Question 11 Mark
If $\int\limits^{\text{a}}_0\frac{1}{4+\text{x}^2}\text{dx}=\frac{\pi}{8}$, find the value of a.
View full question & answer→Question 21 Mark
Find:
$\int \frac{\sin^{2} \text{x} - \cos^{2} \text{x}}{\sin \text{x} \cos \text{x}} \text{dx}$
Answer$\int \frac{\sin^{2} \text{x} - \cos^{2} \text{x}}{\sin \text{x} \cos \text{x}} \text{dx}$
$= \int \frac{\sin^{2}\text{x}}{\sin \text{x} \cos \text{x}} - \int \frac{\cos^{2}\text{x}}{\sin \text{x} \cos \text{x}} \text{dx}$
$ = \int \tan \text{x dx} - \int\cot \text{x dx}$
$= \log | \text{Sec} \text{ x}| - \log |\sin \text{x}| + C$
View full question & answer→Question 31 Mark
Evaluate: $\int\limits_{\text{e}}^{\text{e}^2}\frac{\text{dx}}{\text{x}\log\text{x}}$
View full question & answer→Question 41 Mark
Evaluate: $\int\limits_{2}^{4}\frac{\text{x}}{\text{x}^{2} +1}\text{dx}.$
Answer$\frac{1}{2}(\log17 - \log5)\text{ or } \frac{1}{2}\log\bigg(\frac{17}{5}\bigg).$
View full question & answer→Question 51 Mark
If $\text{f}(\text{x}) = \int\limits_{0}^{\text{x}}\text{t}\sin\text{t }\text{dt},$then write the value of f'(x).
View full question & answer→Question 61 Mark
Evaluate: $\int\limits_{0}^{2}\sqrt{\text{4-x}^{2}}\text{ dx}$ .
View full question & answer→Question 71 Mark
Given $\int e^x(\tan x + 1) \sec x\ dx = e^x f(x) + c$.
Write $f(x)$ satisfying the above.
View full question & answer→Question 81 Mark
Evaluate:
$\int\frac{\text{dx}}{\sqrt{\text{1 - x}^{2}}}$.
View full question & answer→Question 91 Mark
Evaluate:
$\int (ax + b)^3 dx$.
Answer$\frac{1}{\text{4a}}(\text{ax + b})^{4}+\text{c}.$
View full question & answer→Question 101 Mark
Write the value of the following integral:
$\int\limits_\text{-x\2}^\text{x\2}\ \ \ \ \sin^5\ \text{x}\ \ \text{dx}$
View full question & answer→Question 111 Mark
Evaluate:
$\int\text{ }\sec^2\ (7-4\text{x})\ \text{dx}$
Answer$-\frac{1}{4}\tan(7-4\text{x})+\text{c}. $
View full question & answer→Question 121 Mark
Evaluate: $\int\limits_0^{1/\sqrt{2}}\frac{1}{\sqrt{\text{1 - x}^{2}}}\text{dx}.$
View full question & answer→Question 131 Mark
Evaluate:$\int\frac{\cos\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{dx}$
Answer$2\sin\sqrt{\text{x}}+\text{c}.$
View full question & answer→Question 141 Mark
Evaluate:
$\int\limits_0^1\frac{\text{2x}}{\text{1 + x}^{2}}\text{dx}$.
Answer$\Big[\log|1+\text{x}^{2}|\Big]_0^{1}=\log2$.
View full question & answer→Question 151 Mark
Evaluate:
$\int\frac{\text{2 cos x}}{\text{3 sin}^{2}\text{x}}\text{dx}$.
Answer$=-\frac{2}{3\sin\text{x}}+\text{c}\text{ Or }-\frac{2}{3}\text{cosec x + c}.$
View full question & answer→Question 161 Mark
Evaluate:
$\int\limits^{2\pi}_{0} \cos^{5} \text{x dx}. $
Answer$\int\limits^{2\pi}_{0} \cos^{5} \text{x dx} = 2 \int\limits^{\pi}_{0} \cos^{5}\text{x dx}$
$\text{and}\ 2\int^\pi_0\cos^5\text{x}\ \text{dx}=0\Rightarrow\int^{2\pi}_0\cos^5\text{x}\ \text{dx}=0$
View full question & answer→Question 171 Mark
Evaluate: $\int\limits^{3}_{2} 3^{x} \text{d}x.$
Answer$\int\limits^{3}_{2} 3^{\text{x}} \text{dx} = \bigg[\frac{3^\text{x}}{\log 3}\bigg]^{3}_{2} = \frac{18}{\log 3} $
View full question & answer→Question 181 Mark
Write the antiderivative of $\bigg(3\sqrt{\text{x}} + \frac{1}{\sqrt{\text{x}}}\bigg).$
Answer$2\text{x}^{3/2} + 2\sqrt{\text{x}} + \text{c}.$
View full question & answer→Question 191 Mark
Evaluate: $\int\limits^{\pi/2}_0\text{e}^x(\sin x-\cos x)\text{d}x.$
AnswerLet I =$\int\limits^{\pi/2}_0\text{e}^x(\sin x-\cos x)\text{ d}x$
Using f(a) = f(a-x), we have:
$\therefore\ \ \ \int\limits^{\pi/2}_0\text{e}^x(\cos x-\sin x)\text{ d}x\int\limits^{\pi/2}_0\text{e}^x(\cos x+(-\sin x))\text{ d}x$
we know that, $\int\text{e}^x[f(x)+f'(x)]dx=\text{e}^xf(x)+\text{C}$
$\text{Also},\frac{d\cos x}{dx}=-\sin x$
$\text{So, I}=\text{e}^x\cos x^{\pi/2}_0$
$\Rightarrow\text{I}=(\text{e}^{\pi/2}\times0)-(\text{e}^0\times1)=0-1=-1$
$\therefore\ \ \int\limits^{\pi/2}_0\text{e}^x(\sin x-\cos x)\text{ d}x=-1$
View full question & answer→Question 201 Mark
Evaluate: $\int\cos^{-1}(\sin\ x)\ \text{d}x.$
Answer$\frac{\pi\text{x}}{2}-\frac{\text{x}^2}{2}+\text{c}$
View full question & answer→Question 211 Mark
Evaluate, $\int\limits_0^3\frac{\text{dx}}{9 + \text{x}^{2}}.$
View full question & answer→Question 221 Mark
Evaluate: $\int\limits_2^3\frac{1}{\text{x}}\text{dx}.$
Answer$\log\Bigg(\frac{3}{2}\Bigg)$.
View full question & answer→Question 231 Mark
Evaluate: $\int\text{(1-x)}\sqrt{x}\text{dx}.$
Answer$\frac{2}{3}\text{x}^{3/2}-\frac{2}{5}\text{x}^{5/2}+\text{c}$.
View full question & answer→Question 241 Mark
Write the value of $\int\frac{\text{dx}}{\text{x}^{2}+16}$ .
Answer$\frac{1}{4}\tan^{-1}\frac{\text{x}}{4}+\text{c}$.
View full question & answer→Question 251 Mark
Write the value of $\int$ sec x (sec x + tan x) dx.
