrt{\cos\text{x}}}+\text{C}$
View full question & answer→Question 2455 Marks
Evaluvate the following intregals:
$\int\frac{2\text{x}+1}{(\text{x}+1)(\text{x}-2)}\ \text{dx}$
AnswerLet $\int\frac{2\text{x}+1}{(\text{x}+1)(\text{x}-2)}=\frac{\text{A}}{(\text{x}+1)}+\frac{\text{B}}{(\text{x}-2)}$$\Rightarrow2\text{x}+1=\text{A}(\text{x}-2)+\text{B}(\text{x}+1)$
Put $\text{x}=2$
$\Rightarrow5=3\text{B}\Rightarrow\text{B}=\frac{5}{3}$
Put $\text{x}=-1$
$\Rightarrow-1=-3\text{A}\Rightarrow\text{A}=\frac{1}{3}$
so,
$\int\frac{2\text{x}+1}{(\text{x}+1)(\text{x}-2)}\text{ dx}=\frac{1}{3}\int\frac{\text{dx}}{\text{x}+1}+\frac{5}{3}\int\frac{\text{dx}}{\text{x}-2}$
$=\frac{1}{3}\log|\text{x}+1|+\frac{5}{3}\log|\text{x}-2|+\text{C}$
Thus
$\text{I}=\frac{1}{3}\log|\text{x}+1|+\frac{5}{3}\log|\text{x}-2|+\text{C}$
View full question & answer→Question 2465 Marks
If f'(x) = a sin x + b cos x and f'(0) = 4, f(0) = 3, $\text{f}\Big(\frac{\pi}{2}\Big)=5$, find f(x).
Answer$\text{f'(x)}=\text{a}\sin\text{x}+\text{b}\cos\text{x}$
$\text{f'}(0)=4,\text{f}(0)=3$
$\text{f}\Big(\frac{\pi}{2}\Big)=5$
$\text{f'(x)}=\text{a}\sin\text{x + b}\cos\text{x}$
$\int\text{f'(x)}\text{dx}=\int(\text{a}\sin\text{x + b}\cos\text{x})\text{dx}$
$\text{f(x)}=-\text{a}\cos\text{x}+\text{b}\sin\text{x}+\text{C}\ \dots(1)$
Now putting x = 0 in eq. (1)
$\text{f}(0)=-\text{a}\cos0+\text{b}\sin0+\text{C}$
$3=-\text{a}\times1+\text{b}\times0+\text{C}$
$3=-\text{a + C}\ \dots(2)$
Now putting $\text{x}=\frac{\pi}{2}$ in eq. (1)
$\text{f}\Big(\frac{\pi}{2}\Big)=-\text{a}\cos\frac{\pi}{2}+\text{b}\sin\frac{\pi}{2}+\text{C}$
$5=-\text{a}\cos\frac{\pi}{2}+\text{b}\sin\Big(\frac{\pi}{2}\Big)+\text{C}$
$5=-\text{a}\times0+\text{b}\times1+\text{C}$
$5=\text{b + C}\ \dots(3)$
We also have f'(0) = 4
$\text{f'(x)}=\text{a}\sin\text{x + b}\cos\text{x}$
$\text{f'(0)}=\text{a}\sin0+\text{b}\cos0$
$4=\text{a}\times0+\text{b}\times1$
$4=\text{b}\ \dots(4)$
Solving (2), (3) and (4) we get,
$\text{b}=4$
$\text{C}=1$
$\text{a}=-2$
$\therefore\ \text{f(x)}=2\cos\text{x}+4\sin\text{x}+1$
View full question & answer→Question 2475 Marks
Evaluate the following intregals:
$\int\frac{\text{x}^2+\text{x}-1}{(\text{x}+1)^2(\text{x}+2)}\ \text{dx}$
AnswerLet $\int\frac{\text{x}^2+\text{x}-1}{(\text{x}+1)^2(\text{x}+2)}\ \text{dx}=\frac{\text{A}}{(\text{x}+1)}+\frac{\text{B}}{(\text{x}+1)^2}+\frac{\text{C}}{(\text{x}+2)}$$\Rightarrow\text{x}^2+\text{x}-1=\text{A}(\text{x}+1)(\text{x}+2)+\text{B}(\text{x}+2)+\text{C}(\text{x}+1)^2$
$=(\text{A}+\text{C})\text{x}^2+(3\text{A}+\text{B}+2\text{C})\text{x}+(2\text{A}+2\text{B}+\text{C})$
Equating similar terms
$\text{A}+\text{C}=1,3\text{A}+\text{B}+2\text{C}=1,2\text{A}+2\text{B}+\text{C}=-1$
Solving, we get, A = 0, B = -1, C = 1
Thus,
$\text{I}=0\int\frac{\text{dx}}{\text{x}+1}+(-1)\int\frac{\text{dx}}{(\text{x}+1)^2}+1\int\frac{\text{dx}}{(\text{x}+2)}$
$=\frac{1}{\text{x}+1}+\log|\text{x}+2|+\text{C}$
$\text{I}=\frac{1}{\text{x}+1}+\log|\text{x}+2|+\text{C}$
View full question & answer→Question 2485 Marks
Evaluate the following intregals:
$\int\frac{3+4\text{x}-\text{x}^2}{(\text{x}+2)(\text{x}-1)}\ \text{dx}$
Answer$\text{I}=\int\frac{3+4\text{x}-\text{x}^2}{(\text{x}+2)(\text{x}-1)}\ \text{dx}$
