Question 13 Marks
Evaluate:$\int\sin\text{4x}\cos\text{3x dx}$.
AnswerWriting I = $\frac{1}{2}\int2\sin\text{ 4x }\cos\text{ 3x dx}=\frac{1}{2}\int(\sin\text{7x}+\sin\text{x) dx}$$=\frac{1}{2}\Bigg[\frac{-\cos\text{ 7x}}{7}-\cos\text{x}\Bigg]+\text{c }\text{ OR }-\frac{1}{14}\cos\text{ 7x}-\frac{1}{2}\cos\text{ x + c}$.
View full question & answer→Question 23 Marks
Evaluate:$\int\text{x log 2x dx}$.
Answer$\text{I}=\int\log2\text{x}\cdot\text{x dx}=\log2\text{x}\cdot\frac{\text{x}^{2}}{2}-\int\frac{1}{\text{x}}\cdot\frac{\text{x}^{2}}{2}\text{dx}+\text{c}_{1}$= $\frac{\text{x}^{2}}{2}\cdot\log\text{2x}-\frac{1}{2}\int\text{x dx + c}_{1}=\frac{\text{x}^{2}}{2}\cdot\log\text{ 2x}-\frac{\text{x}^{2}}{4}+\text{c}.$
View full question & answer→Question 33 Marks
Evaluate:$\int\frac{\text{2x.tan-1(x}^{2})}{\text{1 + x}^{4}}\text{dx}. $
Answer$\int\frac{\text{2x.tan}^{-1}\text{(x}^{2})}{\text{1 + x}^{4}}\text{dx}=\int\text{t dt }\text{ where tan}^{-1}\text{(x}^{2})=\text{t }\text{ }\therefore\frac{\text{2x}}{\text{1 + x}^{4}}\text{dx}=\text{dt}$
$=\frac{\text{t}^{2}}{2}+\text{c}=\frac{1}{2}\Big[\tan^{-1}\text{(x}^{2})\Big]^{2}+\text{c}.$
View full question & answer→Question 43 Marks
Evaluate:$\int \frac{1 + x^{2}}{1 + x^{4}} \text{dx}$
Answer$\text{I} \int \frac{1 + x^{2}}{1 + x^{4}} \text{dx} = \int \frac{1 +\frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}}}\text{dx} = \int \frac{1 +\frac{1}{x^{2}}}{\bigg( x - \frac{1}{x}\bigg)^{2} + \bigg(\sqrt{2}\bigg)^{2}} \text{dx}$$= \frac{1}{\sqrt{2}} \tan^{-1} \bigg(x-\frac{\frac{1}{x}}{\sqrt{2}}\bigg) + c $
$= \frac{1}{\sqrt{2}} \tan^{-1}\bigg(\frac{x^{2} -1}{\sqrt{2x}}\bigg) + c $
View full question & answer→Question 53 Marks
Evaluate: $\int \cos \text{4 x} \cos 3\text{x dx}$
Answer$\text{I} = \int \cos \text{4 x} \cos 3\text{x dx} = \frac{1}{2} \int(\cos 7x + \cos x) \text{dx}$$= \frac{1}{2} \bigg[ \frac{\sin 7x}{7} + \sin x\bigg] + c$or $\frac{1}{14} \sin 7 x + \frac{1}{2} \sin x + c$
View full question & answer→Question 63 Marks
Evaluate:$\int \frac{\sin (\text{x} - \alpha)}{\sin (\text{x} + \alpha)} \text{dx}$
Answer$I = \int \frac{\sin(x + a)-2\alpha)}{\sin(x + \alpha)}$$= \int \frac{\sin(x + \alpha). \cos 2\alpha - \cos (x + \alpha) .2 \alpha}{\sin(x + \alpha)} dx$
$= \cos 2 \alpha \int dx - \sin 2 \alpha \int \frac{\cos(x + \alpha)}{\sin(x + \alpha)} dx$
$= x \cos 2 \alpha - \sin 2 \alpha \log |\sin (x + \alpha)|+c$
View full question & answer→Question 73 Marks
Evaluate:$\int{\frac{dx} {\sqrt{x^{2} - 3x + 2}}}$
Answer$\int{\frac{dx} {\sqrt{x^{2} - 3x + 2}}}=\int \frac{dx}{\sqrt{\bigg(x - \frac{3}{2}\bigg)^{2} - \bigg(\frac{1}{2}\bigg)^{2}}}$$= \log\bigg| \bigg(x - \frac{3}{2}\bigg) + \sqrt{x^{2} - 3x + 2}\bigg| + c$
View full question & answer→Question 83 Marks
Evaluate the following integrals:
$\int\text{x}^2\cos2\text{x dx}$
AnswerLet $\text{I}=\int\text{x}^2\cos2\text{x}\text{ dx}$
Using integration by parts,
$\text{I}=\text{x}^2\int\cos2\text{x dx}-\int(2\text{x}\int\cos2\text{x dx})\text{dx}$
$=\text{x}^2\frac{\sin2\text{x}}{2}-2\int\text{x}\Big(\frac{\sin2\text{x}}{2}\Big)\text{dx}$
$=\frac{1}{2}\text{x}^2\sin2\text{x}-\int\text{x}\sin2\text{x dx}$
$=\frac{1}{2}\text{x}^2\sin2\text{x}-[\text{x}\int\sin2\text{x dx}-\int(1\int\sin2\text{x dx})\text{dx}]$
$=\frac{1}{2}\text{x}^2\sin2\text{x}-\bigg[\text{x}\Big(\frac{-\cos2\text{x}}{2}\Big)-\int\Big(-\frac{\cos2\text{x}}{2}\Big)\text{dx}\bigg]$
$=\frac{1}{2}\text{x}^2\sin2\text{x}+\frac{\text{x}}{2}\cos2\text{x}-\frac{1}{2}\int(\cos2\text{x})\text{dx}$
$\text{I}=\frac{1}{2}\text{x}^2\sin2\text{x}+\frac{\text{x}}{2}\cos2\text{x}-\frac{1}{4}\sin2\text{x}+\text{C}$
View full question & answer→Question 93 Marks
Evaluate the following integrals:
$\int\frac{1}{\text{x}^2+6\text{x}+13}\text{dx}$
AnswerWe have $\text{x}^2+6\text{x}+13=\text{x}^2+6\text{x}+3^2-3^2+13=(\text{x}+3)^2+4$
Sol, $\int\frac{\text{dx}}{\text{x}^2+6\text{x}+13}=\int\frac{1}{(\text{x}+3)^2+2^2}\text{dx}$
Let $\text{x}+3=\text{t}$ Then $\text{dx = dt}$
Therefore, $\int\frac{\text{dx}}{\text{x}^2+6\text{x}+13}=\int\frac{\text{dt}}{\text{t}^2+2^2}=\frac{1}{2}\tan^{-1}\frac{\text{t}}{2}+\text{c}$ [by 7.4 (3)]