View full question & answer→Question 261 Mark
Evaluate: $\int\frac{\log{\text{x}}}{\text{x}}\text{dx}$ .
Answer$\frac{1}{2}(\log\text{x})^{2}+\text{c}$.
View full question & answer→Question 271 Mark
$\text{If} \int\limits_0^{1} (3x^{2} + 2x + k) dx = 0$ find the value of k.
Answer$\text{If} \int\limits_0^{1} (3x^{2} + 2x + k) dx = 0$$\Rightarrow \bigg[\frac{3x^{3}}{3} + \frac{2x^{2}}{2} + kx\bigg]^{1}_{0} = 0$
$\Rightarrow 1 + 1 + k = 0 \Rightarrow k = -2$
View full question & answer→Question 281 Mark
$\text{Evaluate} \int \frac{\sec^{2} x}{3 + \tan x} \text{dx} $
Answer$ \int \frac{\sec^{2} x}{3 + \tan x} \text{dx} $$\text{Let 3} + \tan x = t$
$\sec^{2} x dx = dt$
$\therefore \int \frac{ \sec^{2} x}{3+ \tan x} dx = \int\frac{dt}{t}$
$= \log |t|+ c$
$= \log|3 + \tan x| + c$
View full question & answer→Question 291 Mark
$\text{Evaluate: } \int\frac{x^{2}}{1 + x^{3}} \text{dx}$
Answer$\text{Let} I \int\frac{x^{2}}{1 + x^{3}} \text{dx}$$\text{Putting 1} + x^{3} = t$
$\Rightarrow 3x^{2}\text{dx = dt}$
$\text{or x}^{2} dx = \frac{dt}{3} $
$\therefore I = \frac{1}{3} \int \frac{dt}{3} = \frac{1}{3}\log |t| + C $
$= \frac{1}{3} \log | 1 + x^{3}| + C$
View full question & answer→Question 301 Mark
$ \text{Evaluate} \int\limits_0^{1} \frac{dx}{1 + x^{2}}$
Answer$\int\limits_0^{1} \frac{dx}{1 + x^{2}}=\big[\tan^{-1}\text{x}\big]^1_0$ $\tan^{-1}\perp-\tan^{-1}0$ $=\frac{\pi}{4}-0$ $=\frac{\pi}{4}$
View full question & answer→Question 311 Mark
Find:$\int\frac{\text{dx}}{\sqrt{9-4\text{x}^2}}$
Answer$\int\frac{\text{dx}}{\sqrt{9-4\text{x}^2}}=\int\frac{\text{dx}}{\sqrt{3^3-(2\text{x})^2}}$
$=\frac{1}{2}\sin^{-1}\Big(\frac{2\text{x}}{3}\Big)+\text{c}$
View full question & answer→Question 321 Mark
Evaluate:
$\int\limits^3_1|2\text{x}-1|\text{dx}$
Answer$\int\limits^3_1|2\text{x}-1|\text{dx}$
$\Rightarrow\int\limits_1^3|2\text{x}-1|\text{dx}=\int\limits^3_1(2\text{x}-1)\text{dx}$
$\Rightarrow\int\limits^3_12\text{x}\text{ dx}-\int\limits^3_1\text{dx}$
$\Rightarrow\Big[\frac{2\cdot\text{x}^2}{2}\Big]^3_1-\big[\text{x}\big]^3_1$
$\Rightarrow[(3)^2-(1)^2]-[(3-1)]$
$\Rightarrow9-1-2=6$
View full question & answer→Question 331 Mark
Integrate the functions in Exercises:
$(4\text{x}+2)\sqrt{\text{x}^2+\text{x}+1}$
Answer$\text{Let I}=\int(4\text{x}+ 2)\sqrt{\text{x}^2+\text{x} +1}\text{ dx}$
$=\int2(2\text{x}+1)\sqrt{\text{x}^2+\text{x}+1}\text{ dx}$
$=\int2\sqrt{\text{x}^2+\text{x}+1}( 2\text{x}+1)\text{ dx} \ \ \ \ \ \ \ ...\text{(i)} $
Putting $\text{ x}^2+\text{x}+1=\text{t} \ \ \ \Rightarrow \ \ \ (2\text{x + 1})=\frac{\text{dt}}{\text{dx}} \ \ \Rightarrow \ \ \ (2\text{x + 1})\text{ dx = dt}$
$\therefore\ \ \ \ \ $From eq. (i), $\text{I}=\int2\sqrt{\text{t}}\text{ dt}=2\int\text{t}^{\frac{1}{2}}\text{ dt}=2\frac{\text{t}^{^3/_2}}{^3/_2}+\text{c}=\frac{4}{3}\text{t}^{^3/_2}+\text{c} $
$=\frac{4}{3}\big(\text{x}^2+\text{x}+1\big)^{^3/_2}+\text{c} $
View full question & answer→Question 341 Mark
Integrate the functions in Exercises:
$\tan^2(2\text{x}-3)$
Answer$\int\tan^2(2\text{x}-3)\text{ dx}=\int\big\{\sec^2(\text{2x}-3)-1\big\}\text{ dx}$
$=\int\sec^2(2\text{x}-3)\text{ dx }-\int1\text{ dx}$
$=\frac{\tan(2\text{x}-3)}{2}-\text{x}+\text{c}$
$=\frac{1}{2}\tan(2\text{x}-3 )+\text{c} $
View full question & answer→Question 351 Mark
Integrate the functions in Exercises:
$\frac{1}{\text{x}-\sqrt{\text{x}}}$
Answer$\text{Let I}=\int\frac{1}{\text{x }-\sqrt{\text{x}}}\text{ dx} \ \ \ \ \ \ \ \dots\text{(i)}$
Putting $\sqrt{ \text{x}}=\text{t}\ \ \ \ \Rightarrow \ \ \ \ \text{x}=\text{t}^2\ \ \ \Rightarrow \ \ \ \frac{\text{dx}}{\text{dt}}=2\text{t}\ \ \ \ \Rightarrow \ \ \ \text{dx}=2\text{t}\text{ dt} $
$\therefore \ \ \ \ $From eq. (i),$\text{ I}=\int\frac{1}{\text{t}^2-\text{t}}\text{2t}\text{ dt} =2\int{\frac{\text{t}}{\text{t(t - 1)}}}\text{ dt}$
$=2\int\frac{1}{(\text{t }-1)}\text{ dt}=2\log\mid\text{t}-1\mid+\text{ c}$
$=2\log\begin{vmatrix}\sqrt{\text{x}} -1\end{vmatrix}+\text{c} $
View full question & answer→Question 361 Mark
Integrate the functions in Exercises:
$\frac{1}{1-\tan\text{x}}$
Answer$\text{Let I}=\int\frac{1}{1-\tan\text{x}}\text{ dx} =\int\frac{1}{1-\frac{\sin\text{x}}{\cos\text{x}}}\text{ dx} $
$=\int\frac{1}{\bigg(\frac{\cos\text{x}-\sin\text{x}}{\cos\text{x}}\bigg)}\text{ dx} =\int\frac{\cos\text{x}}{\cos\text{x}-\sin\text{x}}\text{ dx} $
$=\frac{1}{2}\int\frac{2\cos\text{x}}{\cos\text{x}-\sin\text{x}}\text{ dx} =\frac{1}{2}\int\frac{\cos\text{x}+\cos\text{x}}{\cos\text{x}-\sin\text{x}}\text{ dx} $
Adding and subtracting $\sin\text{x}$ in the numerator,
$=\frac{1}{2}\int\frac{\cos\text{x}-\sin\text{x}+\sin\text{x}+\cos\text{x}} {\cos\text{x}-\sin\text{x}}\text{ dx} $