$=\int-1+\frac{5\text{x}+1}{(\text{x}+2)(\text{x}-1)}\ \text{dx}$
$\Rightarrow\text{I}=-\int\text{dx}+\int\frac{5\text{x}+1}{(\text{x}+2)(\text{x}-1)}\ \text{dx}\ \dots(1)$
Let $\frac{5\text{x}+1}{(\text{x}+2)(\text{x}-1)}=\frac{\text{A}}{\text{x}+2}+\frac{\text{B}}{\text{x}-1}$
$\Rightarrow5\text{x}+1=\text{A}(\text{x}-1)+\text{B}(\text{x}+2)$
put x = 1
$\Rightarrow6=3\text{B}\Rightarrow\text{B}=2$
Put x = -2
$\Rightarrow-9=-3\text{A}\Rightarrow\text{A}=3$
So,
$\text{I}=\int\text{dx}+3\int\frac{\text{dx}}{\text{x}+2}+2\int\frac{\text{dx}}{\text{x}-1}$
$\text{I}=-\text{x}+3\log|\text{x}+2|+2\log|\text{x}-1|+\text{C}$
View full question & answer→Question 2495 Marks
Evaluate the following integrals:
$\int\text{x}\sin\text{x}\cos2\text{x dx}$
Answer$\int\text{x}.\cos2\text{x}\sin\text{x dx}$
$=\frac{1}{2}\int\text{x}(2\cos2\text{x}\sin\text{x})\text{dx}$ $\big[\therefore2\cos\text{A}\sin\text{B}=\sin(\text{A+B})-\sin(\text{A}-\text{B})\big]$
$=\frac{1}{2}\int\text{x}(\sin3\text{x}-\sin\text{x})\text{dx}$
$=\frac{1}{2}\int\text{x}\sin3\text{x dx}-\frac{1}{2}\int\text{x}\sin\text{x dx}$
$=\frac{1}{2}\int\text{x}\sin3\text{x dx}-\frac{1}{2}\int\text{x}\sin\text{x dx}$
$=\frac{1}{2}\Big[\text{x}\int\sin3\text{x dx}-\int\Big\{\frac{\text{x}}{\text{dx}}(\text{x})\int\sin3\text{x dx}\Big\}\text{dx}\Big]\\-\frac{1}{2}\Big[\text{x}\int\sin\text{x dx}-\int\Big\{\frac{\text{d}}{\text{dx}}(\text{x})\int\sin\text{x dx}\Big\}\text{dx}\Big]$
$=\frac{1}{2}\Big[\text{x}\Big(\frac{-\cos3\text{x}}{3}\Big)-\int1\Big(\frac{-\cos3\text{x}}{3}\Big)\text{dx}\Big]\\-\frac{1}{2}\big[\text{x}(-\cos\text{x})-\int1(-\cos\text{x})\text{dx}\big]$
$=\frac{1}{2}\Big[\text{x}\Big(\frac{-\cos3\text{x}}{3}\Big)+\frac{1}{9}\sin3\text{x}\big]-\frac{1}{2}\big[\text{x}(-\cos\text{x})+\sin\text{x}\big]$
$=-\frac{\text{x}\cos3\text{x}}{6}+\frac{\sin3\text{x}}{18}+\frac{\text{x}\cos\text{x}}{2}-\frac{\sin\text{x}}{2}+\text{C}$
View full question & answer→Question 2505 Marks
Evaluate the following integrals:
$\int\frac{\sqrt{1+\text{x}^2}}{\text{x}^4}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sqrt{1+\text{x}^2}}{\text{x}^4}\text{ dx}$
Let $\text{x}=\tan\theta$
On differentiating both sides, we get
$\text{dx}=\sec^2\theta\text{ d}\theta$
$\therefore\ \text{I}=\int\frac{\sqrt{1+\tan^2\theta}}{\tan^4\theta}\sec^2\theta\text{ d}\theta$
$=\int\frac{\sec^{3}\theta}{\tan^4\theta}\text{ d}\theta$
$=\int\frac{\cos\theta}{\sin^4\theta}\text{ d}\theta$
$=\int\cot\theta\text{cosec}^3\theta\text{ d}\theta$
Let $\text{cosec}^3\theta=\text{t}$
On differentiating both sides, we get
$-3\text{cosec}^3\theta\cot\theta\text{ d}\theta=\text{dt}$
$\therefore\ \text{I}=-\frac{1}{3}\int\cot\theta\text{cosec}^3\theta\frac{\text{dt}}{\text{cosec}^3\theta\cot\theta}$
$=-\frac{\text{t}}{3}+\text{C}$
$=-\frac{1}{3}(\text{cosec}^3\theta)+\text{C}$
$=-\frac{1}{3}(\text{cosec}(\tan^{-1}\text{x}))^3+\text{C}$
$=-\frac{1}{3}\bigg(\text{cosec}\Big(\text{cosec}^1\frac{\sqrt{1+\text{x}^2}}{\text{x}}\Big)\bigg)^3+\text{C}$
$=-\frac{1}{3}\bigg(\frac{\sqrt{1+\text{x}^2}}{\text{x}}\bigg)^3+\text{C}$
Hence, $\int\frac{\sqrt{1+\text{x}^2}}{\text{x}^4}\text{ dx}=-\frac{1}{3}\bigg(\frac{\sqrt{1+\text{x}^2}}{\text{x}}\bigg)^3+\text{C}$
View full question & answer→