$=\frac{1}{2}\tan^{-1}\frac{\text{x}+3}{2}+\text{C}$
View full question & answer→Question 103 Marks
Evalute the following integrals:
$\int\frac{\cot\text{x}}{\log\sin\text{x}}\text{dx}$
AnswerNote: Here we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$
Let $\text{I}=\int\frac{\cot\text{x}}{\log\sin\text{x}}\text{dx}$
Putting $\log\sin\text{x}=\text{t}$
$\Rightarrow\cot\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\cot\text{x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\log|\text{t}|+\text{C}$
$=\log|\log\sin\text{x}|+\text{C}\ \big[\because\text{t}=\log\sin\text{x}\big]$
View full question & answer→Question 113 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 123 Marks
Evaluate the following integrals:
$\int^\limits\frac{\pi}{4}_{0}\sin^32\text{t}\cos2\text{t}\text{ dt}$
AnswerLet $\text{I}\int^\limits\frac{\pi}{4}_{0}\sin^32\text{t}\cos2\text{t}\text{ dt}$ Then,Let $\sin2\text{t}=\text{u}$ Then, $2\cos2\text{t dt} =\text{du}$
When $\text{t}=0,\text{u}=0$ and $\text{t}=\frac{\pi}{4},\text{u}=1$
$\therefore\ \text{I}=\frac{1}{2}\int^\limits{1}_{0}\text{u}^3\text{ du}$
$\Rightarrow\text{I}=\frac{1}{2}\Big[\frac{\text{u}^4}{4}\Big]^1_0$
$\Rightarrow\text{I}=\frac{1}{2}\Big(\frac{1}{4}-0\Big)$
$\Rightarrow\text{I}=\frac{1}{8}$
View full question & answer→Question 133 Marks
Integrate the rational function in exercise:
$\frac{3\text{x}-1}{(\text{x}+2)^2}$
AnswerLet, $\text{I}=\int\frac{3\text{x}-1}{(\text{x}+2)^2}\text{dx}\dots(\text{i})$Putting x + 2 = t
⇒ x = t - 2
$\frac{\text{dx}}{\text{dt}}=1$
⇒ dx = dt
Putting this value in eq. (i),
$\text{I}=\int\frac{3(\text{t}-2)-1}{(\text{t})^2}\text{dt}=\int\frac{3\text{t}-6-1}{\text{t}^2}\text{dt}=\int\frac{3\text{t}-7}{\text{t}^2}\text{dt}$
$=\int\Bigg(\frac{3\text{t}}{\text{t}^2}-\frac{7}{\text{t}^2}\Bigg)\text{dt}=\int\Bigg(\frac{3}{\text{t}}-\frac{7}{\text{t}^2}\Bigg)\text{dt}=3\int\frac{1}{\text{t}}\text{dt}-7\int\text{t}^{-2}\text{dt}$
$=3\text{log}|\text{t}|-7\frac{\text{t}^{-1}}{-1}+\text{c}=3\text{log}|\text{t}|+\frac{7}{\text{t}}+\text{c}=3\text{log}|\text{x}+2|+\frac{7}{\text{x}+2}+\text{c}$
View full question & answer→Question 143 Marks
Evalute the following integrals:
$\int\frac{\cos2\text{x}}{(\cos\text{x}+\sin\text{x})^2}\text{dx}$
Answer Let $\text{I}=\int\frac{\cos2\text{x}}{(\cos\text{x}+\sin\text{x})^2}\text{dx}$$=\int\frac{\cos^2\text{x}-\sin^2\text{x}}{(\cos\text{x}+\sin\text{x})^2}\text{dx}$
$=\int\frac{\cos\text{x}-\sin\text{x}}{\cos\text{x}+\sin\text{x}}\text{dx}$
Putting $\cos\text{x}+\sin\text{x}=\text{t}$
$\Rightarrow-\sin\text{x}+\cos\text{x}=\frac{\text{dt}}{\text{dt}}$
$\Rightarrow(\cos\text{x}-\sin\text{x})\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\text{ln}|\text{t}|+\text{C}$
$=\text{ln}|\cos\text{x}+\sin\text{x}|+\text{C}\ \big[\because\text{t}=\cos\text{x}+\sin\text{x}\big]$
View full question & answer→Question 153 Marks
Evalute the following integrals:
$\int\frac{1}{\sqrt{1-\cos2\text{x}}}\text{dx}$
Answer$\int\frac{1}{\sqrt{1-\cos2\text{x}}}\text{dx}$$\int\frac{1}{\sqrt{2\sin^2\text{x}}}\text{dx}\ \big[\because 1-\cos 2\text{x}=2\sin^2\text{x}\big]$
$=\frac{1}{\sqrt{2}}\int\text{cosec x dx}$
$=\frac{1}{\sqrt{2}}\text{ln}|\text{cosec x}-\cot\text{x}|=\text{C}$
$=\frac{1}{\sqrt{2}}\text{ln}\Big|\frac{1}{\sin\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}\Big|+\text{C}$
$=\frac{1}{\sqrt{2}}\text{ln}\bigg|\frac{2\sin^2\frac{\text{x}}{2}}{\sin\text{x}}\bigg|+\text{C} \Big[\because 1-\cos\text{x}=2\sin^2\frac{\text{x}}{2}\Big]$
$=\frac{1}{\sqrt{2}}\text{ln}\Bigg|\frac{2\sin^2\frac{\text{x}}{2}}{2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}\Bigg|+\text{C}\ \Big[\because\sin\text{x}=2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}\Big]$
$=\frac{1}{\sqrt{2}}\text{ln}\Big|\tan\frac{\text{x}}{2}\Big|+\text{C}$
View full question & answer→Question 163 Marks
If $f'(x) = 8x^3 - 2x, f'(2) = 8,$ find $f'(x).$
AnswerWe have,
$\text{f}'\text{(x)}=8\text{x}^3-2\text{x}$
$\Rightarrow\text{f}'\text{(x)}=\int\text{f}'\text{(x)dx}=\int(8\text{x}^3-2\text{x})\text{dx}$
$\Rightarrow\text{f}'\text{(x)}=\int(8\text{x}^3-2\text{x})\text{dx}$
$=\int8\text{x}^3\text{dx}-\int2\text{ x dx}$