$=\frac{1}{2}\int\frac{(\cos\text{x}-\sin\text{x})+(\sin\text{x}+\cos\text{x})}{\cos\text{x}-\sin\text{x}}\text{ dx} $
$=\frac{1}{2}\int\bigg(\frac{\cos\text{x}-\sin\text{x}}{\cos\text{x}-\sin\text{x}}-\frac{\sin\text{x}+\cos\text{x}}{\cos\text{x}-\sin\text{x}}\bigg)\text{ dx} $
$=\frac{1}{2}\int\bigg(1-\frac{\sin\text{x}+\cos\text{x}}{\cos\text{x}-\sin\text{x}}\bigg)\text{ dx} $
$\frac{1}{2}\bigg[\int1\text{ dx}-\int\frac{-\sin\text{x}-\cos\text{x}}{\cos\text{x}-\sin\text{x}}\text{ dx}\bigg]$
$=\frac{1}{2}\big[\text{x}-\log\begin{vmatrix}\cos\text{x} - \sin \text{x} \end{vmatrix}\big]+\text{c} $
View full question & answer→Question 371 Mark
Integrate the functions in Exercises:
$\frac{\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}$
Answer$\text{Let I}=\int\frac{\sin^{-1}\text{x}}{\sqrt{1 - \text{x}^2}}\text{ dx}\ \ \ \ \ \ ...\text{(i)} $Putting $\sin^{-1}\text{x}=\text{t}\ \ \ \Rightarrow\ \ \ \ \frac{1}{\sqrt{1 - \text{x}^2}}=\frac{\text{dt}}{\text{dx}}\ \ \ \ \Rightarrow \ \ \ \ \frac{\text{dx}}{\sqrt{1 - \text{x}^2}}=\text{ dt} $
$\therefore\ \ \ \ \ $From eq. (i), $\text{I}=\int\text{t }\text{ dt}=\frac{\text{t}^2}{2}+\text{c}= \frac{1}{2}\big(\sin^{-1}\text{x}\big)^2+\text{c}$
View full question & answer→Question 381 Mark
Integrate the functions in Exercises:
$\frac{2\cos\text{x}-3\sin\text{x}}{6\cos\text{x}+4\sin\text{x}}$
Answer$\text{Let I }=\int\frac{2\cos\text{x}-3\sin\text{x}}{6\cos\text{x}+4\sin\text{x}}\text{ dx}=\int\frac{2\cos\text{x}-3\sin\text{x}}{2(2\sin\text{x}+3\cos\text{x})}\text{ dx} $
$=\frac{1}{2}\int\frac{2\cos\text{x}-3\sin\text{x}}{2\sin\text{x}+3\cos\text{x}}\text{ dx} \ \ \ \ \ \ ...\text{(i)} $
Putting $2\sin\text{x}+3\cos\text{x}=\text{t}\ \ \ \ \Rightarrow \ \ \ \ 2\cos\text{x}-3\sin\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow \ \ \ \ (2\cos\text{x}-3\sin\text{x})\text{ dx}=\text{dt} $
$ \therefore \ \ \ \ $From eq. (i), $\text{I}=\frac{1}{2}\int\frac{\text{dt}}{\text{t}} =\frac{1}{2}\log\begin{vmatrix}t\end{vmatrix}+\text{c}$
$=\frac{1}{2}\log\begin{vmatrix}2\sin\text{x}+3\cos \text{x}\end{vmatrix}+\text{c}$
View full question & answer→Question 391 Mark
Integrate the functions in Exercises:
$\sec^2(7-4\text{x})$
Answer$\int\sec^2(7 -4\text{x})\text{ dx}$
$=\frac{\tan(7 - 4\text{x})}{-4\rightarrow\text{Coeff. of x}}+\text{c} $
$=\frac{-1}{4}\tan(7 - 4\text{x})+\text{c} $
View full question & answer→Question 401 Mark
Integrate the functions in Exercises:
$\frac{\text{x}}{\sqrt{\text{x}+4}},\text{x}>0$
Answer$ \int\frac{\text{x}}{\sqrt{\text{x + 4}}}\text{ dx}= \int\frac{\text{x}+4-4}{\sqrt{\text{x}+4}}\text{ dx}$
$=\int\frac{\text{x}+4}{\sqrt{\text{x}+ 4}}-\frac{4}{\sqrt{\text{x}+4}}\text{ dx}$
$=\int\sqrt{\text{x}+4}\text{ dx}-4\int\frac{1}{\sqrt{\text{x + 4}}}\text{ dx} $
$=\int(\text{x}+4)^{\frac{1}{2}}\text{ dx}-4\int(\text{x}+ 4)^{\frac{-1}{2}}\text{ dx} $
$=\frac{(\text{x}+4)^{\frac{3}{2}}}{\frac{3}{2}(1)}-\frac{4(\text{x}+4)^{\frac{1}{2}}}{\frac{1}{2}(1)}+\text{c}$
$=\frac{2}{3}(\text{x}+4)^{\frac{3}{2}}-8(\text{x}+ 4)^{\frac{1}{2}}+\text{c}$
$=\frac{2}{3}\sqrt{\text{x}+4}\bigg(\frac{\text{x}+4}{3}-4\bigg)+\text{c} $
$=\frac{2}{3}\sqrt{\text{x}+4}\bigg(\frac{\text{x}+4-12}{3}\bigg)+\text{c} $
$=\frac{2}{3}\sqrt{\text{x}+4}(\text{x}-8)+\text{c} $
View full question & answer→Question 411 Mark
Integrate the functions in Exercises:
$\text{e}^{2\text{x}+3}$
Answer$\int\text{e}^{2\text{x}+3}\text{ dx}=\frac{\text{e}^{2\text{x}+3}}{2\rightarrow \text{Coeff. of x }}+\text{c} \ \ \ \ \ \ \ \ \because \int\text{e}^{\text{ax}+\text{b}}\text{ dx}=\frac{\text{e}^{\text{ax}+\text{b}}}{\text{a}} $
$=\frac{1}{2}\text{e}^{2\text{x}+3}+\text{c} $
View full question & answer→Question 421 Mark
Find the following integrals in Exercises. $\int\frac{\text{x}^3 + 5\text{x}^2-4}{\text{x}~^2}\text{ dx}$
Answer$\int\frac{\text{x}^3+5\text{x}^2-4}{\text{x}^2}\text{ dx}$
$=\int\bigg(\frac{\text{x}^3}{\text{x}^2}+\frac{\text{5x}^2}{\text{x}^2}-\frac{\text{4}}{\text{x}^2}\bigg)\text{ dx}$
$=\int(\text{x}+\text{5}-4\text{x}^{-2})\text{ dx}$
$=\int\text{x}\text{ dx}+\int5\text{ dx}-\int4\text{x}^{-2}\text { dx} $
$=\int\text{x}\text{ dx}+5\int1\text{ dx}-4\int\text{x}^{-2}\text{ dx} $
$=\frac{\text{x}^2}{2}+5\text{x}+\frac{4}{\text{x}}+\text{c}$
View full question & answer→Question 431 Mark
Find an anti derivative (or integral) of the following function by the method of inspection.$(\text{ax + b})^2$
Answer$\because \ \ \ \ \ \ \ \ $$\text{ }\frac{\text{d}}{\text{dx}}(\text{ ax }+\text{b})^{3}=3(\text{ ax }+\text{b})^{2}\text{ }\frac{\text{d}}{\text{dx}}(\text{ax }+\text{b})=3(\text{ ax }+\text{b})^{2}\text{ a}$
