Question 2013 Marks
By using the properties of definite integral, evaluate the integral in Exercise:
$\int\limits_{0}^{\frac{\pi}{2}}\frac{\sin^{\frac{3}{2}}\text{x}\ \text{dx}}{\sin^{\frac{3}{2}}\text{x}+\cos^{\frac{3}{2}}\text{x}}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{\frac{\pi}{2}}\frac{\sin^{\frac{3}{2}}\text{x}\ }{\sin^{\frac{3}{2}}\text{x}+\cos^{\frac{3}{2}}\text{x}}\text{dx}$ $\Rightarrow\ \ \text{I}=\int\limits_{0}^{\frac{\pi}{2}}\frac{\sin^{\frac{3}{2}}\bigg(\frac{\pi}{2}-\text{x}\bigg)}{\sin^{\frac{3}{2}}\bigg(\frac{\pi}{2}-\text{x}\bigg)+\cos^{\frac{3}{2}}\bigg(\frac{\pi}{2}-\text{x}\bigg)}\text{dx}\ \ \bigg[\because\int\limits_{0}^{\text{a}}\text{f}\text{(x)}\ \text{dx}=\int\limits_{0}^{\text{a}}\text{f}(\text{a}-\text{x})\text{dx}=\bigg]$ $\Rightarrow\ \ \text{I}=\int\limits_{0}^{\frac{\pi}{2}}\frac{\cos^{\frac{3}{2}}\text{x}}{\cos^{\frac{3}{2}}\text{x}+\sin^{\frac{3}{2}}\text{x}}\text{dx}$Adding eq. (i) and (ii),
$21=\int\limits_{0}^{\frac{\pi}{2}}\bigg[\frac{\sin^{\frac{3}{2}}\text{x}}{\sin^{\frac{3}{2}}\text{x}+\cos^{\frac{3}{2}}\text{x}}+\frac{\cos^{\frac{3}{2}}\text{x}}{\cos^{\frac{3}{2}}\text{x}+\sin^{\frac{3}{2}}\text{x}}\bigg]\text{dx}=\int\limits_{0}^{\frac{\pi}{2}}\bigg[\frac{\sin^{\frac{3}{2}}+\cos^{\frac{3}{2}}\text{x}}{\sin^{\frac{3}{2}}\text{x}+\cos^{\frac{3}{2}}\text{x}}\bigg]\text{dx}$
$\Rightarrow\ \ 21=\int\limits_{0}^{\frac{\pi}{2}}1\ \text{dx}=\bigg(\text{x}^{\frac{\pi}{2}}_{0}\bigg)\ \Rightarrow\ \ 21=\frac{\pi}{2}\ \Rightarrow\text{I}=\frac{\pi}{4}$
View full question & answer→Question 2023 Marks
Evaluate the following integrals:
$\int\frac{1-\cos\text{x}}{1+\cos\text{x}}\text{dx}$
Answer$\int\Big(\frac{1-\cos\text{x}}{1+\cos\text{x}}\Big)\text{dx}$
$=\int\frac{(1-\cos\text{x})^2}{1-\cos^2\text{x}}\text{dx}$
$=\int\frac{1+\cos^2\text{x}-2\cos\text{x}}{\sin^2\text{x}}\text{dx}$
$=\int\Big(\frac{1}{\sin^2\text{x}}+\frac{\cos^2\text{x}}{\sin^2\text{x}}-\frac{2\cos\text{x}}{\sin^2\text{x}}\Big)\text{dx}$
$=\int(\text{cosec}^2\text{x}+\cot^2\text{x}-2\cot\text{x}\text{ cosec x})\text{dx}$
$=\int(\text{cosec}^2\text{x}+\text{cosec}^2\text{x}-1-2\cot\text{x cosec x})\text{dx}$
$=\int(2\text{cosec}^2\text{x}-1-2\cot\text{x cosec x})\text{dx}$
$=\int2\text{cosec}^2\text{x dx}-\int1\text{dx}-\int2\cot\text{x cosec x }\text{dx}$
$=-2\cot\text{x}-\text{x}+2\text{cosec x}+\text{C}$
$=2(\text{cosec x}-\cot\text{x})-\text{x + C}$
View full question & answer→Question 2033 Marks
Find the integrals of the functions in Exercises:
$\frac{\cos\text{x}-\sin\text{x}}{1+\sin2\text{x}}$
Answer$\frac{\cos\text{x}-\sin\text{x}}{1+\sin2\text{x}}=\frac{\cos\text{x}-\sin\text{x}}{(\sin^2\text{x}+\cos^2\text{x})+2\sin\text{x}\cos\text{x}}$$\ \ \ \ \ \ \ \big[\sin^2\text{x}+\cos^2\text{x}=1;\ \sin2\text{x}=2\sin\text{x }\cos\text{x}\big]$
$=\frac{\cos\text{x}-\sin\text{x}}{(\sin\text{x}+\cos\text{x})^2}$
$\text{Let }\sin\text{x}+\cos{\text{x}}=\text{t}$
$\therefore(\cos\text{x}-\sin\text{x})\text{dx}=\text{dt}$
$\Rightarrow\int\frac{\cos\text{x}-\sin\text{x}}{1+\sin2\text{x}}\text{ dx}=\int\frac{\cos\text{x}-\sin\text{x}}{(\sin\text{x}+\cos\text{x})^2}\text{ dx}$
$=\int\frac{\text{dt}}{\text{t}^2}$
$=\int\text{t}^{-2}\text{dt}$
$=-\text{t}^{-1}+\text{C}$
$=-\frac{1}{\text{t}}+\text{C}$
$=\frac{-1}{\sin\text{x}+\cos\text{x}}+\text{C}$
View full question & answer→Question 2043 Marks
Evaluate the following integrals:
$\int\sqrt{4\text{x}^2-5}\text{dx}$
Answer$\text{I}=\int\sqrt{4\text{x}^2-5}\text{dx}$
$=\int\sqrt{4\Big(\text{x}^2-\frac{5}{4}\Big)}\text{dx}$
$=2\int\sqrt{\text{x}^2-\Big(\frac{\sqrt5}{2}}\Big)^2\text{dx}$
$=2\Big[\frac{\text{x}}{2}\sqrt{\text{x}^2-\frac{5}{4}}-\frac{5}{8}\int\Big|\text{x}+\sqrt{\text{x}^2-\frac{5}{4}}\Big|\Big]+\text{C}$
$\Big[\because\ \int\sqrt{\text{x}^2-\text{a}^2}\text{dx}=\frac{1}{2}\text{x}\sqrt{\text{x}^2-\text{a}^2}\frac{1}{2}\text{a}^2\int\Big|\text{x}+\sqrt{\text{x}^2-\text{a}^2}\Big|+\text{C}\Big]$
$=\text{x}\sqrt{\text{x}^2-\frac{5}{4}}-\frac{5}{4}\int\bigg|\text{x}+\sqrt{\text{x}^2-\frac{5}{4}}\bigg|+\text{C}$
View full question & answer→Question 2053 Marks
Integrate the function in Exercise:$\frac{\text{x}^{2}+\text{x}+1}{(\text{x}+1)^{2}(\text{x}+2)}$
Answer$\text{Let}\frac{\text{x}^{2}+\text{x}+1}{(\text{x}+1)^{2}(\text{x}+2)}=\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}^{2}+2\text{x}+1)$
$\Rightarrow{\text{x}^{2}+\text{x}+1}=\text{A}(\text{x}^{2}+3\text{x}+2)+\text{B}(\text{x}+2)+\text{C}(\text{x}^{2}+2\text{x}+1)$
$\Rightarrow{\text{x}^{2}+\text{x}+1}=\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 the cofficients of $\text{x}^{2},\text{x,}$ and constant term, we obtain
$\text{A}+\text{C}=1$
$3\text{A}+\text{B}+2\text{C}=1$
$2\text{A}+2\text{B}+\text{C}=1$
On solving these equations, we obtain