$=\frac{8\text{x}^4}{4}-\frac{2\text{x}^2}{2}+\text{C}$
$=2\text{x}^4-\text{x}^2+\text{C}$
$\Rightarrow\text{f}'\text{(x)}=2\text{x}^4-\text{x}^2+\text{C}\ \dots(1)$
Since, $\text{f}'(2)=8$
$\therefore\ \text{f}'(2)=2(2)^4-(2)^2+\text{C}=8$
$\Rightarrow32-4+\text{C}=8$
$\Rightarrow28+\text{C}=8$
$\Rightarrow\text{C}=-20$
Putting $C = -20$ in eq. $(1),$ we get
$\text{f}'\text{(x)}=2\text{x}^4-\text{x}^2-20$
Hence, $\text{f}'\text{(x)}=2\text{x}^4-\text{x}^2-20$
View full question & answer→Question 173 Marks
Evaluate the following integrals:$\int\frac{\sin\text{x}-\cos\text{x}}{\sqrt{\sin2\text{x}}}\text{ dx}$
AnswerTo evaluate the following integral follow the steps:
$\int\frac{\sin\text{x}-\cos\text{x}}{\sqrt{\sin2\text{x}}}\text{ dx}$
$=\int\frac{\sin\text{x}-\cos\text{x}}{\sqrt{(\sin\text{x}+\cos\text{x})^2-1}}\text{ dx}$
Let $\sin\text{x}+\cos\text{x}=\text{t}$ therefore $(\cos\text{x}-\sin\text{x})\text{ dx}=\text{dt}$
Now,
$\int\frac{\sin\text{x}-\cos\text{x}}{\sqrt{(\sin\text{x}+\cos\text{x})^2-1}}\text{ dx}=-\int\frac{\text{dt}}{\sqrt{\text{t}^2-1}}$
$=-\ln\Big|\text{t}+\sqrt{\text{t}^2-1}\Big|+\text{C}$
$=-\ln\Big|\sin\text{x}+\cos\text{x}+\sqrt{\sin2\text{x}}\Big|+\text{C}$
View full question & answer→Question 183 Marks
Evaluate the following integrals:$\int\frac{\text{x}\cos^{-1}\text{x}}{\sqrt{1-\text{x}^3}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}\cos^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}$
Let the first function be $\cos^{-1}\text{x}$ and second function be $\frac{\text{x}}{\sqrt{1-\text{x}^2}}\cdot$
First we find the integral of the second function, i.e, $\int\frac{\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}.$
Put $\text{t}=1-\text{x}^2.$ Then $\text{dt}=-2\text{x dx}$
Therefore,
$\int\frac{\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}=-\frac{1}{2}\int\frac{1}{\sqrt{\text{t}}}\text{dt}$
$=-\sqrt{\text{t}}$
$=-\sqrt{1-\text{x}^2}$
Hence, using integration by parts, we get
$\int\frac{\text{x}\cos^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}=(\cos^{-1}\text{x})\int\frac{\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}-\int\bigg[\bigg(\frac{\text{d}(\cos^{-1}\text{x})}{\text{dx}}\bigg)\int\bigg(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}\bigg)\bigg]\text{dx}$
$=(\cos^{-1})\big(-\sqrt{1-\text{x}^2}\big)-\int\Big(\frac{-1}{\sqrt{1-\text{x}^2}}\Big)\big(-\sqrt{1-\text{x}^2}\big)\text{dx}$
$=-\sqrt{1-\text{x}^2}\cos^{-1}\text{x}-\text{x+C}$
Hence, $\int\frac{\text{x}\cos^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}=-\sqrt{1-\text{x}^2}\cos^{-1}\text{x}-\text{x+C}$
View full question & answer→Question 193 Marks
Evaluate the following integrals:$\int\text{e}^{\text{x}}\big(\cot\text{x}-\text{cosec}^2\text{x}\big)\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\big(\cot\text{x}-\text{cosec}^2\text{x}\big)\text{dx}$
$=\int\text{e}^{\text{x}}\cot\text{x dx}-\int\text{e}^{\text{x}}\text{cosec}^2\text{x dx}$
Integration by parts
$=\text{e}^{\text{x}}\cot\text{x}-\int\text{e}^{\text{x}}\Big(\frac{\text{d}}{\text{dx}}\cot\text{x}\Big)\text{dx}-\int\text{e}^{\text{x}}\text{cosec}^2\text{x dx}$
$=\text{e}^{\text{x}}\cot\text{x}+\int\text{e}^{\text{x}}\text{cosec}^2\text{x dx}-\int\text{e}^{\text{x}}\text{cosec}^{2}\text{x dx}$
$=\text{e}^{\text{x}}\cot\text{x+C}$
$\int\text{e}^{\text{x}}\big(\cot\text{x}-\text{cosec}^2\text{x}\big)\text{dx}=\text{e}^\text{x}\cot\text{x}+\text{C}$
View full question & answer→Question 203 Marks
Evaluate the following integrals:$\int\frac{1}{\text{x}\sqrt{4-9(\log\text{x})^2}}\text{ dx}$
Answer$\int\frac{\text{dx}}{\text{x}\sqrt{4-9(\log\text{x})^2}}$
Let $\log\text{x}=\text{t}$
$\Rightarrow\frac{1}{\text{x}}\text{ dx}=\text{dt}$
Now, $\int\frac{\text{dx}}{\text{x}\sqrt{4-9(\log\text{x})^2}}$
$=\int\frac{\text{dt}}{\sqrt{4-9\text{t}^2}}$
$=\int\frac{\text{dt}}{\sqrt{2^2-(3\text{t})^2}}$
$=\frac{1}{3}\sin^{-1}\Big(\frac{3\text{t}}{2}\Big)+\text{C}$
$=\frac{1}{3}\sin^{-1}\Big(\frac{3\log\text{x}}{2}\Big)+\text{C}$
View full question & answer→Question 213 Marks
Evaluate the following integrals:$\int\frac{\text{x}+\sin\text{x}}{1+\cos\text{x}}\text{dx}$
Answer$\int\Big(\frac{\text{x}+\sin\text{x}}{1+\cos\text{x}}\Big)\text{dx}$