$\ \ \ \ \ \Rightarrow \ \ \ \ \ \text{ }\frac{\text{1}}{\text{3a}}\text{ }\frac{\text{d}}{\text{dx}}(\text{ax}+\text{b})^{3}=(\text{ax}+\text{b})^{2}$
$ \ \ \ \ \ \Rightarrow \ \ \ \ \ \text{ }\frac{\text{d}}{\text{dx}}\bigg[\text{ }\frac{\text{1}}{\text{3a}}(\text{ax}+\text{b})^{3}\bigg]=(\text{ax}+\text{b})^{2}$
$\therefore \ \ \ \ \ $An anti-derivative of $(\text{ax}+\text{b})^{2} \text{ is }\frac{\text{1}}{\text{3a}}(\text{ax}+\text{b})^{3}.$
View full question & answer→Question 441 Mark
Integrate the functions in Exercises:
$\sqrt{\text{ax+b}}$
Answer$\int\sqrt{\text{ax+b}}\text{ dx}=\int(\text{ax+b})^{\frac{1}{2}}\text{ dx} $ $=\frac{(\text{ax}+\text{b})^{\frac{1}{2}+1}}{\bigg(\frac{1}{2}+1\bigg)\text{a}\rightarrow\text{Coeff. of x}}+\text{c} =\frac{(\text{ax}+\text{b})^{\frac{3}{2}}}{\frac{3}{2}\text{a}}+\text{c} $ $=\frac{2}{3\text{a}}(\text{ax}+\text{b})^{\frac{3}{2}}+\text{c} $
View full question & answer→Question 451 Mark
Integrate the functions in Exercises:
$\sin\text{x}\sin(\cos\text{x)}$
AnswerPutting $\cos \text{x}=\text{t}\ \ \ \ \ \Rightarrow \ \ \ \ \ -\sin\text{x}=\frac{\text{dt}}{\text{dx}}\ \ \ \ \ \Rightarrow\ \ \ \ -\sin\text{x}\text{ dx = dt}$ $\therefore\ \ \ \ \ \int\sin\text{x}\sin(\cos\text{x})\text{ dx}$ $=-\int\sin(\cos\text{x})(-\sin\text{x dx})$ $=-\int\sin\text{t dt}=-(-\cos\text{t})+\text{c}$ $=\cos\text{t}+\text{c}=\cos(\cos\text{x})+\text{c} $
View full question & answer→Question 461 Mark
Find the following integrals in Exercises: $\int(2\text{x}^2-3\sin\text{x}+5\sqrt{\text{x}})\text{ dx}$
Answer$\int(2\text{x}^2-3\sin\text{x}+5\sqrt{\text{x}})\text{ dx}$ $=2\int\text{x}^2\text{ dx}-3\int\sin\text{x}\text{ dx}+5\int\text{x}^\frac{1}{2}\text{ dx} $ $=2\frac{\text{x}^3}{3}-3(-\cos\text{x})+5\frac{\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\text{c} $ $2\frac{\text{x}^3}{3}+3\cos\text{x}+5\frac{\text{x}^\frac{3}{2}}{\frac{3}{2}}+\text{c} $ $=2\frac{\text{x}^3}{3}+3\cos\text{x}+\frac{10}{3}\text{x}^\frac{3}{2}+\text{c} $
View full question & answer→Question 471 Mark
Integrate the functions in Exercises:
$\frac{\cos\text{x}}{\sqrt{1+\sin\text{x}}}$
Answer$\text{Let I}=\int\frac{\cos\text{x}}{\sqrt{1+\sin\text{x}}}\text{dx} $
Putting $1 +\sin\text{x}=\text{t}\ \ \ \ \Rightarrow\ \ \ \ \cos\text{x}=\frac{\text{dt}}{\text{dx}}\ \ \ \ \Rightarrow\ \ \ \cos\text{ x dx = dt} $
$\therefore \ \ \ \ \ $From eq. (i), $\text{I}=\int\frac{\text{dt}}{\sqrt{\text{t}}}=\int\text{t}^{\frac{-1}{2}}\text{ dt}=\frac{\text{t}^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+\text{c}$
$=\frac{\text{t}^{^1/_2}}{^1/_2}+\text{c} =2\sqrt{t} +\text{c}=2\sqrt{1+\sin\text{x}}+\text{c}$
View full question & answer→Question 481 Mark
Integrate the functions in Exercises:
$\frac{\cos\sqrt{\text{x}}}{\sqrt{\text{x}}}$
Answer$\text{ Let I}=\int\frac{\cos\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx} \ \ \ \ \ \ \ \ \ ...\text{(i)} $
Putting $\sqrt{\text{x}}=\text{t} \ \ \ \Rightarrow \ \ \ \ \text{x}=\text{t}^2 \ \ \ \Rightarrow \ \ \ \ \frac{\text{dx}}{\text{dt}}=2\text{t}\ \ \ \Rightarrow \ \ \ \ \ \text{dx}=\text{2t}\text{ dt} $
$\therefore \ \ \ \ $From eq. (i), $\text{I}=\int\frac{\cos\text{t}}{\text{t}}2\text{t dt}=2\int\cos\text{t}\text{ dt}$
$=2\sin\text{t}+\text{c}=2\sin\sqrt{\text{x}}+\text{c} $
View full question & answer→Question 491 Mark
Find the following integrals in Exercises: $\int\frac{2-3\sin\text{x}}{\cos^2\text{x}}\text{ dx}$
Answer$\int\frac{2-3\sin\text{x}}{\cos^2\text{x}}\text{ dx}=\int\bigg(\frac{2}{\cos^2\text{x}}-\frac{3\sin\text{x}}{\cos^2\text{x}}\bigg)\text{ dx} $
$= \int(2\sec^2\text{x}-3\tan\text{x }\sec\text{x})\text{ dx} $
$=2\int\sec^2\text{x}\text{ dx}-3\int\sec\text{x}\tan\text{x}\text{ dx} $
$=2\tan\text{x}-3\sec\text{x}+\text{c} $
View full question & answer→Question 501 Mark
Integrate the functions in Exercises:
$\cot\text{x}\log\sin\text{x}$
Answer$\text{Let I}=\int\cot\text{x }\log\sin\text{x}\text{ dx} \ \ \ \ \ \ \ ...\text{(i)} $
Putting $\log\sin\text{x}=\text{t}\ \ \ \Rightarrow\ \ \ \frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})=\frac{\text{dt}}{\text{dx}}\ \ \ \Rightarrow \ \ \ \frac{1}{\sin\text{x}}\cos\text{x}=\frac{\text{dt}}{\text{dx}} $
$\Rightarrow\ \ \ \cot\text{ x dx = dt} $
$\therefore\ \ \ \ $From eq. (i), $\text{I}=\int\text{t dt}=\frac{\text{t}^2}{2}+\text{c}=\frac{1}{2}(\log \sin\text{x})^2+\text{c}$
View full question & answer→Question 511 Mark
Find the following integrals in Exercises: $\int(2\text{x}-3\cos\text{x}+\text{e}^\text{x})\text{ dx}$