$\text{A}=-2,\text{B}=1,$ and $\text{C}=3$
From equation (1), we obtain
$\frac{\text{x}^{2}+\text{x}+1}{\text{(x}+1)^{2}\text{(x}+2)}=\frac{-2}{\text{(x}+1)}+\frac{3}{\text{(x}+2)}+\frac{1}{\text{(x}+1)^{2}}$
$\int\frac{\text{x}^{2}+\text{x}+1}{\text{(x}+1)^{2}\text{(x}+2)}\text{dx}=-2\int\frac{1}{\text{x}+1}\text{dx}+3\int\frac{1}{\text{(x}+2)}\text{dx}+\int\frac{1}{\text{(x}+1)^{2}}\text{dx}$
$=-2\log|\text{x}+1|+3\log|\text{x}+2|-\frac{1}{\text{(x}+1)^{2}}\text{dx}$
View full question & answer→Question 2063 Marks
Evaluate the definite integral in Exercise:$\int_{1}^{2}(4\text{x}^{3}-5\text{x}^{2}+6\text{x}+9)\text{dx}$
Answer$\text{Let}=\int\limits_{1}^{2}(4\text{x}^{3}-5\text{x}^{2}+6\text{x}+9)\text{dx}$$\int(4\text{x}^{3}-5\text{x}^{2}+6\text{x}+9)\text{dx}=4\bigg(\frac{\text{x}^{4}}{4}\bigg)-5\bigg(\frac{\text{x}^{3}}{3}\bigg)+6\bigg(\frac{\text{x}^{2}}{2}\bigg)+9\text{(x)}$
$=\text{x}^{4}-\frac{5\text{x}^{3}}{3}+3\text{x}^{2}+9\text{x}=\text{F}\text{(x)}$
By second fundamental theorem of calculus, we obtain
$\text{I}=\text{F}(2)-\text{F}(1)$
$\text{I}=\Bigg\{2^{4}-\frac{5.(2)^{3}}{3}+3(2)^{2}+9(2)\Bigg\}-\bigg\{(1)^{4}-\frac{5(1)^{3}}{3}+3(1)^{2}+9(1)\bigg\}$
$=\bigg(16-\frac{40}{3}+12+18\bigg)-\bigg(1-\frac{5}3{+3+9}\bigg)$
$=16-\frac{40}{3}+12+18-1+\frac{5}{3}-3-9$
$=33-\frac{35}{3}$
$=\frac{99-35}{3}$
$=\frac{64}{3}$
View full question & answer→Question 2073 Marks
Evaluate the definite integral in Exercise:
$\int^{\frac{\pi}{4}}_{0}(2\sec^{2}\text{x}+\text{x}^{3}+2)\text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{\frac{\pi}{4}}(2\sec^{2}\text{x}+\text{x}^{3}+2)\text{dx}$ $\int\limits(2\sec^{2}\text{x}+\text{x}^{3}+2)\text{dx}=2\tan\text{x}+\frac{\text{x}^{4}}{4}+2\text{x}=\text{F}\text{(x)}$ By second fundamental theorem of calculus, we obtain $\text{I}=\text{F}\bigg(\frac{\pi}{4}\bigg)-\text{F}(0)$ $=\left\{\Bigg(2\tan\frac{\pi}{4}+\frac{1}{4}\bigg(\frac{\pi}{4}\bigg)^{4}+2\bigg(\frac{\pi}{4}\bigg)\Bigg)-(2\tan0+0+0)\right\}$ $=2\tan\frac{\pi}{4}+\frac{\pi^{4}}{4^{5}}+\frac{\pi}{2}$$=2+\frac{\pi}{2}+\frac{\pi^{4}}{1024}$
View full question & answer→Question 2083 Marks
Write the anti-derivative of $\Big(3\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)$
AnswerLet $\text{I}=\int\Big(3\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)\text{dx}$
$\text{I}=3\sqrt{\text{x}}\text{ dx}+\int\frac{\text{dx}}{\sqrt{\text{x}}}$
$=3\frac{\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{\text{x}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\text{C}$
$=3\frac{\text{x}^{\frac{3}{2}}}{\frac{3}{2}}+\frac{\text{x}^{\frac{1}{2}}}{\frac{1}{2}}+\text{C}$
$=2\times3\times\frac{\text{x}^\frac{3}{2}}{3}+2\times\frac{\text{x}^{\frac{1}{2}}}{1}+\text{C}$
$=2\Big(\text{x}^{\frac{3}{2}}+\text{x}^{\frac{1}{2}}\Big)+\text{C}$
View full question & answer→Question 2093 Marks
Evalute the following integrals:
$\int\frac{1}{\cos3\text{x}-\cos\text{x}}\text{dx}$
Answer$\int\frac{1}{\cos3\text{x}-\cos\text{x}}\text{dx}$
$=\int\frac{1}{4\cos^2\text{x}-4\cos\text{x}}\text{dx}\big[\because\cos3\text{x}=\text{4}\cos^3\text{x}\cos\text{x}\big]$
$=\int\frac{1}{4\cos\text{x}(\cos^2\text{x}-1)}\text{dx}$
$=\frac{-1}{4}\int\frac{1}{\cos\text{x}\sin^2\text{x}}\text{dx}$
$=\frac{-1}{4}\int\Big(\frac{\sin^2\text{x}+\cos^2\text{x}}{\cos\text{x}\sin^2\text{x}}\text{dx}\Big)$
$=\frac{-1}{\text{4}}\Big[\int\sec\text{ x dx}+\int\cot\text{x cosec x dx}$
$=\frac{-1}{4}\big(\text{ln }|\sec\text{x}+\tan\text{x}|-\text{cosec x}\big)+\text{C}$
$=\frac{1}{4}\big(\text{cosec x}-\text{ln}|\sec\text{x}+\tan\text{x}|\big)+\text{C}$
View full question & answer→Question 2103 Marks
Prove the following Exercise:
$\int^{1}_{0}\sin^{-1}\text{x dx}=\frac{\pi}{2}-1$
Answer$\text{Let I}=\int^{1}\limits_{0}\sin^{-1}\text{x dx}$
$\text{I}=\int^{1}\limits_{0}\sin^{-1}\text{x.}\ 1.\text{ dx}$
Integrating by Parts, we obtain
$\text{I}=\Big[\sin^{-1}\text{x}.\text{x}\Big]^{1}_{0}-\int^{1}\limits_{0}\frac{1}{\sqrt{1-\text{x}^{2}}}.\text{x dx}$
$=\Big[\text{x}\sin^{-1}\text{x}\Big]^{1}_{0}+\frac{1}{2}\int^{1}\limits_{0}\frac{(-2\text{x)}}{\sqrt{1-\text{x}^{2}}}\text{dx}$
$\text{Let} 1-\text{x}^{2}=\text{t}\Rightarrow-2\text{x dx}=\text{dt}$
when $\text{x}=0,\text{t}=1$ and when $\text{x}=1,\text{t}=0$
$\text{I}=\Big[\text{x}\sin^{-1}\text{x}\Big]^{1}_{0}+\frac{1}{2}\int^{0}\limits_{1}\frac{\text{dt}}{\sqrt{\text{t}}}$
$=\Big[\text{x}\sin^{-1}\text{x}\Big]^{1}_{0}+\frac{1}{2}\Big[2\sqrt{\text{t}}\Big]^{0}_{1}$
$=\sin^{-1}(1)+\Big[-\sqrt{1}\Big]$
$=\frac{\pi}{2}-1$
Hence, the given result is Proved.
View full question & answer→Question 2113 Marks
If $\int\text{e}^{\text{x}}(\tan\text{x}+1)\sec\text{x dx}=\text{e}^{\text{x}}\text{ f}(\text{x})+\text{C},$ then write the value of f(x).