$=\int\Big[\frac{\text{x}}{1+\cos\text{x}}+\frac{\sin\text{x}}{1+\cos\text{x}}\Big]\text{dx}$
$=\int\bigg[\frac{\text{x}}{2\cos^{2}\frac{\text{x}}{2}}+\frac{2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}\bigg]\text{dx}$
$=\frac{1}{2}\int\text{x}\cdot\sec^2\frac{\text{x}}{2}\text{dx}+\int\tan\frac{\text{x}}{2}\text{dx}$
$=\frac{1}{2}\bigg[\text{x}\cdot\frac{\tan\big(\frac{\text{x}}{2}\big)}{\frac{1}{2}}-\int1\times2\tan\big(\frac{\text{x}}{2}\big)\text{dx}\bigg]+\frac{\log\big|\sec\frac{\text{x}}{2}\big|}{\frac{1}{2}}+\text{C}$
$=\text{x}\tan\big(\frac{\text{x}}{2}\big)-\frac{\log\big|\sec\frac{\text{x}}{2}\big|}{\frac{1}{2}}+\log\frac{\big|\sec\frac{\text{x}}{2}\big|}{\frac{1}{2}}+\text{C}$
$=\text{x}\tan\big(\frac{\text{x}}{2}\big)+\text{C}$
View full question & answer→Question 223 Marks
Evaluate the following integrals:$\int\log_{10}\text{x dx}$
AnswerLet $\text{I}=\int\log_{10}\text{x dx}$
$=\int\frac{\log\text{x}}{\log10}\text{dx}$
$=\frac{1}{\log10}\int1\times\log\text{x dx}$
Using integration by parts,
$=\frac{1}{\log10}\Big[\log\text{x}\int\text{dx}-\int\Big(\frac{1}{\text{x}}\int\text{dx}\Big)\text{dx}\Big]$
$=\frac{1}{\log10}\Big[\text{x}\log\text{x}-\int\big(\frac{\text{x}}{\text{x}}\big)\text{dx}\Big]$
$=\frac{1}{\log10}[\text{x}\log\text{x}-\text{x}]$
$\text{I}=\frac{\text{x}}{\log10}(\log\text{x}-1)$
View full question & answer→Question 233 Marks
Evaluate the following integrals:$\int\frac{1}{\sqrt{16-6\text{x}-\text{x}^2}}\text{ dx}$
Answer$\int\frac{\text{dx}}{\sqrt{16-6\text{x}-\text{x}^2}}$
$=\int\frac{\text{dx}}{\sqrt{16-(\text{x}^2+6\text{x})}}$
$=\int\frac{\text{dx}}{\sqrt{16-(\text{x}^2+6\text{x}+3^2-3^2)}}$
$=\int\frac{\text{dx}}{\sqrt{16+9-(\text{x}+3)^2}}$
$=\int\frac{\text{dx}}{\sqrt{5^2-(\text{x}+3)^2}}$
$=\sin^{-1}\Big(\frac{\text{x}+3}{5}\Big)+\text{C}$
View full question & answer→Question 243 Marks
Evaluate the following integrals:$\int\frac{\text{x}+7}{3\text{x}^2+25\text{x}+28}\text{ dx}$
Answer$\text{I}=\int\frac{\text{x}+7}{3\text{x}^2+25\text{x}+28}\text{ dx}$ $=\int\frac{\text{x}+7}{3\text{x}^2+21\text{x}+4\text{x}+28}\text{ dx}$ $=\int\frac{\text{x}+7}{3\text{x}(\text{x}+7)+4(\text{x}+7)}\text{ dx}$ $=\int\frac{\text{x}+7}{(3\text{x}+4)(\text{x}+7)}\text{ dx}$ $=\int\frac{1}{(3\text{x}+4)}\text{ dx}$$=\frac{1}{3}\ln|3\text{x}+4|+\text{C}$
View full question & answer→Question 253 Marks
Evaluate the following integrals:$\int\text{e}^{\text{x}}(\tan\text{x}-\log\cos\text{x})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}(\tan\text{x}-\log\cos\text{x})\text{dx}$$=\int\text{e}^{\text{x}}\tan\text{x dx}-\int\text{e}^{\text{x}}\log\cos\text{x dx}$
Integrating by parts
$=\int\text{e}^{\text{x}}\tan\text{x dx}-\Big\{\text{e}^{\text{x}}\log\cos\text{x}-\int\text{e}^{\text{x}}\Big(\frac{\text{d}}{\text{dx}}\log\cos\text{x}\Big)\text{dx}\Big\}$
$=\int\text{e}^{\text{x}}\tan\text{x dx}-\big\{\text{e}^{\text{x}}\log\cos\text{x}+\int\text{e}^{\text{x}}\tan\text{x dx}\big\}$
$=\int\text{e}^{\text{x}}\tan\text{x dx}-\text{e}^{\text{x}}\log\cos\text{x}-\int\text{e}^{\text{x}}\tan\text{x dx}+\text{C}$
$=-\text{e}^{\text{x}}\log\cos\text{x}+\text{C}$
$=\text{e}^{\text{x}}\log\sec\text{x}+\text{C}$ $\big[\because\log\sec\text{x}=-\log\cos\text{x}\big]$
View full question & answer→Question 263 Marks
Evaluate the following integrals:$\int\frac{1}{\text{x}^{\frac{2}{3}}\sqrt{\text{x}^{\frac{2}{3}}-4}}\text{ dx}$
Answer$\int\frac{\text{dx}}{\text{x}^{\frac{2}{3}}\sqrt{\text{x}^{\frac{2}{3}}-2^2}}$
$=\int\frac{\text{dx}}{\text{x}^{\frac{2}{3}}\sqrt{\Big(\text{x}^{\frac{1}{3}}\Big)^2-2^2}}$
Let $\text{x}^{\frac{1}{3}}=\text{t}$
$\Rightarrow\frac{1}{3}\text{x}^{\frac{-2}{3}}\text{ dx}=\text{dt}$
$\Rightarrow\frac{1}{3\text{x}^{\frac{2}{3}}}\text{ dx}=\text{dt}$
$\Rightarrow\frac{\text{dx}}{\text{x}^{\frac{2}{3}}}=3\text{dt}$
Now, $\int\frac{\text{dx}}{\text{x}^{\frac{2}{3}}\sqrt{\text{x}^{\frac{2}{3}}-2^2}}$
$=3\int\frac{\text{dt}}{\sqrt{\text{t}^2-2^2}}$
$=3\log\Big|\text{t}+\sqrt{\text{t}^2-2^2}\Big|+\text{C}$
$=3\log\Bigg|\text{x}^{\frac{1}{3}}+\sqrt{\text{x}^{\frac{2}{3}}-4}\Bigg|+\text{C}$
View full question & answer→Question 273 Marks