Answer$\int(2\text{x}-3\cos\text{x}+\text{e}^\text{x})\text{ dx}$
$=\int2\text{x}\text{ dx}-\int3 \cos\text{x dx}+\int\text{e}^\text{x}\text{ dx}$
$=2\int\text{x}\text{ dx}-3\int\cos\text {x}\text{ dx}+\int\text{e}^{\text{ x}}\text{ dx} $
$=2\frac{\text{x}^2}{2}-3\sin\text{x}+\text{e}^{\text{x}}+\text{c} $
$=\text{x}^2-3\sin\text{x }+\text{e}^{\text{x}}+\text{c} $
View full question & answer→Question 521 Mark
Find the following integrals in Exercises: $\int\frac{\sec^2\text{x}}{\text{cosec}^2\text{x}}\text{ dx}$
Answer$\int\frac{\sec^2\text{x}}{\text{cosec}^2\text{x}}\text{ dx}$ $=\int\frac{\frac{1}{\cos^2\text{x}}}{\frac{1}{\sin^2\text{x}}}\text{ dx} $
$=\int\frac{\sin^2\text{x}}{\cos^2\text{x}}\text{ dx}=\int\tan^2\text{x}\text{ dx} $
$=\int(\sec^2\text{x}-1)\text{ dx}$
$\int\sec^2\text{x}\text{ dx}-\int1\text{ dx}=\tan\text{x}- \text{x}+\text{c} $
View full question & answer→Question 531 Mark
Find an anti derivative (or integral) of the following function by the method of inspection.
$\cos3\text{x}$
Answer$\because \ \ \ \ \ \ \frac{\text{d}}{\text{dx}}(\sin3\text{x})=3\cos3\text{x}$$\ \ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ \ \frac{1}{3}\frac{\text{d}}{\text{dx}}(\sin3\text{x})=(\cos3\text{x})$
$\ \ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ \ \frac{\text{d}}{\text{dx}} \bigg(\frac{1}{3}\sin3\text{x} \bigg)=\cos3\text{x}$ $\therefore \ \ \ \ \ \ \ \ \ \ $An anti-derivative of $ \cos 3\text{x} \text{ is}\frac{1}{3}\sin3\text{x}$
View full question & answer→Question 541 Mark
Find an anti derivative (or integral) of the following function by the method of inspection. $\text{e}^{2\text{x}}$
Answer$\because \ \ \ \ \ \ $$\frac{\text{d}}{\text{dx}}\text{e}^{2\text{x}}=\text{e}^ {2\text{x}}\text{ }\frac{\text{d}}{\text{dx}}(2\text{x})=2\text{e}^{2\text{x}}$ $\ \ \ \ \ \ \ \Rightarrow \ \ \ \ \frac{\text{1}}{\text{2}}\frac{\text{d}}{\text{dx}}\text{e}^{2\text{x}}=\text{e}^{2\text{x}}$ $ \ \ \ \ \ \ \Rightarrow \ \ \ \ \frac{\text{d}}{\text{dx}}\bigg(\text{ }\frac{\text{1}}{\text{2}}\text{e}^{2\text{x}} \bigg)=\text{e}^{2\text{x}}$$\because \ \ \ \ \ \ $An anti-derivative of $\text{e}^{2\text{x}} \text { is } \frac{1}{2}\text{e}^{2\text{x}}. $
View full question & answer→Question 551 Mark
Integrate the functions in Exercises:
$\frac{1}{1+\cot\text{x}}$
Answer$\text{Let I}=\int\frac{1}{1+\cot\text{x}}\text{ dx} =\int\frac{1}{1+\frac{\cos\text{x}}{\sin\text{x}}}\text{ dx} $
$=\int\frac{1}{\bigg(\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}}\bigg)}\text{ dx} =\int\frac{\sin\text{x}}{\sin\text{x}+\cot\text{x}}\text{ dx} $
$=\frac{1}{2}\int\frac{2\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}=\frac{1}{2}\int\frac{\sin\text{x}+\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx} $
Adding and subtracting $\cos\text{x}$ in the numerator,
$=\frac{1}{2}\int\frac{\sin\text{x}+\cos\text{x}-\cos\text{x}+\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx} $
$=\frac{1}{2}\int\frac{(\sin\text{x}+\cos\text{x})-(\cos\text{x}-\sin\text{x})}{\sin\text{x}+\cos\text{x}}\text{ dx} $
$=\frac{1}{2}\int\bigg(\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}+\cos\text{x}}-\frac{\cos\text{x}-\sin\text{x}}{\sin \text{x}+\cos\text{x}}\bigg)\text{ dx} $
$=\frac{1}{2}\int\bigg(1-\frac{\cos\text{x}-\sin\text{x}}{\sin\text{x}+\cos\text{x}}\bigg)\text{ dx} $
$=\frac{1}{2}\bigg[\int1\text{ dx}-\int\frac{\cos\text{x}-\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}\bigg]=\frac{1} {2}[\text{x}-\text{I}_{1}]$$\ \ \ \ \ \text{where }\text{I}_{1}=\int\frac{\cos\text{x}-\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx} \ \ \ ...\text{(i)}$
Putting $\sin\text{ x + cos x = t}\ \ \ \Rightarrow\ \ \ \cos\text{x}-\sin\text{x}=\frac{\text{dt}}{\text{dx}}\ \ \ \Rightarrow\ \ \ \ (\cos\text{x}-\sin\text{x})\text{ dx = dt} $
$ \therefore\ \ \ \ \ $$\text{I}_1=\int\frac{\text{dt}}{\text{t}}=\log\begin{vmatrix}\text{t}\end{vmatrix}=\log\begin{vmatrix}\sin\text{x}+\cos\text{x}\end{vmatrix} $
Putting this value in eq. (i), we get required integral,
$=\frac{1}{2}[\text{x}-\log\begin{vmatrix}\sin\text{x + cosx}\end{vmatrix}]+\text{c} $
View full question & answer→Question 561 Mark
Find the following integrals in Exercises:
$\int(\text{ax}^2+\text{bx}+\text{c})\text { dx}$
Answer$\int(\text{ax}^2+\text{bx}+\text{c)}\text{ dx}$
$=\int\text{ax}^2 \text{ dx}+\int\text{bx}\text{ dx}+\int\text{c}\text{ dx}$
$=\text{a}\int\text{x}^2\text{ dx}+\text{b}\int\text{x}+\text{dx}+c\int1\text{ dx}$
$=\text{a}\frac{\text{x}^3}{3}+\text{b}\frac{\text{x}^2}{2}+\text{cx}+\text{c}_{1}$
where $\text{c}_{1}$ is the constant of integration.