AnswerSince, $\int\text{e}^{\text{x}}\big(\text{f(x)}+\text{f}(\text{x})\big)\text{dx}=\text{e}^{\text{x}}\text{ f}(\text{x})+\text{C}$
We can write the expression as
$\int\text{e}^{\text{x}}(\tan\text{x}+1)\sec\text{x dx}=\text{e}^{\text{x}}\text{ f}(\text{x})+\text{C}$
$\int\text{e}^{\text{x}}(\sec\text{x}+\tan\text{x}\sec\text{x})\text{dx}=\text{f}(\text{x})\text{e}^{\text{x}}+\text{C}$
Comparing with equation $\int\text{e}^{\text{x}}(\sec\text{x}+\tan\text{x}\sec\text{x})\text{dx}=\text{f}(\text{x})\text{e}^{\text{x}}+\text{C}$ with standard equation we have
$\text{f(x)}=\sec\text{x},\text{f}'(\text{x})=\sec\text{x}\tan\text{x}$
Therefore,
$\int\text{e}^{\text{x}}(\sec\text{x}+\tan\text{x}\sec\text{x})\text{dx}=\text{e}^{\text{x}}\sec\text{x}+\text{C}$
Thus, $\text{f(x)}=\sec\text{x}$
View full question & answer→Question 2123 Marks
Evaluate the following integrals:$\int\text{e}^{\text{x}}\sec\text{x}(1+\tan\text{x})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\sec\text{x}(1+\tan\text{x})\text{dx}$
$=\int\text{e}^{\text{x}}(\sec\text{x}+\sec\text{x}\tan\text{x})\text{dx}$
Here,$\text{f(x)}=\sec\text{x}$ Put$\text{e}^{\text{x}}\text{f(x)}=\text{t}$
$\Rightarrow\text{f}'(\text{x})=\sec\text{x}\tan\text{x}$
Let $\text{e}^{\text{x}}\sec\text{x}=\text{t}$
Diff. both sides e.r.t.x
$\text{e}^{\text{x}}\sec\text{x}+\text{e}^{\text{x}}\sec\text{x}\tan\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^{\text{x}}(\sec\text{x}+\tan\text{x})\text{dx = dt}$
$\therefore\int\text{e}^{\text{x}}(\sec\text{x}+\sec\text{x}\tan\text{x})\text{dx}=\int\text{dt}$
$=\text{t}+\text{C}$
$=\text{e}^{\text{x}}\sec\text{x}+\text{C}$
View full question & answer→Question 2133 Marks
Find the integrals of the functions in Exercises:
$\frac{\cos2\text{x}}{(\cos\text{x}+\sin\text{x})^2}$
Answer$\frac{\cos2\text{x}}{(\cos\text{x}+\sin\text{x})^2}=\frac{\cos2\text{x}}{\cos^2\text{x}+\sin^2\text{x}+2\sin\text{x}\cos\text{x}}=\frac{\cos2\text{x}}{1+\sin2\text{x}}$
$\therefore\int\frac{\cos2\text{x}}{(\cos\text{x}+\sin\text{x})^2}\text{ dx}=\int\frac{\cos2\text{x}}{(1+\sin2\text{x})}\text{dx}$
$\text{Let }1+\sin2\text{x}=\text{t}$
$\Rightarrow2\cos2\text{x}\text{ dx}=\text{dt}$
$\therefore\int\frac{\cos2\text{x}}{(\cos\text{x}+\sin\text{x})^2}\text{dx}=\frac{1}{2}\int\frac{1}{\text{t}}\text{dt}$
$=\frac{1}{2}\log|\text{t}|+\text{C}$
$=\frac{1}{2}\log|1+\sin2\text{x}|+\text{C} $
$=\frac{1}{2}\log|(\sin\text{x}+\cos\text{x})^2|+\text{C} $
$=\log|\sin\text{x}+\cos\text{x}|+\text{C}$
View full question & answer→Question 2143 Marks
Find the integrals of the functions in Exercises:
$\sin3\text{x}\cos4\text{x}$
AnswerIt is know that, $\sin\text{A}\cos\text{B}=\frac{1}{2}\big\{\sin(\text{A+B})+\sin(\text{A}-\text{B}) \big\} $
$\therefore\int\sin3\text{x}\cos4\text{x dx}=\frac{1}{2}\int\big\{\sin(3\text{x}+4\text{x})+\sin(\text{3x}-\text{4x}) \big\}\text{ dx} $
$=\frac{1}{2}\int\big\{\sin\text{7x}+\sin(-\text{x}) \big\}\text{ dx} $
$=\frac{1}{2}\int\big\{\sin\text{7x}-\sin\text{x} \big\}\text{ dx} $
$=\frac{1}{2}\int\sin\text{7x dx}-\frac{1}{2}\int\sin\text{x} \text{ dx} $
$=\frac{1}{2}\bigg(\frac{-\cos7\text{x}}{7}\bigg)-\frac{1}{2}(-\cos\text{x}) +\text{ C} $
$=\frac{-\cos7\text{x}}{14}+\frac{\cos\text{x}}{2}+\text{C} $
View full question & answer→Question 2153 Marks
Evalute the following integrals:
$\int\frac{\text{e}^{\text{x}-1}+\text{x}^{\text{e}-1}}{\text{e}^\text{x}+\text{x}^\text{e}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{e}^{\text{x}-1}+\text{x}^{\text{e}-1}}{\text{e}^\text{x}+\text{x}^\text{e}}\text{dx}$
Putting $e^x + x^e = t$
$\Rightarrow\text{e}^\text{x}+\text{ex}^{\text{e}-1}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}\big(\text{e}^{\text{x}-1}+\text{x}^{\text{e}-1}\big)=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\big(\text{e}^{\text{x}-1}+\text{x}^{\text{e}-1}\big)\text{dx}=\frac{\text{dt}}{\text{e}}$
$\therefore\text{I}=\frac{1}{\text{e}}\int\frac{1}{\text{t}}\text{dt}$
$=\frac{1}{\text{e}}\text{ ln}|\text{t}|+\text{C}$
$=\frac{1}{\text{e}}\text{ ln}\Big|\text{e}^\text{x}+\text{x}^\text{e}\Big|+\text{C}\big[\because\text{t}=\text{e}^\text{x}+\text{x}^\text{e}\big]$
View full question & answer→Question 2163 Marks
Evaluate the definite integral in Exercise:
$\int^{1}_{0}\frac{\text{dx}}{\sqrt{1+\text{x}}-\sqrt{\text{x}}}$
Answer$\text{Let I}=\int^{1}\limits_{0}\frac{\text{dx}}{\sqrt{1+\text{x}}-\sqrt{\text{x}}}$
$\text{I}=\int^{1}\limits_{0}\frac{1}{\Big(\sqrt{1+\text{x}}-\sqrt{\text{x}}\Big)}\times\frac{\Big(\sqrt{1+\text{x}}+\sqrt{\text{x}}\Big)}{\Big(\sqrt{1+\text{x}}+\sqrt{\text{x}}\Big)}\text{dx}$
$=\int^{1}_{0}\frac{\sqrt{1+\text{x}}+\sqrt{\text{x}}}{1+\text{x}-\text{x}}\text{dx}$
$=\int^{1}_{0}{\sqrt{1+\text{x}}\text{dx}+\int^{1}\limits_{0}\sqrt{\text{x}}}\ \text{dx}$
$=\bigg[\frac{2}{3}\big(1+\text{x}\big)^{\frac{3}{2}}\bigg]^{1}_{0}+\bigg[\frac{2}{3}(\text{x)}^{\frac{3}{2}}\bigg]^{1}_{0}$
$=\frac{2}{3}\bigg[\big(2)^{\frac{3}{2}}-1\bigg]+\frac{2}{3}[1]$
$=\frac{2}{3}(2)^{\frac{3}{2}}$
$=\frac{2.2\sqrt{2}}{3}$
$=\frac{4\sqrt{2}}{3}$
View full question & answer→Question 2173 Marks
Evaluate the following integrals:
$\int_{0}^\limits{1}\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\text{dx}$
Answer$\int_{0}^\limits{1}\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\text{dx}$
$=\int_{0}^\limits{1}2\tan^{-1}\text{x dx}$
$=2\Big[\text{x}\tan^{-1}\text{x}\Big]^1_0-2\int_{0}^\limits{1}\frac{\text{x}}{1+\text{x}^2}\text{ dx}$
$=2\Big[\text{x}\tan^{-1}\text{x}\Big]^1_0-\Big[\log\big(1+\log\text{x}^2\big)\Big]$
$=2\frac{\pi}{4}-0-\log2+0$
$=\frac{\pi}{2}-\log2$
View full question & answer→Question 2183 Marks
Evaluate the following integrals:
$\int\frac{\sin(2+3\log\text{x})}{\text{x}}\text{ dx}$
Answer$\int\frac{\sin(2+3\log\text{x})}{\text{x}}\text{ dx}$ Let $2+3\log\text{x}=\text{t}$ $\Rightarrow\frac{3}{\text{x}}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\frac{\text{dx}}{\text{x}}=\frac{\text{dt}}{3}$ Now, $\int\frac{\sin(2+3\log\text{x})}{\text{x}}\text{ dx}$$=\frac{1}{3}\int\sin\text{t dt}$
$=\frac{1}{3}[-\cos\text{t}]+\text{C}$
$=-\frac{1}{3}\cos(2+3\log\text{x})+\text{C}$
View full question & answer→Question 2193 Marks
Evaluate the following integrals:
$\int\cos\text{x}\sqrt{4-\sin^2\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\cos\text{x}\sqrt{4-\sin^2\text{x}}\text{dx}$
Let $\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x dx}=\text{dt}$
$\Rightarrow\text{I}=\int\sqrt{4-\text{t}^2}\text{dt}$
$=\int\sqrt{2^2-\text{t}^2}\text{dt}$
$=\frac{\text{t}^2}{2}\sqrt{2^2-\text{t}^2}+\frac{4}{2}\sin^{-1}\frac{\text{t}}{2}+\text{C}$
$\therefore\ \text{I}=\frac{1}{2}\sin\text{x}\sqrt{4-\sin^2\text{x}}+2\sin^{-1}\Big(\frac{\sin\text{x}}{2}\Big)+\text{C}$
View full question & answer→Question 2203 Marks
Evaluate the following integrals:
$\int\sqrt{\text{x}-\text{x}^2}\text{dx}$
Answer$\int\sqrt{\text{x}-\text{x}^2}\text{dx}$
$=\int\sqrt{-(\text{x}^2-\text{x})}\text{dx}$
$=\int\sqrt{-\bigg\{\text{x}^2-\text{x}+\Big(\frac{1}{2}\Big)^2-\Big(\frac{1}{2}\Big)^2\bigg\}}\text{dx}$
$=\int\sqrt{\Big(\frac{1}{2}\Big)^2-\Big(\text{x}-\frac{1}{2}\Big)^2}\text{dx}$
$=\Big(\frac{\text{x}-\frac{1}{2}}{2}\Big)\sqrt{\text{x}-\text{x}^2}+\frac{1}{8}\sin^{-1}\bigg(\frac{\text{x}-\frac{1}{2}}{\frac{1}{2}}\bigg)+\text{C}$
$\Big[\because\ \int\sqrt{\text{a}^2-\text{x}^2}\text{dx}=\frac{1}{2}\text{x}\sqrt{\text{a}^2-\text{x}^2}+\frac{1}{2}\text{a}^2\sin^{-1}\frac{\text{x}}{\text{a}}=\text{C}\Big]$
$=\Big(\frac{2\text{x}-1}{4}\Big)\sqrt{\text{x}-\text{x}^2}+\frac{1}{8}\sin^{-1}(2\text{x}-1)+\text{C}$
View full question & answer→Question 2213 Marks
Evaluate the following integrals:
$\int\cot \text{x}. \text{log}\ \sin\text{x dx}$
Answer$\int\cot \text{x}. \text{log}\ \sin\text{x dx}$
$\text{Let}\ \text{log}\sin\text{x} =\text{t}$
$\Rightarrow\frac{1}{\sin\text{x}}\times \cos\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\cot\text{x dx} =\text{dt}$
$\text{Now},\int\cot\text{x}. \log\sin\text{x dx} $
$=\int\text{t}.\text{dt}$
$=\frac{\text{t}^{2}}{2}+\text{C}$
$=\frac{(\text{log}\begin{vmatrix}\sin\text{x}\end{vmatrix})^{2}}{2}+\text{C}$
View full question & answer→Question 2223 Marks
Write a value of $\int\frac{1+\cot\text{x}}{\text{x}+\log\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1+\cot\text{x}}{\text{x}+\log\sin\text{x}}\text{ dx}$
Let $\text{x}+\log\sin\text{x}=\text{t}$
$(1+\cot\text{x})\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
Hence,
$\text{I}=\log|\text{x}+\log\sin\text{x}|+\text{C}$
View full question & answer→Question 2233 Marks
Evaluate the following integrals:
$\int^\limits{9}_4\frac{\sqrt{\text{x}}}{\big(30-\text{x}^{\frac{3}{2}}\big)^2}\text{ dx}$
AnswerLet $\text{I}=\int^\limits{9}_4\frac{\sqrt{\text{x}}}{\big(30-\text{x}^{\frac{3}{2}}\big)^2}\text{ dx}$ Then,
Let $\Big(30-\text{x}^{\frac{3}{2}}\Big)=\text{t}$ Then, $-\frac{3}{2}\sqrt{\text{x}}\text{ dx}=\text{dt}$
When $\text{x}=4,\text{t}=22$ and $\text{x}=9,\text{t}=3$
$\therefore\ \text{I}=\int^\limits{3}_{22}-\frac{2}{3}\frac{1}{\text{t}^2}\text{ dt}$
$\Rightarrow\text{I}=\frac{2}{3}\Big[\frac{1}{\text{t}}\Big]^3_{22}$
$\Rightarrow\text{I}=\frac{2}{3}\Big(\frac{1}{3}-\frac{1}{22}\Big)$
$\Rightarrow\text{I}=\frac{19}{99}$
View full question & answer→Question 2243 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{1}\Big(\text{xe}^{2\text{x}}+\sin\frac{\pi\text{x}}{2}\Big)\text{dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\Big(\text{xe}^{2\text{x}}+\sin\frac{\pi\text{x}}{2}\Big)\text{dx}$ Then,
$\text{I}=\int_{0}^\limits{1}\text{xe}^{2\text{x}}\text{ dx}+\int_{0}^\limits{1}\sin\frac{\pi\text{x}}{2}\text{ dx}$
Integrating first term by parts,
$\text{I}=\Big[\text{x }\frac{\text{e}^{2\text{x}}}{2}\Big]^1_0-\int_{0}^\limits{1}1\frac{\text{e}^{2\text{x}}}{2}\text{ dx}+\bigg[-\frac{\cos\frac{\pi\text{x}}{2}}{\frac{\pi}{2}}\bigg]_0^1$
$\Rightarrow\text{I}=\Big[\text{x }\frac{\text{e}^{2\text{x}}}{2}\Big]^1_0-\Big[\frac{\text{e}^{2\text{x}}}{4}\Big]^1_0-\frac{2}{\pi}\Big[\cos\frac{\pi\text{x}}{2}\Big]^1_0$
$\Rightarrow\text{I}=\frac{\text{e}^{2}}{2}-\frac{\text{e}^{2}}{4}+\frac{1}{4}+\frac{2}{\pi}$
$\Rightarrow\text{I}=\frac{\text{e}^{2}}{4}+\frac{1}{4}+\frac{2}{\pi}$
View full question & answer→Question 2253 Marks
Integrate the function in Exercise.