Evaluate the following integrals:$\int\text{e}^{\text{x}}\big[\sec\text{x}+\log(\sec\text{x}+\tan\text{x})\big]\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\big[\sec\text{x}+\log(\sec\text{x}+\tan\text{x})\big]\text{dx}$
$=\int\text{e}^{\text{x}}\sec\text{x dx}+\int\text{e}^{\text{x}}\log(\sec\text{x}+\tan\text{x})\text{dx}$
Integrating by parts
$=\int\text{e}^{\text{x}}\sec\text{x dx}+\text{e}^{\text{x}}\log(\sec\text{x}\tan\text{x})-\int\text{e}^{\text{dx}}\Big\{\frac{\text{d}}{\text{dx}}\log(\sec\text{x}+\tan\text{x})\Big\}\text{dx}$
$=\int\text{e}^{\text{x}}\sec\text{x dx}+\text{e}^{\text{x}}\log(\sec\text{x}+\tan\text{x})-\int\text{e}^{\text{x}}\sec\text{x dx}$
$=\text{e}^{\text{x}}\log(\sec\text{x}+\tan\text{x})+\text{C}$
View full question & answer→Question 283 Marks
Write a value of $\int\log_\text{e}\text{x}\text{ dx}$
Answer$\int\log_\text{e}\text{x}\text{ dx}$
$=\int1\cdot\log_\text{e}\text{x dx}$
$=\log_\text{e}\text{x}\int1\text{ dx}-\int\Big\{\frac{\text{d}}{\text{dx}}(\log_\text{e}\text{x})\int1\text{dx}\Big\}\text{dx}$
$=\log_\text{e}\text{x}\int1\cdot\text{dx}-\int\frac{1}{\text{x}}\cdot\text{x dx}$
$=\log_\text{e}\text{x}\cdot\text{x}-\int\text{dx}$
$=\text{x}\log_\text{e}\text{x}-\text{x}+\text{C}$
$=\text{x}(\log_\text{e}\text{x}-1)+\text{C}$
View full question & answer→Question 293 Marks
Evaluate the following integrals:$\int\text{e}^{\text{}x}\Big(\frac{1+\sin\text{x}}{1+\cos\text{x}}\Big)\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{}x}\Big(\frac{1+\sin\text{x}}{1+\cos\text{x}}\Big)\text{dx}$
$=\int\text{e}^{\text{x}}\Big(\frac{1}{1+\cos\text{x}}+\frac{\sin\text{x}}{1+\cos\text{x}}\Big)\text{dx}$
$=\int\text{e}^{\text{x}}\bigg(\frac{1}{2\cos\frac{\text{x}}{2}}+\frac{2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}\bigg)\text{dx}$
$=\int\text{e}^{\text{}x}\big(\frac{1}{2}\sec^2\frac{\text{x}}{2}+\tan\frac{\text{x}}{2}\big)\text{dx}$
Putting $\text{e}^{\text{x}}\tan\frac{\text{x}}{2}=\text{t}$
Diff. both sides w.r.t.x
$\text{e}^{\text{x}}.\tan\big(\frac{\text{x}}{2}\big)+\text{e}^{\text{x}}\times\frac{1}{2}\sec^{2}\frac{\text{x}}{2}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^{\text{x}}\big[\tan\frac{\text{x}}{2}+\frac{1}{2}\sec^2\big(\frac{\text{x}}{2}\big)\big]\text{dx}=\text{dt}$
$\therefore\int\text{e}^{\text{x}}\big(\frac{1}{2}\sec^2\frac{\text{x}}{2}+\tan\frac{\text{x}}{2}\big)\text{dx}=\int\text{dt}$
$=\text{t}+\text{C}$
$=\text{e}^{\text{x}}\tan\big(\frac{\text{x}}{2}\big)+\text{C}$
View full question & answer→Question 303 Marks
Write a value of $\int\frac{\text{a}^{\text{x}}}{3+\text{a}^{\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{a}^{\text{x}}}{3+\text{a}^{\text{x}}}\text{ dx}$
Let $3+\text{a}^{\text{x}}=\text{t}$
$\text{a}^{\text{x}}\log\text{a dx}=\text{dt}$
$\text{a}^{\text{x}}\text{dx}=\frac{\text{dt}}{\log\text{a}}$
$\text{I}=\int\frac{\text{dt}}{\log\text{a}\cdot\text{t}}=\frac{1}{\log\text{a}}\log\text{t}+\text{C}$
$\text{I}=\frac{1}{\log\text{a}}\log(3+\text{a}^{\text{x}})+\text{C}$
View full question & answer→Question 313 Marks
Evaluate the following integrals:$\int\frac{\cos2\text{x}}{\sqrt{\sin^22\text{x}+8}}\text{ dx}$
Answer$\int\frac{\cos(2\text{x}).\text{dx}}{\sqrt{\sin^22\text{x}+8}}$
Let $\sin(2\text{x})=\text{t}$
$\Rightarrow\cos(2\text{x})\times2.\text{dx}=\text{dt}$
$\Rightarrow\cos(2\text{x}).\text{dx}=\frac{\text{dt}}{2}$
Now, $\int\frac{\cos(2\text{x}).\text{dx}}{\sqrt{\sin^22\text{x}+8}}$
$=\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}^2+\big(2\sqrt2\big)^2}}$
$=\frac{1}{2}\log\Big|\text{t}+\sqrt{\text{t}^2+8}\Big|+\text{C}$
$=\frac{1}{2}\log\Big|\sin(2\text{x})+\sqrt{\sin^2(2\text{x})+8}\Big|+\text{C}$
View full question & answer→Question 323 Marks
Evaluate the following integrals:$\int\frac{1}{\sqrt{(1-\text{x}^2)\big\{9+\big(\sin^{-1}\text{x}\big)^2\big\}}}\text{ dx}.$
AnswerLet $\text{I}=\int\frac{1}{\sqrt{(1-\text{x}^2)\Big[9+\big(\sin^{-1}\text{x}\big)^2\Big]}}\text{ dx}$
Let $\sin^{-1}\text{x}=\text{t}$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^2}}\text{ dx}=\text{dt}$
$\Rightarrow\text{I}=\int\frac{\text{dt}}{\sqrt{(3)^2+\text{t}^2}}$