View full question & answer→Question 571 Mark
Integrate the functions in Exercises:
$\frac{1}{\cos^2\text{x}(1-\tan\text{x})^2}$
Answer$\text{Let I}=\int\frac{1}{\cos^2\text{x}(1-\tan\text{x})^2}\text{ dx}=\int\frac{\sec^2\text{x}}{(1 - \tan\text{x})^2}\text{ dx}$
$=-\int\frac{-\sec^2\text{x}}{(1 - \tan\text{x})^2}\text{ dx}\ \ \ \ \ \ \ \ \ \ ...\text{(i)} $
Putting $1-\tan\text{x = t}\ \ \Rightarrow\ \ \ \ -\sec^2\text{x}=\frac{\text{dt}}{\text{dx}}\ \ \ \Rightarrow\ \ \ \ -\sec^2\text{x}\text{ dx}=\text{dt} $
$\therefore \ \ \ \ $From eq. (i), $\text{I}=-\int\frac{\text{dt}}{\text{t}^2}=-\int\text{t}^{-2}\text{ dt}=\frac{-\text{t}^{-1}}{-1}+\text{c}=\frac{1}{\text{t}}+\text{c}$
$=\frac{1}{1 - \tan \text{x}}+\text{c} $
View full question & answer→Question 581 Mark
Find an anti derivative (or integral) of the following function by the method of inspection.
$\sin2\text{x}-4\text{ e}^{3\text{x}}$
Answer$\because \ \ \ \ \ \ \ \text{ }\frac{\text{d}}{\text{dx}}(\cos2\text{x})=-2\sin2\text{x}$ $\Rightarrow\frac{1}{-2}\text{ }\frac{\text{d}} {\text{dx}}(\cos2\text{x})=\sin2\text{x}$ $\Rightarrow\text{ }\frac{\text{d}}{\text{dx}}\bigg(\frac{-1}{2}\cos2\text{x} \bigg)=\sin2\text{x} \ \ \ ....\text{(i)}$Again$ \ \ \ \ \ \ \ \ \ \text{ }\frac{\text{d}}{\text{dx}}\text{e}^{3\text{x}}=3\text{e}^{3\text{x}}$
$\ \ \ \ \Rightarrow\ \ \ \ \ \text{ }\frac{\text{d}}{\text{dx}}\bigg (\text{ }\frac{\text{1}}{\text{3}}\text{e}^{3\text{x}}\bigg)=\text{e}^{3\text{x}}$
$\frac{\text{d}}{\text{dx}}\bigg( \text{ }\frac{\text{-4}}{\text{3}}\text{e}^{3\text{x}}\bigg)=-4\text{e}^{3\text{x } } \ \ \ \ \ \ \ \text{[Multiplying both sides by -4 ]} \ \ \ \ \ \ \ ...\text{(ii)}$ Adding eq. (i) and (ii), we get
$\Rightarrow \ \ \ \ \text{ }\frac{\text{d}}{\text{dx}}\bigg(\text{ }\frac{\text{-1}}{\text{2}}\cos2\text{x}\bigg)+\text{ }\frac {\text{d}}{\text{dx}}\bigg(\text{ }\frac{\text{-4}}{\text{3}}\text{e}^{3\text{x}}\bigg)= \sin2\text{x}+\bigg(-4\text{e}^ {3\text{x}}\bigg)$
$\Rightarrow\text{ }\frac{\text{d}}{\text{dx}}\bigg(\text{ }\frac{\text{-1}}{\text{2}}\cos2\text{x}-\text{ }\text{ }\frac {\text{4}}{\text{3}}\text{e}^{3\text{x}}\bigg)= \sin2\text{x}-4\text{e}^{3\text{x}}$ $\therefore\ \ \ \ \ \ \ \ $An anti-derivative of $\sin2\text{x}-4\text{e}^{3\text{x}} \text{ is }\frac{-1}{2} \cos2\text{x}-\frac{4}{3} \text{e}^{3\text{x}}.$
View full question & answer→Question 591 Mark
Find the following integrals in Exercises: $\int\sqrt{\text{x}}(3\text{x}^2+2\text{x+3})\text{ dx}$
Answer$\int\sqrt{\text{x}}(3\text{x}^2+2\text{x+3})\text{ dx}$
$=\int\text{x}^\frac{1}{2}(3\text{x}^2+2\text{x+3})\text{ dx}$
$=\int\bigg(3\text{x}^\frac{5}{2}+2\text{x}^{\frac{3}{2}}+3\text{x}^\frac{1}{2}\bigg)\text{ dx}$
$=\int3\text{x}^{\frac{5}{2}}\text{ dx }+\int2\text{x}^{\frac{3}{2}}\text{ dx}+\int3\text{x}^{\frac{1}{2}}\text{ dx}$
$=3\frac{\text{x}^{\frac{5}{2}+1}}{\frac{5}{2}+1}+2 \frac { \text{x}^{\frac{3}{2}+1}}{\frac{3}{2}+1}+3 \frac{\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\text{c} $
$=3\frac{\text{x}^{\frac{7}{2}}}{\frac{7}{2}}+2 \frac { \text{x}^{\frac{5}{2}}}{\frac{5}{2}}+3 \frac{\text{x}^{\frac{3}{2}}}{\frac{3}{2}}+\text{c} $
$\frac{6}{7}\text{x}^\frac{7}{2 }+ \frac{4}{5}\text{x}^\frac{5}{2 }+ 2\text{x}^\frac{3}{2}+\text{c} $
View full question & answer→Question 601 Mark
Integrate the functions in Exercises:
$\frac{\sqrt{\tan\text{x}}}{\sin\text{x}\cos\text{x}}$
Answer$\text{Let I}=\int\frac{\sqrt{\tan\text{x}}}{\sin\text{x}\cos\text{x}}\text{ dx}=\int\frac{\sqrt{\tan\text{x}}}{\frac{\sin\text{x}}{\cos\text{x}}\cos\text{x}.\cos\text{x}}\text{ dx} $ $=\int\frac{\sqrt{\tan\text{x}}}{{\tan\text{x}\cos^2\text{x}}}\text{ dx} =\int\frac{\sec^2\text{x}}{\sqrt{\tan\text{x}}}\text{ dx} \ \ \ \ ...\text{(i)}$ Putting $\tan\text{x}=\text{t}\ \ \ \ \Rightarrow \ \ \ \ \sec^2\text{x}=\frac{\text{dt}}{\text{dx}}\ \ \ \Rightarrow \ \ \ \ \sec^2\text{x}\text{ dx = dt}$ $\therefore \ \ \ \ \ $From eq. (i), $\text{I}=\int\frac{\text{dt}}{\sqrt{\text{t}}}=\int\text{t}^{\frac{-1}{2}}\text{dt}=\frac{\text{t}^{^1/_2}}{^1/_2}+\text{c}$$=2\sqrt{\text{t}}+\text{c}=2\sqrt{\tan\text{x}}+\text{c}$
View full question & answer→Question 611 Mark
Evaluate $\int\frac{1-\sin\text{x}}{\cos^2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1-\sin\text{x}}{\cos^2\text{x}}\text{ dx}$
$=\int\sec^2\text{x dx}-\int\sec\text{x}\tan\text{x dx}$
$\text{I}=\tan\text{x}-\sec\text{x}+\text{C}$