$\sqrt{\text{x}^2+3\text{x}}$
Answer$\int\sqrt{\text{x}^2+3\text{x}}\text{dx}=\int\sqrt{\text{x}^2+3\text{x}+\Bigg(\frac{3}{2}\Bigg)^2-\Bigg(\frac{3}{2}\Bigg)^2}\text{dx}=\int\sqrt{\Bigg(\text{x}+\frac{3}{2}\Bigg)^2-\Bigg(\frac{3}{2}\Bigg)^2}\text{dx}$
$=\frac{\text{x}+\frac{3}{2}}{2}\sqrt{\Bigg(\text{x}+\frac{3}{2}\Bigg)^2-\Bigg(\frac{3}{2}\Bigg)^2}-\frac{\Bigg(\frac{3}{2}\Bigg)^2}{2}\text{log}\Bigg|\text{x}+\frac{3}{2}+\sqrt{\Bigg(\text{x}+\frac{3}{2}\Bigg)^2-\Bigg(\frac{3}{2}\Bigg)^2}\Bigg|+\text{c}$
$\Bigg[\therefore\int\sqrt{\text{x}^2-\text{a}^2}\text{dx}=\frac{\text{x}}{2}\sqrt{\text{x}^2-\text{a}^2}-\frac{\text{a}^2}{2}\text{log}\Bigg|\text{x}+\sqrt{\text{x}^2-\text{a}^2}\Bigg|\Bigg]$
$=\frac{2\text{x}+3}{4}\sqrt{\text{x}^2+3\text{x}}-\frac{9}{8}\text{log}\Bigg|\text{x}+\frac{3}{2}+\sqrt{\text{x}^2+3\text{x}}\Bigg|+\text{c}$
View full question & answer→Question 2263 Marks
Integrate the function in Exercise:
$\frac{\text{x}\cos^{-1}\text{x}}{\sqrt{1-\text{x}^2}}$
AnswerLet $\text{I}=\frac{\text{x}\cos^{-1}\text{x}}{\sqrt{1-\text{x}^2}}$ $\text{I}=\frac{-1}{2}\int\frac{-2\text{x}}{\sqrt{1-\text{x}^2}}.\cos^{-1}\text{x}\text{dx}$ Taking $\cos^{-1}$ x as first function and $\Bigg(\frac{-2\text{x}}{\sqrt{1-\text{x}^2}}\Bigg)$ as second function and integrating by parts, we obtain. $\text{I}=\frac{-1}{2}\Bigg[\cos^{-1}\text{x}\int\frac{-2\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}-\int\Bigg\{\Bigg(\frac{\text{d}}{\text{dx}}\cos^{-1}\text{x}\Bigg)\int\frac{-2\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}\Bigg\}\text{dx}\Bigg]$$=\frac{-1}{2}\Bigg[\cos^{-1}\text{x}.2\sqrt{1-\text{x}^2}-\int\frac{-1}{\sqrt{1-\text{x}^2}}.2\sqrt{1-\text{x}^2}\text{dx}\Bigg]\frac{-1}{2}\Bigg[\cos^{-1}\text{x}.2\sqrt{1-\text{x}^2}-\int\frac{-1}{\sqrt{1-\text{x}^2}}.2\sqrt{1-\text{x}^2}\text{dx}\Bigg]$
$=\frac{-1}{2}\Big[2\sqrt{1-\text{x}^2}\cos^{-1}\text{x}-\int2\text{dx}\Big]$ $=\frac{-1}{2}\Big[2\sqrt{1-\text{x}^2}\cos^{-1}\text{x}+2\text{x}\Big]+\text{C}$ $=-\Big[\sqrt{1-\text{x}^2}\cos^{-1}\text{x}+\text{x}\Big]+\text{C}$
View full question & answer→Question 2273 Marks
Evaluate the following integrals:
$\int\limits^\text{b}_{\text{a}}\frac{\text{f(x)}}{\text{f(x)}+\text{f}(\text{a}+\text{b}-\text{x})}\text{ dx}$
AnswerLet $\text{I}=\int\limits^\text{b}_{\text{a}}\frac{\text{f(x)}}{\text{f(x)}+\text{f}(\text{a}+\text{b}-\text{x})}\text{ dx}\ ....(\text{i})$
$=\int\limits^\text{b}_{\text{a}}\frac{\text{f}(\text{a}+\text{b}-\text{x})}{\text{f}(\text{a}+\text{b}-\text{x})+\text{f}(\text{a}+\text{b}-\text{a}-\text{b}+\text{x})}\text{ dx}$
$=\int\limits^\text{b}_{\text{a}}\frac{\text{f}(\text{a}+\text{b}-\text{x})}{\text{f}(\text{a}+\text{b}-\text{x})+\text{f(x)}}\text{ dx}$
$\therefore\ \text{I}=\int\limits^\text{b}_{\text{a}}\frac{\text{f}(\text{a}+\text{b}-\text{x})}{\text{f(x)}+\text{f}(\text{a}+\text{b}-\text{x})}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^\text{b}_{\text{a}}\bigg[\frac{\text{f(x)}}{\text{f(x)}+\text{f}(\text{a}+\text{b}-\text{x})}+\frac{\text{f}(\text{a}+\text{b}-\text{x})}{\text{f(x)}+\text{f}(\text{a}+\text{b}-\text{x})}\bigg]\text{dx}$
$=\int\limits^\text{b}_{\text{a}}\frac{\text{f(x)}+\text{f}(\text{a}+\text{b}-\text{x})}{\text{f(x)}+\text{f}(\text{a}+\text{b}-\text{x})}\text{ dx}$
$=\big[\text{x}\big]^{\text{b}}_\text{a}$
$=\text{b}-\text{a}$
Hence, $\text{I}=\frac{\text{b}-\text{a}}{2}$
View full question & answer→Question 2283 Marks
Integrate the function in exercise.
$\text{x}^2 \ \text{e}^\text{x}$
AnswerLet $\text{I}=\int\text{x}\sin3\text{x dx}$
Taking $x^2$ as first function and ex as second function and integrating by parts, we obtain.
$\text{I}=\text{x}\int\sin3\text{x dx}-\int\Bigg\{\Big(\frac{\text{d}}{\text{dx}}\text{x}\Big)\int\sin3\text{x dx}\Bigg\}$
$=\text{x}\Big(\frac{-\cos3\text{x}}{3}\Big)-\int1.\Big(\frac{-\cos3\text{x}}{3}\Big)\text{dx}$
$=\frac{-\text{x}\cos3\text{x}}{3}+\frac{1}{3}\int\cos3\text{x} \ \text{dx}$
$=\frac{-\text{x}\cos3\text{x}}{3}+\frac{1}{9}\sin3\text{x} \ \text{dx}+\text{C}$
View full question & answer→Question 2293 Marks
Write a value of $\int\text{e}^{\text{x}}\sec\text{x}(1+\tan\text{x})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\sec\text{x}(1+\tan\text{x})\text{dx}$ $=\int\text{e}^{\text{x}}(\sec\text{x}+\sec\text{x}\tan\text{x})\text{dx}$ Let $\text{e}^{\text{x}}\sec\text{x}=\text{t}$ $(\text{e}^{\text{x}}\sec\text{x}+\text{e}^{\text{x}}\sec\text{x}\tan\text{x})\text{dx}=\text{dt}$ $\text{e}^{\text{x}}\sec\text{x}(1+\tan\text{x})\text{dx}=\text{dt}$ $\therefore\text{I}=\int\text{dt}$ $=\text{t}+\text{C}$$=\text{e}^{\text{x}}\sec\text{x}+\text{C}$ $(\because\text{t}=\text{e}^{\text{x}}\sec\text{x})$
View full question & answer→Question 2303 Marks
Evalute the following integrals:
$\int\frac{1+\tan\text{x}}{1-\tan\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1+\tan\text{x}}{1-\tan\text{x}}\text{dx}$
$=\int\bigg(\frac{1+\frac{\sin\text{x}}{\cos\text{x}}}{1-\frac{\sin\text{x}}{\cos\text{x}}}\bigg)\text{dx}$
$=\int\Big(\frac{\cos\text{x}+\sin\text{x}}{\cos\text{x}-\sin\text{x}}\Big)\text{dx}$
Putting $\cos\text{x}-\sin\text{x}=\text{t}$
$\Rightarrow(-\sin\text{x}-\cos\text{x})\text{dx}=\text{dt}$
$\Rightarrow(\sin\text{x}+\cos\text{x})\text{dx}=-\text{dt}$
$\therefore\text{I}=-\int\frac{1}{\text{t}}\text{dt}$