$\Rightarrow\text{I}=\log\Big|\text{t}+\sqrt{9+\text{t}^2}\Big|+\text{C}$ $\Big[\text{Since }\int\frac{1}{\sqrt{\text{a}^2+\text{x}^2}}\text{ dx}=\log\Big|\text{x}+\sqrt{\text{a}^2+\text{x}^2}\Big|+\text{C}\Big]$
$\text{I}=\log\Big|\sin^{-1}\text{x}+\sqrt{9+\big(\sin^{-1}\text{x}\big)^2}\Big|+\text{C}$
View full question & answer→Question 333 Marks
Write a value of $\int\frac{\sin2\text{x}}{\text{a}^2\sin^2\text{x}+\text{b}^2\cos^2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sin2\text{x}}{\text{a}^2\sin^2\text{x}+\text{b}^2\cos^2\text{x}}\text{ dx}$
Let $\text{a}^2\sin^2\text{x}+\text{b}^2\cos^2\text{x}=\text{t}$
$\big(2\text{a}^2\sin\text{x}\cos\text{x}-2\text{b}^2\cos\text{x}\sin\text{x}\big)\text{dx}=\text{dt}$
$2(\text{a}^2-\text{b}^2)\sin\text{x}\cos\text{x dx}=\text{dt}$
$(\text{a}^2-\text{b}^2)\sin2\text{x}\text{ dx}=\text{dt}$
$\text{I}=\frac{1}{\text{a}^2-\text{b}^2}\int\frac{\text{dt}}{\text{t}}$
$=\frac{1}{\text{a}^2-\text{b}^2}\log\text{t}+\text{C}$
$\text{I}=\frac{1}{\text{a}^2-\text{b}^2}\log\big(\text{a}^2\sin^2\text{x}+\text{b}^2\cos^2\text{x}\big)+\text{C}$
View full question & answer→Question 343 Marks
Evaluate the following integrals:$\int\frac{\sin2\text{x}}{\sqrt{\cos^4\text{x}-\sin^2\text{x}+2}}\text{ dx}$
Answer$\int\frac{\sin2\text{x}}{\sqrt{\cos^4\text{x}-\sin^2\text{x}+2}}\text{ dx}$
Let $\text{t}=\cos^2\text{x}\rightarrow-\text{dt}=2\cos\text{x}\sin\text{x}\text{ dx}$
$\int\frac{\sin2\text{x}}{\sqrt{\cos^4\text{x}-\sin^2\text{x}+2}}\text{ dx}$
$=\int\frac{-1}{\sqrt{\text{t}^2-(1-\text{t})+2}}\text{ dt}$
$=\int\frac{-1}{\sqrt{\text{t}^2+\text{t}+1}}\text{ dt}$
$=\int\frac{-1}{\sqrt{\text{t}^2+\text{t}+\frac{1}{4}+\frac{3}{4}}}\text{ dt}$
$=\int\frac{-1}{\sqrt{\big(\text{t}+\frac{1}{2}\big)^2+\frac{3}{4}}}\text{ dt}$
$=-\log\Big|\Big(\text{t}+\frac{1}{2}\Big)+\sqrt{\text{t}^2+\text{t}+1}\Big|$
$=-\log\Big|\Big(\cos^2\text{x}+\frac{1}{2}\Big)+\sqrt{\cos^4\text{x}+\cos^2\text{x}+1}\Big|+\text{C}$
View full question & answer→Question 353 Marks
Write a value of $\int\frac{\cos\text{x}}{\sin\text{x}\log\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\cos\text{x}}{\sin\text{x}\log\sin\text{x}}\text{ dx}$
$\int\frac{\cot\text{x}}{\log\sin\text{x}}\text{ dx}$
Let $\log\sin\text{x}=\text{t}$
$\cot\text{x dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log\text{t}+\text{C}$
$=\log(\log\sin\text{x})+\text{C}$
View full question & answer→Question 363 Marks
Write a value of $\int(\text{e}^{\text{x}\log_\text{e}\text{a}}+\text{e}^{\text{a}\log_\text{e}\text{x}})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}\log_\text{e}\text{a}}+\text{e}^{\text{a}\log_\text{e}\text{x}}\text{ dx}$
$=\int(\text{e}\log\text{a}^{\text{x}}+\text{e}\log\text{x}^{\text{a}})\text{dx}$
$=\int(\text{a}^{\text{x}}+\text{x}^{\text{a}})\text{dx}$
$=\int\text{a}^{\text{x}}\text{dx}+\int\text{x}^{\text{a}}\text{dx}$
$=\frac{\text{a}^{\text{x}}}{\log_\text{e}\text{a}}+\frac{\text{x}^{\text{a}+1}}{\text{a}+1}+\text{C}$
$\therefore\ \text{I}=\frac{\text{a}^{\text{x}}}{\log_\text{e}\text{a}}+\frac{\text{x}^{\text{a}+1}}{\text{a}+1}+\text{C}$
View full question & answer→Question 373 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 383 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 393 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 403 Marks
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 413 Marks
Evaluate the following integrals:$\int2\text{x}^3\text{e}^{\text{x}^{2}}\text{dx}$
Answer$\int2\text{x}^3\cdot\text{e}^{\text{x}^{2}}\text{dx}$
$=\int\text{x}^2\cdot\big(\text{e}^{\text{x}^2}\big)\cdot2\text{x dx}$
Let $\text{x}^2=\text{t}$
$\Rightarrow2\text{x dx = dt}$
$=\int\text{t}\cdot\text{e}^{\text{t}}\text{dt}$
$=\text{t}\cdot\text{e}^{\text{t}}-\int1\cdot\text{e}^{\text{t}}\text{dt}$
$=\text{t e}^{\text{t}}-\text{e}^{\text{t}}+\text{C}$
$=\text{x}^2\text{e}^{\text{x}^{2}}-\text{e}^{\text{x}^{2}}+\text{C}$
$=\text{e}^{\text{x}^2}(\text{x}^2-1)+\text{C}$
View full question & answer→Question 423 Marks
Evaluate the following integrals:$\int\frac{\log(\log\text{x})}{\text{x}}\text{dx}$
Answer$\int\frac{\log(\log\text{x})}{\text{x}}\text{dx}$
Taking log log x as the first function and $\frac{1}{\text{x}}$ as the second function.