View full question & answer→Question 621 Mark
Integrate the functions in Exercises:
$\text{x}\sqrt{1+2\text{x}^2}$
Answer$\text{Let}\text{ I}=\int\text{x}\sqrt{1+2\text{x}^2}\text{ dx} =\frac{1}{4}\int\sqrt{1+2\text{x}^2}(4\text{x dx}) \ \ \ \ \ \ ....\text{(i)} $
Putting $ 1+2\text{x}^2=\text{t}\ \ \ \ \Rightarrow \ \ \ \ \ 4\text{x}=\frac{\text{dt}}{\text{dx}}\ \ \ \ \ \Rightarrow \ \ \ \ 4\text{x}\text{ dx}=\text{dt} $
$\therefore\ \ \ \ \ $From eq. (i), $\text{I}=\frac{1}{4}\int\sqrt{\text{t}}\text{ dt}=\frac{1}{4}\int\text{t}^{\frac{1}{2}}\text{ dt}$
$=\frac{1}{4}\frac{\text{t}^{^3/_2}}{{^3/_2}}+\text{c}=\frac{1}{4}\dot\ \frac{2}{3}\text{t}^{^3/_2 } +\text{c} $
$=\frac{1}{6}\big(1 + 2\text{x}^2\big)^{^3/_2}+\text{c} $
View full question & answer→Question 631 Mark
Integrate the functions in Exercises:
$(\text{x}^3-1)^{\frac{1}{3}}\text{ x}^5$
Answer$\text{Let I}=\int(\text{x}^3 - 1)^{\frac{1}{3}}\text{x}^5\text{ dx}$ $=\int(\text{x}^3 - 1\big)^{\frac{1}{3}}\text{x}^3\text{x}^2\text{ dx}=\frac{1}{3}\int(\text{x}^3 - 1)^{\frac{1}{3}}\text{ x}^3(3\text{x}^2\text{ dx})\ \ \ \ \ \ .....\text{(i)}$ Putting $\text{ x}^3-1=\text{t}\ \ \ \ \Rightarrow \ \ \ \ \ \text{ x}^3=\text{t}+1\ \ \ \Rightarrow \ \ \ \ \text{3x}^2=\frac{\text{dt}}{\text{dx}}\ \ \ \ \Rightarrow \ \ \ \ 3\text{x}^2\text{ dx}=\text{dt} $ From eq. (i), $\text{I}=\frac{1}{3}\int\text{t}^{\frac{1}{3}}(\text{t}+1)\text{ dt}$ $=\frac{1}{3}\int\bigg(\text{t}^{\frac{4}{3}}+\text{t}^{\frac{1}{3}}\bigg)\text{ dt}=\frac{1}{3}\bigg(\int\text{t}^{\frac{4} {3}}\text{ dt}+\int\text{t}^{\frac{1}{3}}\text{ dt}\bigg) $$=\frac{1}{3}\Bigg(\frac{\text{t}^{\frac{7}{3}}}{\frac{7}{3}}+\frac{\text{t}^{\frac{4}{3}}}{\frac{4}{3}}\Bigg)+\text{ c} =\frac{1}{3}\bigg(\frac{3}{7}\text{ t}^{\frac{7}{3}}+\frac{3}{4}\text{ t}^{\frac{4}{3}}\bigg)+\text{c}$
$=\frac{1}{7}\text{t}^{\frac{7}{3}}+\frac{1}{4}\text{t}^{\frac{4}{3}}+ \text{c} $ $=\frac{1}{7}(\text{x}^3 - 1)^{\frac{7}{3}}+\frac{1}{4}(\text{x}^3 - 1)^{\frac{4}{3}}+\text{c} $
View full question & answer→Question 641 Mark
Integrate the functions in Exercises:
$\frac{\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}}{\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}}$
Answer$\text{Let I}=\int\frac{\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}}{\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}} \text{ dx}=\frac{1}{2}\int\frac{\text 2\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)}{\text {e}^{\text{2x}}+\text{e}^{-\text{2x}}}\text{ dx} \ \ \ \ \ \ ...\text{(i)} $
Putting$\text{ e}^{\text{2x}}+\text{e}^{-2\text{x}}=t\ \ \ \Rightarrow \ \ \ \ \text{ e}^{\text{2x}}\frac{\text{d}}{\text{dx }} 2\text{x} +\text{e}^{-2\text{x}}\frac{\text{d}}{\text{dx}}(-2\text{x})=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^{2\text{x}}.2-2\text{e}^{-2\text{x}}=\frac{\text{dt}}{\text{dx}} \ \ \ \ \Rightarrow\ \ \ \ \ 2\big(\text{e}^{\text{2x}}-\text{e}^{-\text{2x}}\big)\text{ dx}=\text{dt} $
$\therefore \ \ \ \ \ $From eq. (i), $\text{I}=\frac{1}{2}\int\frac{\text{dt}}{\text{t}} $
$=\frac{1}{2}\log\begin{vmatrix}t\end{vmatrix}+\text{c}=\frac{1}{2}\log\begin{vmatrix}\text{e}^\text {2x}+\text{e}^{-\text{2x}}\end{vmatrix}+\text{c} $
$=\frac{1}{2}\log\big(\text{e}^\text{2x}+\text{e}^{-\text{2x}}\big)+\text{c} $
View full question & answer→Question 651 Mark
Integrate the functions in Exercises:
$\frac{1}{\text{x}+\text{x}\log\text{x}}$
AnswerPutting $1 + \log \text{x}=\text{t}\ \ \ \ \ \Rightarrow\ \ \ \ \ \ \frac{1}{\text{x}}=\frac{\text{dt}}{\text{dx}}\ \ \ \ \ \Rightarrow\ \ \ \ \ \frac{\text{dx}}{\text{x}}=\text{dt} $
$\therefore\ \ \ \ \ \int \frac{1}{\text{x + x log x }}\text{ dx}$
$\int \frac{1}{\text{1 + log x }}\frac{\text{dx}}{\text{x}}==\int\frac{1}{\text{t}}\text{ dt}=\log\mid\text{t}\mid+\text{c}$
$=\log\mid\text{1+ log x}\mid+\text{ c}$
View full question & answer→Question 661 Mark
Evaluate the following integrals:
$\int\log_\text{x}\text{xdx}$
Answer$\int\log_\text{x}\text{xdx}$
$=\int1.\text{dx}$
$=\text{x}+\text{c}$
View full question & answer→Question 671 Mark
Evaluate $\int\sec^2(7-4\text{x})\text{dx}$
Answer$\int\sec^2(7-4\text{x})\text{dx}$
$=\frac{\tan(7-4\text{x})}{-4}+\text{C}$ $(\because\sec^2\text{x}=\tan\text{x}+\text{C})$
View full question & answer→Question 681 Mark
Find an anti derivative (or integral) of the following function by the method of inspection.