$=-\text{ln}|\text{t}|+\text{C}$
$=-\text{ln}|\cos\text{x}-\sin\text{x}|+\text{C}\ \big[\because\text{t}\cos\text{x}-\sin\text{x}\big]$
View full question & answer→Question 2313 Marks
By using the properties of definite integral, evaluate the integral in Exercise:$ \int^{\frac{\pi}{2}}_{\frac{-\pi}{2}}\sin^{2}\text{x}\ \text{dx}$
Answer$ \text{Let}\ \text{I}=\int^{\frac{\pi}{2}}\limits_{\frac{-\pi}{2}}\sin^{2}\text{x}\ \text{dx}=2\int\limits_{0}^{\frac{\pi}{2}}\sin^{2}\text{x}\ \text{dx}..\text{(i)}\Big[\because\int^{\text{a}}_{-\text{a}}\text{f}\text{(x)}\ \text{dx}=2\int^{\text{a}}_{0}\text{f}\text{(x)}\text{dx},$when$\ \text{f(x)}$ is even function$\Big]$ $\Rightarrow\ \ \text{I}=2\int\limits_{0}^{\frac{\pi}{2}}\sin^{2}\bigg(\frac{\pi}{2}-\text{x}\bigg)\text{dx}\ \ \ \Big[\because\int^{\text{a}}\limits_{0}\text{f}\text{(x)}\ \text{dx}=\int\limits_{0}^{\text{a}}\text{f}\text{(a}-\text{x)}\text{dx}=\Big]$
$\Rightarrow\ \ \text{I}=2\int^{\frac{\pi}{2}}\limits_{0}\cos^{2}\text{x}\ \text{dx}$
Adding eq. (i) and (ii),
$21=2\int^{\frac{\pi}{2}}\limits_{0}\big(\sin^{2}\text{x}+\cos^{2}\text{x}\big)\text{dx}=2\int^{\frac{\pi}{2}}\limits_{0}1\text{dx}=2\text{(x)}^{\frac{\pi}{2}}_{0}=2.\frac{\pi}{2}=\pi$
$\Rightarrow\ \ \ \text{I}=\frac{\pi}{2}$
View full question & answer→Question 2323 Marks
Evaluate the following integrals:$\int\text{x}\sin\text{x}\cos\text{x dx}$
AnswerLet $\text{I}=\int\text{x}\sin\text{x}\cos\text{x dx}$
$=\int\frac{\text{x}}{2}(2\sin\text{x} \cos\text{x})\text{dx}$
$=\frac{1}{2}\int\text{x}\sin2\text{x dx}$
Using integration by parts,
$=\frac{1}{2}[\text{x}\int\sin2\text{x dx}-\int(1\times\int\sin2\text{x dx})\text{dx}]$
$=\frac{1}{2}\Big[\text{x}\Big(\frac{-\cos2\text{x}}{2}\Big)-\int\Big(\frac{-\cos2\text{x}}{2}\Big)\text{dx}\Big]$
$=-\frac{1}{4}\text{x}\cos2\text{x}+\frac{1}{4}\int\cos2\text{x dx}$
$\text{I}=-\frac{1}{4}\text{x}\cos2\text{x}+\frac{1}{8}\sin2\text{x}+\text{C}$
View full question & answer→Question 2333 Marks
Evaluate the following integrals:
$\int^\limits4_1\text{f(x)}\text{dx},$ Where $\text{f(x)}=\begin{cases}4\text{x}+3,&\text{if }\ 1\leq\text{x}\leq2\\3\text{x}+5,&\text{if }\ 2\leq\text{x}\leq4\end{cases}$
AnswerWe have,
$\int^\limits4_1\text{f(x)}\text{dx},$ Where $\text{f(x)}=\begin{cases}4\text{x}+3,&\text{if }\ 1\leq\text{x}\leq2\\3\text{x}+5,&\text{if }\ 2\leq\text{x}\leq4\end{cases}$
$\text{I}=\int^\limits4_1\text{f(x)}\text{dx}$
$\Rightarrow\text{I}=\int^\limits2_1\text{f(x)}\text{dx}+\int^\limits4_2\text{f(x)}\text{dx}$ [Additive property]
$\Rightarrow\text{I}=\int^\limits2_1(4\text{x}+3)\text{dx}+\int^\limits4_2(3\text{x}+5)\text{dx}$
$\Rightarrow\text{I}=\Big[2\text{x}^2+3\text{x}\Big]^2_1+\Big[\frac{3\text{x}^2}{2}+5\text{x}\Big]^4_2$
$\Rightarrow\text{I}=8+6-2-3+24+20-6-10$
$\Rightarrow\text{I}=37$
View full question & answer→Question 2343 Marks
Evalute the following integrals:
$\int\frac{10\text{x}^9+10^\text{x}\log_\text{e}10}{10^\text{x}+\text{x}^{10}}\text{dx}$
AnswerLet $\text{I}=\int\frac{10\text{x}^9+10^\text{x}\log_\text{e}10}{10^\text{x}+\text{x}^{10}}\text{dx}\ .....(\text{i})$
Let $10^\text{x}+\text{x}^{10}=\text{t}$ then,
$\text{d}(10^\text{x}+\text{x}^{10})=\text{dt}$
$\Rightarrow(10^\text{x}\log_\text{e}10+10\text{x}^9)\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{(10\text{x}^9+10^\text{x}\log_\text{e}10)}$
Putting $10^x + x^{10} = t$ and $\text{dx}=\frac{\text{dt}}{10\text{x}^9+10^\text{x}\log_\text{e}10}$ in equation (i), we get,
$\text{I}=\frac{10\text{x}^9+10^\text{x}\log_\text{e}10}{\text{t}}\times\frac{\text{dt}}{10\text{x}^9+10^\text{x}\log_\text{e}10}$
$=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|10^\text{x}+\text{x}^{10}|+\text{C}$
$\therefore\text{I}=\log|10^\text{x}+\text{x}^{10}|+\text{C}$
View full question & answer→Question 2353 Marks
Integrate the function in Exercise:$\frac{\sin^{8}-\cos^{8}\text{x}}{1-2\sin^{2}\text{x}\cos^{2}\text{x}}$
Answer$\frac{\sin^{8}\text{x}-\cos^{8}\text{x}}{1-2\sin^{2}\text{x}\cos^{2}\text{x}}=\frac{\Big(\sin^{4}\text{x}+\cos^{4}\text{x)}(\sin^{4}\text{x}-\cos^{4}\text{x}\Big)}{\sin^{2}\text{x}+\cos^{2}\text{x}-\sin^{2}\text{x}\cos^{2}\text{x}-\sin^{2}\text{x}\cos^{2}\text{x}}$
$=\frac{\Big(\sin^{4}\text{x}+\cos^{4}\text{x}\Big)\Big(\sin^{2}\text{x}+\cos^{2}\text{x}\Big)\Big(\sin^{2}\text{x}-\cos^{2}\text{x}\Big)}{\Big(\sin^{2}\text{x}-\sin^{2}\text{x}\cos^{2}\text{x}\Big)+\Big(\cos^{2}\text{x}-\sin^{2}\text{x}\cos^{2}\text{x}\Big)}$
$=\frac{\Big(\sin^{4}\text{x}+\cos^{4}\text{x}\Big)\Big(\sin^{2}\text{x}-\cos^{2}\text{x}\Big)}{\Big(\sin^{2}\text{x}\Big(1-\cos^{2}\text{x}\Big)+\cos^{2}\text{x}\Big(1-\sin^{2}\text{x}\Big)}$
$=\frac{-\Big(\sin^{4}\text{x}+\cos^{4}\text{x}\Big)\Big(\cos^{2}\text{x}-\sin^{2}\text{x}\Big)}{\Big(\sin^{4}\text{x}+\cos^{4}\text{x}\Big)}$
$=-\cos2\text{x}$
$\therefore\int\frac{\sin^{8}\text{x}-\cos^{8}\text{x}}{1-2\sin^{2}\text{x}\cos^{2}\text{x}}\text{dx}=\int-\cos2\text{x}\ \text{dx}=-\frac{\sin2\text{x}}{2}+\text{C}$
View full question & answer→Question 2363 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{2}\frac{1}{\sqrt{3+2\text{x}-\text{x}^2}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{2}\frac{1}{\sqrt{3+2\text{x}-\text{x}^2}}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{2}\frac{1}{\sqrt{-\text{x}^2+2\text{x}-1+1+3}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{2}\frac{1}{\sqrt{-(\text{x}-1)^2+4}}\text{ dx}$