$=\log \log\text{x}\int\frac{1}{\text{x}}\text{dx}-\int\Big\{\frac{\text{d}}{\text{dx}}\log(\log\text{x})\int\frac{1}{\text{x}}\text{dx}\Big\}\text{dx}$
$=\log\text{x}.\log(\log\text{x})-\int\frac{1}{\text{x}\log\text{x}}(\log\text{x})\text{dx}$
$=\log\text{x}.\log(\log\text{x})-\int\frac{1}{\text{x}}\text{dx}$
$=\log\text{x}.\log(\log\text{x})-\log\text{x}+\text{C}$
$=\log\text{x}[\log(\log\text{x})-1]+\text{C}$
View full question & answer→Question 433 Marks
Evaluate the following integrals:$\int\frac{\text{x}+5}{3\text{x}^2+13\text{x}-10}\text{ dx}$
Answer$\text{I}=\int\frac{\text{x}+5}{3\text{x}^2+13\text{x}-10}\text{ dx}$
$=\int\frac{\text{x}+5}{3\text{x}^2+15\text{x}-2\text{x}-10}\text{ dx}$
$=\int\frac{\text{x}+5}{3\text{x}(\text{x}+5)-2(\text{x}+5)}\text{ dx}$
$=\int\frac{\text{x}+5}{(3\text{x}-2)(\text{x}+5)}\text{ dx}$
$=\int\frac{1}{3\text{x}-2}\text{ dx}$
$\therefore\ \text{I}=\frac{1}{3}\int|3\text{x}-2|+\text{C}$
View full question & answer→Question 443 Marks
Evaluate the following integrals:$\int\sec^{-1}\sqrt{\text{x}}\text{dx}$
Answer$\int1.\sec^{-1}\sqrt{\text{x}}\text{dx}$
$=\sec^{-1}\sqrt{\text{x}}\int1\text{dx}-\int\Big\{\frac{\text{d}}{\text{dx}}\Big(\sec^{-1}\sqrt{\text{x}}\Big)\int1\text{dx}\Big\}\text{dx}$
$=\sec^{-1}\sqrt{\text{x}}.\text{x}-\int\frac{1}{\sqrt{\text{x}}\sqrt{1-\text{x}}}\times\frac{1}{2\sqrt{\text{x}}}\times\text{x dx}$
$=\text{x}\sec^{-1}\sqrt{\text{x}}-\frac{1}{2}\int(1-\text{x})^{-\frac{1}{2}\text{dx}}$
$=\text{x}\sec^{-1}\text{x}-\frac{1}{2}\Bigg[\frac{(1-\text{x})^{-\frac{1}{2}+1}}{\big(-\frac{1}{2}+1\big)(-1)}\Bigg]+\text{C}$
$=\text{x}\sec^{-1}\text{x}+(1-\text{x})^{\frac{1}{2}}+\text{C}$
View full question & answer→Question 453 Marks
Write a value of $\int\text{e}^{2\text{x}^2+\ln\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\text{e}^{2\text{x}^2+\ln\text{x}}\text{ dx}$
$=\int\text{e}^{2\text{x}^2}\cdot\text{e}^{\ln{\text{x}}}\text{dx}$
$=\int\text{x}\cdot\text{e}^{2\text{x}^2}\text{dx}$ $\big[\because\text{e}^{\ln\text{x}}=\text{x}\big]$
$=\int\text{x}\cdot\big(\text{e}^{\text{x}^2}\big)\text{dx}$
Let $\text{e}^{\text{x}^2}=\text{t}$
$\text{e}^{\text{x}^2}\cdot2\text{x dx}=\text{dt}$
$\therefore\ \frac{1}{2}\int\text{t dt}$
$=\frac{1}{2}\frac{\text{t}^2}{2}+\text{C}$
$=\frac{1}{4}\text{e}^{2\text{x}^2}+\text{C}$
$\therefore\ \text{I}=\frac{1}{4}\text{e}^{2\text{x}^2}+\text{C}$
View full question & answer→Question 463 Marks
Evaluate the following integrals:$\int\text{e}^{\text{x}}(\log\text{x}+\frac{1}{2})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}(\log\text{x}+\frac{1}{2})\text{dx}$
Here, $\text{f(x)}=\log\text{x}$
$\Rightarrow\text{f}'\text{(x)}=\frac{1}{\text{x}}$
Put $\text{e}^{\text{x}}\text{f(x)}=\text{t}$
$\Rightarrow\text{e}^{\text{x}}\log\text{x}=\text{t}$
Diff. both sides w.r.t x
$\text{e}^{\text{x}}\log\text{x}+\text{e}^{\text{x}}\frac{1}{\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^{\text{x}}(\log\text{x}+\frac{1}{\text{x}})\text{dx = dt}$
$\therefore\int\text{e}^{\text{x}}\big[\log\text{x}+\frac{1}{\text{x}}\big]\text{dx}=\int\text{dt}$
$\Rightarrow\text{I}=\text{t}+\text{C}$
$=\text{e}^{\text{x}}\log\text{x}+\text{C}$
View full question & answer→Question 473 Marks
$\int\frac{1}{\sqrt{\text{x+a}}+\sqrt{\text{x+b}}}\text{dx}$
AnswerLet $\text{l}=\int\frac{1}{\sqrt{\text{x+a}}+\sqrt{\text{x+b}}}\text{dx}. $ Then,
$\text{I}=\int\frac{1}{\sqrt{\text{x+a}}+\sqrt{\text{x+b}}}\times\frac{\sqrt{\text{x+a}}-\sqrt{\text{x+b}}}{\sqrt{\text{x+a}}-\sqrt{\text{x+b}}}\times\text{dx}$
$=\int\frac{\sqrt{\text{x+a}}-\sqrt{\text{x+b}}}{\text{x+a}-\text{x-b}}\times\text{dx}$
$=\int\frac{\sqrt{\text{x+a}}-\sqrt{\text{x+b}}}{\text{a}-\text{b}}\times\text{dx}$
$=\frac{1}{\text{a}-\text{b}}\bigg[\frac{2}{3}(\text{x+a})^{\frac{3}{2}}-\frac{2}{3}(\text{x+b})^{\frac{3}{2}}\bigg]+\text{c}$
$=\frac{2}{3(\text{a}-\text{b})}\Big[(\text{x+a})^{\frac{3}{2}}-(\text{x+b})^{\frac{3}{2}}\Big]+\text{c}$
$ \text{I}=\frac{2}{3(\text{a}-\text{b})}\Big[(\text{x+a})^{\frac{3}{2}}-(\text{x+b})^{\frac{3}{2}}\Big]+\text{c}$
View full question & answer→Question 483 Marks
Evaluate the following integrals:$\int\sqrt{\text{cosec}\text{x}-1}\text{ dx}$
Answer$\int\sqrt{\text{cosec}\text{x}-1}\text{ dx}$
$=\int\sqrt{\frac{1}{\sin\text{x}}-1}\text{ dx}$
$=\int\frac{\sqrt{1-\sin\text{x}}}{\sqrt{\sin\text{x}}}\text{ dx}$
$=\int\frac{\sqrt{(1-\sin\text{x})(1+\sin\text{x})}}{\sqrt{\sin\text{x}(1+\sin\text{x})}}\text{ dx}$
$=\int\frac{\sqrt{1-\sin^2\text{x}}}{\sqrt{\sin^2\text{x}+\sin\text{x}}}\text{ dx}$
$=\int\frac{\cos\text{x}\text{ dx}}{\sqrt{\sin^2\text{x}+\sin\text{x}}}$
Let $\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x}\text{ dx}=\text{dt}$
Now, $=\int\frac{\cos\text{x}\text{ dx}}{\sqrt{\sin^2\text{x}+\sin\text{x}}}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}^2+\text{t}}}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}^2+\text{t}+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}}$
$=\int\frac{\text{dt}}{\sqrt{\big(\text{t}+\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}}$
$=\log\Big|\Big(\text{t}+\frac{1}{2}\Big)+\sqrt{\big(\text{t}+\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}\Big|+\text{C}$
$=\log\Big|\text{t}+\frac{1}{2}+\sqrt{\text{t}^2+\text{t}}\Big|+\text{C}$
$=\log\Big|\sin\text{x}+\frac{1}{2}\sqrt{\sin^2\text{x}+\sin\text{x}}\Big|+\text{C}$
View full question & answer→Question 493 Marks
Evaluate the following integrals:$\int\frac{1}{\sqrt{5-4\text{x}-2\text{x}^2}}\text{ dx}$
Answer$\int\frac{1}{\sqrt{5-4\text{x}-2\text{x}^2}}$
$=\int\frac{\text{dx}}{\sqrt{2\big[\frac{5}{2}-2\text{x}-\text{x}^2}\big]}$
$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{5}{2}-2\text{x}-\text{x}^2}}$
$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{5}{2}(\text{x}^2+2\text{x})}}$
$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{5}{2}-(\text{x}^2+2\text{x}+1-1)}}$
$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{5}{2}-(\text{x}+1)^2+1}}$
$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{7}{2}-(\text{x}+1)^2}}$
$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\Big(\frac{\sqrt7}{\sqrt3}\Big)^2-(\text{x}+1)^2}}$
$=\frac{1}{\sqrt2}\sin^{-1}\Big(\frac{(\text{x}+1)\sqrt2}{\sqrt7}\Big)+\text{C}$
$=\frac{1}{\sqrt2}\sin^{-1}\Big(\sqrt{\frac{2}{7}}(\text{x}+1)\Big)+\text{C}$
View full question & answer→Question 503 Marks
Evaluate the following integrals:$\int\frac{\log(\text{x}+2)}{(\text{x}+2)^2}\text{dx}$
AnswerLet $\text{I}=\int\frac{\log(\text{x}+2)}{(\text{x}+2)^2}\text{dx}$
Let $\frac{1}{\text{x}+2}=\text{t}$
$-\frac{1}{(\text{x}+2)^2}\text{dx = dt}$
$\text{I}=-\int\log\big(\frac{1}{\text{t}}\big)\text{dt}$
$=-\int\log\text{t}^{-1}\text{dt}$
$=-\int1\times\log\text{t dt}$
Using integration by parts,
$\text{I}=\log\text{t}\int\text{dt}-\int\big(\frac{1}{\text{t}}\int\text{dt}\big)\text{dt}$
$=\text{t}\log\text{t}-\int\Big(\frac{1}{\text{t}}\times\text{t}\Big)\text{dt}$
$=\text{t}\log\text{t}-\int\text{dt}$
$=\text{t}\log\text{t}-\text{t+C}$
$=\frac{1}{\text{x}+2}\big(\log(\text{x}+2)^{-1}-1\big)+\text{C}$
$\text{I}=\frac{-1}{\text{x}+2}-\frac{\log(\text{x}+2)}{\text{x}+2}+\text{C}$
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