$\Rightarrow\text{I}=\Big[\sin^{-1}\frac{(\text{x}-1)}{2}\Big]^{2}_0$
$\Rightarrow\text{I}=\sin^{-1}\frac{1}{2}-\sin^{-1}\Big(-\frac{1}{2}\Big)$
$\Rightarrow\text{I}=2\sin^{-1}\frac{1}{2}$
$\Rightarrow\text{I}=2\times\frac{\pi}{6}$
$\Rightarrow\text{I}=\frac{\pi}{3}$
View full question & answer→Question 2373 Marks
$\int\frac{1}{1+\cos3\text{x}}\text{dx}$
Answer$\int\frac{\text{dx}}{1+\cos3\text{x}}$
$=\int\frac{(1-\cos3\text{x})}{(1+\cos3\text{x})(1-\cos3\text{x})}\text{dx}$
$=\int\Big(\frac{1-\cos3\text{x}}{1-\cos^23\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{1-\cos3\text{x}}{\sin^23\text{x}}\Big)\text{dx}$
$=\int\text{cosec}^23\text{x}\text{ dx}-\int\text{cosec}3\text{x}\cot3\text{x dx}$
$=-\frac{\cot3\text{x}}{3}+\frac{\text{cosec}3\text{x}}{3}+\text{c}$
$=\frac{1}{3}[\text{cosec}3\text{x}-\cot3\text{x}]+\text{c}$
$=\frac{1}{3}\bigg[\frac{1}{\sin3\text{x}}-\frac{\cos3\text{x}}{\sin3\text{x}}\bigg]+\text{c}$
$=\frac{1}{3}\bigg[\frac{1-\cos3\text{x}}{\sin3\text{x}}\bigg]+\text{c}$
View full question & answer→Question 2383 Marks
Evaluate the integral in Exercise:
$\int^{2}_{1}\bigg(\frac{1}{\text{x}}-\frac{1}{2\text{x}^{2}}\bigg)\text{e}^{2\text{x}}\text{dx}$
Answer$\text{Let}$
$\text{I}=\int_{1}^{2}\bigg(\frac{1}{\text{x}}-\frac{1}{2\text{x}^{2}}\bigg)\text{e}^{2\text{x}}\text{dx}$
$=\int^{2}_{1}\frac{1}{\text{x}}\text{e}^{2\text{x}}\text{dx}+\int\frac{-1}{2\text{x}^{2}}\text{e}^{2\text{x}}\text{dx}$
$=\bigg[\frac{1}{\text{x}}\frac{\text{e}^{2\text{x}}}{2}\bigg]^{2}_{1}-\int^{2}_{1}\frac{-1}{\text{x}^{2}}\frac{\text{e}^{2\text{x}}}{2}\text{dx}+\int^{2}_{1}\frac{-1}{2\text{x}^{2}}\text{e}^{2\text{x}}\text{dx}$ [integrating by parts]
$=\frac{1}{2}\bigg[\frac{\text{e}^{2\text{x}}}{\text{x}}\bigg]^{2}_{1}-\int^{2}_{1}\frac{-1}{2\text{x}^{2}}{\text{e}^{2\text{x}}}{}\text{dx}+\int^{2}_{1}\frac{-1}{2\text{x}^{2}}\text{e}^{2\text{x}}\text{dx}$
$=\frac{1}{2}\bigg[\frac{\text{e}^{4}}{2}-\frac{\text{e}^{2}}{1}\bigg]=\frac{1}{4}\big(\text{e}^{4}-2\text{e}^{2}\big)=\frac{1}{4}\big(\text{e}^{2}-2\big)$
View full question & answer→Question 2393 Marks
$\int(5\text{x}+3)\sqrt{2\text{x}-1}\text{ dx}$
Answer$\text{Let I}=\int(5\text{x}+3)\sqrt{2\text{x}-1}\text{ dx}$
$\text{Putting}\ \ 2\text{x}-1=\text{t}$
$\Rightarrow2\text{x}=\text{t}+1$
$\Rightarrow\text{x}=\frac{\text{t}+1}{2}$
$\&\ 2\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{2}$
$\therefore\text{I}=\int\Big[5\Big(\frac{\text{t}+1}{2}\Big)+3\Big]\times\sqrt{\text{t}}\times\frac{\text{dt}}{2}$
$=\int\Big(\frac{5\text{t}}{5}+\frac{5}{2}+3\Big)\times\frac{\sqrt{\text{t}}\text{ dt}}{2}$
$=\frac{1}{4}\int(5\text{t}+11)\text{t}^\frac{1}{2}\text{ dt}$
$=\frac{1}{4}\int(5\text{t}^\frac{3}{2}+11\text{t}^\frac{1}{2})\text{ dt}$
View full question & answer→Question 2403 Marks
Integrate the function in Exercise:
$\tan^{-1}\text{x}$
AnswerLet $\text{I}=\int1.\tan^{-1}\text{x dx}$
Taking $\tan^{-1} \text{x}$ as first function and 1 as second function and integrating by parts, we obtain.
$\text{I}=\tan^{-1}\text{x}\int1.\text{dx}-\int\Bigg\{\Bigg(\frac{\text{d}}{\text{dx}}\tan^{-1}\text{x}\Bigg)\int1. \ \text{dx}\Bigg\}\text{dx}$
$=\tan^{-1}\text{x}.\text{x}-\int\frac{1}{1+\text{x}^2}.\text{x dx}$
$=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\int\frac{2\text{x}}{1+\text{x}^2}\text{dx}$
$=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\text{log}|1+\text{x}^2|+\text{C}$
$=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\text{log}(1+\text{x}^2)+\text{C}$
View full question & answer→Question 2413 Marks
Evaluate the following integrals:
$\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}^2}-\frac{2}{\text{x}^3}\Big)\text{dx}$
Answer$\text{I}=\int\text{e}^{\text{x}}\big(\text{x}^{-2}-2\text{x}^{-3}\big)\text{dx}$
$=\int\text{e}^{\text{x}}\text{x}^{-2}\text{dx}-2\int\text{e}^{\text{x}}\text{x}^{-3}\text{dx}$
Integration by parts
$=\text{e}^{\text{x}}\text{x}^{-2}-\int\text{e}^{\text{x}}\Big(\frac{\text{d}}{\text{dx}}\big(\text{x}^{-2}\big)\Big)\text{dx}-2\int\text{e}^{\text{x}}\text{x}^{-3}\text{dx}$
$=\text{e}^{\text{x}}\text{x}^{-2}+2\int\text{e}^{\text{x}}\text{x}^{-3}\text{dx}-2\int\text{e}^{\text{x}}\text{x}^{-3}\text{dx}$
$=\frac{\text{e}^{\text{x}}}{\text{x}^2}+\text{C}$
$\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}^2}-\frac{2}{\text{x}^3}\Big)\text{dx}=\frac{\text{e}^\text{x}}{\text{x}^2}+\text{C}$
View full question & answer→Question 2423 Marks
Write a value of $\int\frac{\cos\text{x}}{3+2\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\cos\text{x}}{3+2\sin\text{x}}\text{ dx}$
Let $3+2\sin\text{x}=\text{t}$
$2\cos\text{x dx}=\text{dt}$
$\therefore\ \text{I}=\frac{1}{2}\int\frac{\text{dt}}{\text{t}}$
$=\frac{1}{2}\log\text{t}+\text{C}$
$=\frac{1}{2}\log(3+2\sin\text{x})+\text{C}$
$\therefore\ \int\frac{\cos\text{x}}{3+2\sin\text{x}}\text{ dx}=\frac{1}{2}\log(3+2\sin\text{x})+\text{C}$
View full question & answer→Question 2433 Marks
Write a value of $\int\frac{\sin\text{x}-\cos\text{x}}{\sqrt{1+\sin2\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sin\text{x}-\cos\text{x}}{\sqrt{1+\sin2\text{x}}}\text{ dx}$
$=\int\frac{\sin\text{x}-\cos\text{x}}{\sqrt{\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}+\cos\text{x}}}\text{ dx}$
$=\int\frac{\sin\text{x}-\cos\text{x}}{\sqrt{(\sin\text{x}+\cos\text{x})^2}}\text{ dx}$
$=\int\frac{\sin\text{x}-\cos\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$
Let $\sin\text{x}+\cos\text{x}=\text{t}$
$(\cos\text{x}-\sin\text{x})\text{dx}=\text{dt}$
$\therefore\ \text{I}=-\int\frac{\text{dt}}{\text{t}}$
$=-\log\text{t}+\text{C}$
$\text{I}=\log|\sin\text{x}+\cos\text{x}|+\text{C}$
View full question & answer→Question 2443 Marks
Integrate the function in Exercise:
$\frac{\sec^2\text{x}}{\sqrt{\tan^2\text{x}+4}}$
Answer$\text{Ler I}=\int\frac{\sec^2\text{x}}{\sqrt{\tan^2\text{x}+4}}\text{ dx}\ \ \ \ ...\text{(i)}$
$\text{Putting }\tan\text{x}=\text{t}\ \ \Rightarrow\ \ \sec^2\text{x}=\frac{\text{dt}}{\text{dx}}\ \ \Rightarrow\ \ \sec^2\text{x}\text{ dx}=\text{dt}$
$\therefore\ \ \text{From eq. (i), }\text{ I}=\int\frac{\text{dt}}{\sqrt{\text{t}^2+4}}=\int\frac{1}{\sqrt{\text{t}^2+(2)^2}}\text{ dt}$
$=\log\bigg|\text{t}+\sqrt{\text{t}^2+(2)^2}\bigg|+\text{c}$
$=\log\Big|\tan\text{x}+\sqrt{\tan^2\text{x}+4}\Big|+\text{c}$
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Write a value of $\int\frac{1}{1+\text{e}^{\text{x}}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{1+\text{e}^{\text{x}}}\text{dx}$
Dividing and multiplying by $e^x$
$=\frac{\text{e}^{-\text{x}}}{\text{e}^{-\text{x}}+1}\text{dx}$
Let $\text{e}^{-\text{x}}+1=\text{t}$
$-\text{e}^{-\text{x}}\text{dx}=\text{dt}$
$\therefore\ \text{I}=-\int\frac{\text{dt}}{\text{t}}$
$=-\log|\text{t}|+\text{C}$
$\therefore\ \text{I}=-\log|1+\text{e}^{-\text{x}}|+\text{C}$
View full question & answer→Question 2463 Marks
Evaluate the definite integral in Exercise:
$\int_{0}^{2}\frac{6\text{x}+3}{\text{x}^{2}+4}\text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{2}\frac{6\text{x}+3}{\text{x}^{2}+4}\text{dx}$$\int\frac{6\text{x}+3}{\text{x}^{2}+4}\text{dx}=3\int\frac{2\text{x}+1}{\text{x}^{2}+4}\text{dx}$
$=3\int\frac{2\text{x}}{\text{x}^{2}+4}\text{dx}+3\int\frac{1}{\text{x}^{2}+4}\text{dx}$
$=3\log(\text{x}^{2}+4)+\frac{3}{2}\tan^{-1}\frac{\text{x}}{2}=\text{F}\text{(x)}$
By second fundamental theorem of calculus, we obtain
$\text{I}=\text{F}(2)-\text{F}(0)$
$=\left\{3\log(2^{2}+4)+\frac{3}{2}\tan^{-1}\bigg(\frac{2}{2}\bigg)\right\}-\left\{3\log(0+4)+\frac{3}{2}\tan^{-1}\bigg(\frac{0}{2}\bigg)\right\}$
$=3\log8+\frac{3}{2}\tan^{-1}1-3\log4-\frac{3}{2}\tan^{-1}0$
$=3\log8+\frac{3}{2}\bigg(\frac{\pi}{4}\bigg)-3\log4-0$
$=3\log\bigg(\frac{8}{4}\bigg)+\frac{3\pi}{8}$
$=3\log2+\frac{3\pi}{8}$
View full question & answer→Question 2473 Marks
Evaluate the following integrals:
$\int\log(\text{x}+1)\text{dx}$
AnswerLet $\text{I}=\int\log(\text{x}+1)\text{dx}$
$=\int1\times\log(\text{x}+1)\text{dx}$
Using integration by parts,
$\text{I}=\log(\text{x}+1)\int1\text{dx}-\int\Big(\frac{1}{\text{x}+1}\times\int1\text{dx}\Big)\text{dx+C}$
$=\text{x}\log(\text{x}+1)-\int\Big(\frac{\text{x}}{\text{x}+1}\Big)\text{dx+C}$
$=\text{x}\log(\text{x}+1)-\int\Big(1-\frac{1}{\text{x}+1}\Big)\text{dx+C}$
$\text{I}=\text{x}\log(\text{x}+1)-\text{x}+\log(\text{x}+1)+\text{C}$
View full question & answer→Question 2483 Marks
Evaluate the following integrals:
$\int\big(2-3\text{x}\big)\big(3+2\text{x}\big)\big(1-2\text{x}\big)\text{dx}$
Answer$\int\big(2-3\text{x}\big)\big(3+2\text{x}\big)\big(1-2\text{x}\big)\text{dx}$
$=\int\big(6+4\text{x}-9\text{x}-6\text{x}^2\big)\big(1-2\text{x}\big)\text{dx}$
$=\int\big(-6\text{x}^2-5\text{x}+6\big)\big(1-2\text{x}\big)\text{dx}$
$=\int\big(-6\text{x}^2+12\text{x}^3-5\text{x}+10\text{x}^2+6-12\text{x}\big)\text{dx}$
$=\int\big(4\text{x}^2+12\text{x}^3-17\text{x}+6\big)\text{dx}$
$=\int\big(12\text{x}^3+4\text{x}^2-17\text{x}+6\big)\text{dx}$
$=\frac{12}{4}\text{x}^4+\frac{4}{3}\text{x}^3-\frac{17}{2}\text{x}^2+6\text{x}+\text{C}$
$=3\text{x}^4+\frac{4}{3}\text{x}^3-\frac{17}{2}\text{x}^2+6\text{x}+\text{C}$
View full question & answer→Question 2493 Marks
Find the integrals of the functions in Exercises:
$\sin^3\text{x}\cos^3\text{x}$
Answer$\text{Let}\text{ I}=\int\sin^3\text{x}\cos^3\text{x}\text{ dx}$
$=\int\cos^3\text{x }\sin^2\text{x }\sin\text{x}\text{ dx}$
$=\int\cos^3\text{x}\big(1-\cos^2\text{x}\big)\sin\text{x}\text{ dx}$
$\text{Let}\cos\text{x}=\text{t}$
$\Rightarrow -\sin\text{x}\text{ dx}=\text{dt}$
$\Rightarrow\text{I}=-\int\text{t}^3\big(1-\text{t}^2\big)\text{ dt}$
$=-\int\big(\text{t}^3-\text{t}^5\big)\text{dt}$
$=-\Bigg\{\frac{\text{t}^4}{4}-\frac{\text{t}^6}{6}\Bigg\}+\text{C}$
$=-\Bigg\{\frac{\cos^4\text{x}}{4}-\frac{\cos^6\text{x}}{6}\Bigg\}+\text{C}$
$=\frac{\cos^6\text{x}}{6}-\frac{\cos^4\text{x}}{4}+\text{C}$
View full question & answer→Question 2503 Marks
Evaluate the following integrals:
$\int_{0}^\limits{1}\frac{2\text{x}}{1+\text{x}^4}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\frac{2\text{x}}{1+\text{x}^4}\text{ dx}$
Let $\text{x}^2=\text{t}$ Then, $2\text{xdx}=\text{dt}$
When $\text{x}=10,\text{t}=0$ and $\text{x}=1,\text{t}=1$
$\therefore\ \text{I}=\int_{0}^\limits{1}\frac{2\text{x}}{1+\text{x}^4}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{1}\frac{1}{1+\text{t}^2}\text{ dt}$
$\Rightarrow\text{I}=\big[\tan^{-1}\text{t}\big]^1_0$
$\Rightarrow\text{I}=\tan^{-1}1-\tan^{-1}0$
$\Rightarrow\text{I}=\frac{\pi}{4}$
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