$\sin2\text{x}$
Answer$\because\text{ }\frac{\text{d}}{\text{dx}}(\cos2\text{x})=-2\sin2\text{x} $$\Rightarrow \frac{1}{-2}\frac{\text{d}}{\text{dx}}(\cos2\text{x})=\sin2{\text{x}} $
$\Rightarrow \text{ }\frac{\text{d}}{\text{dx}}\bigg( \frac{-1}{2}\cos2\text{x}\bigg)=\sin2\text{x}$ $\therefore \ \ \ \ \ \ \ $An anti-derivative of $\sin2\text{x is}\frac{-1}{2}\cos2\text{x}.$
View full question & answer→Question 691 Mark
Find the following integrals in Exercises: $\int\sec\text{x}(\sec\text{x}+\tan\text{x})\text{ dx}$
Answer$\int\sec\text{x}(\sec\text{x}+\tan\text{x})\text{ dx}$
$=\int(\sec^2\text{x}+\sec\text{x }\tan\text{x})\text{ dx} $
$=\int\sec^2\text{x}\text{ dx}+\int\sec\text{x}\tan\text{x}\text{ dx} $
$=\tan\text{x}+\sec\text{x}+\text{c}$
View full question & answer→Question 701 Mark
Integrate the functions in Exercises:
$\frac{1}{\text{x}(\log\text{x})^\text{m}},\text{x}>0$
Answer$\text{Let I}=\int\frac{1}{\text {x}(\log \text{x})^\text{m}} \text{ dx}=\int\frac{\frac{1}{\text{x}}\text{dx}}{(\log \text{x)}^\text{m}} \ \ \ \ ....\text{(i)} $
Putting $\log\text{x}=\text{t}\ \ \ \Rightarrow \ \ \ \ \frac {1}{\text{x}}=\frac{\text{dt}}{\text {dx}}\ \ \ \ \Rightarrow \ \ \ \ \frac{\text{dx}}{\text{x}}=\text{dt}$
$\therefore \ \ \ \ \ $ From eq. (i), $\text{I}=\int\frac{\text{dt}}{\text{t}^{\text{m}}}=\int\text{t}^\text{-m}\text{ dt}=\frac{\text{t}^{-\text{m}+1}}{-\text {m + 1}}+\text{c}$
$=\frac{(\log\text{x})^{1-\text {m}}}{1-\text{m}}+\text{c} $
View full question & answer→Question 711 Mark
Integrate the functions in Exercises:
$\frac{\text{x}^3\sin\big(\tan^{-1}\text{x}^4\big)}{1+\text{x}^8}$
Answer$\text{Let I}=\int\frac{\text{x}^3\sin{(\tan^{-1}\text{x}^4)}}{1 +\text{x}^8}\text{ dx} $
$=\frac{1}{4}\int\sin(\tan^{-1}\text{x}^4)\dot\ \frac{4\text{x}^3}{1 + \text{x}^8}\text{ dx} \ \ \ \ ...\text{(i)} $
Putting $\tan^{-1} \text{x}^4=\text{t}\ \ \ $$\Rightarrow\ \ \ \frac{1}{1+(\text{x}^4)^2}\frac{\text{d}}{\text{dx}}\text{x}^4=\frac{\text{dt}}{\text{dx}}\ \ \ $$\Rightarrow \ \ \ \frac{4\text{x}^3}{1+\text{x}^8}\text{ dx = dt} $
$\therefore \ \ \ $From eq. (i), $\text{ I}=\frac{1}{4}\int\sin\text{t}\text{ dt}=\frac{-1}{4}\text{cos t}+\text{c}$
$=\frac{-1}{4}\cos(\tan^{-1}\text{x}^4)+\text{c}$
View full question & answer→Question 721 Mark
Find the following integrals in Exercises: $\int(1-\text{x})\sqrt{\text{x}}\text{ dx}$
Answer$\int(1-\text{x})\sqrt{\text{x }}\text{ dx} =\int\big(\sqrt{\text{x}}-\text{x}\sqrt{\text{x}}\big)\text{ dx}$
$=\int\bigg(\text{x}^\frac{1}{2}-\text{x}^\frac{3}{2}\bigg)\text{ dx}=\int\text{x}^\frac{1}{2}\text{ dx}-\int\text{x}^\frac{3}{2}\text{ dx} $
$=\frac{\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{\text{x}^{\frac{3}{2}+1}}{\frac{3}{2}+1}+\text{c} =\frac{\text{x}^{\frac{3}{2}}}{\frac{3}{2}}+ \frac{\text{x}^{\frac{5}{2}}}{\frac{5}{2}}$
$=\frac{2}{3}\text{x}^\frac{3}{2}-\frac{2}{5}\text{x}^{\frac{5}{2}}+\text{c}$
View full question & answer→Question 731 Mark
Integrate the functions in Exercises:
$\sin(\text{ax+b})\cos(\text{ax+b})$
Answer$\int\sin(\text{ax+b})\cos(\text{ax+b})\text{ dx}$
$=\frac{1}{2}\int2\sin(\text{ax+b})\cos(\text{ax+b})\text { dx} $
$=\frac{1}{2}\int\sin2(\text{ax+b})\text{ dx} $
$=\frac{1}{2}\int\sin(2\text{ax+2b})\text{ dx} $
$=\frac{1}{2}\frac{[-\cos(2\text{ax}+2\text{b})]}{2\text{a}\rightarrow\text{Coeff. of x}}+\text{c} $
$=\frac{-1}{4\text{a}}\cos2(\text{ax+b})+\text{c} $
View full question & answer→Question 741 Mark
Find the following integrals in Exercises. $\int\frac{\text{x}^3+3\text{x}+4}{\sqrt{\text{x}}}\text{ dx}$
Answer$\int\frac{\text{x}^3+3\text{x}+4}{\sqrt{\text{x}}}\text{ dx}$
$=\int\Bigg(\frac{\text{x}^3}{\text{x}^{\frac{1}{2}}}+\frac{3\text{x}}{\text{x}^{\frac{1}{2}}}+\frac{4}{\text{x}^{\frac{1}{2}}}\Bigg)\text{ dx} $
$=\int\bigg(\text{x}^{3-\frac{1}{2}}+3\text{x}^{1-\frac {1}{2}}+4\text{x}^{\frac{-1}{2}}\bigg)\text{ dx}$
$=\int\bigg(\text{x}^{\frac{5}{2}}+3\text{x}^{\frac{1}{2}}+4\text{x}^{\frac{-1}{2}}\bigg)\text{ dx}$
$=\int\text{x}^\frac{5}{2}\text{ dx}+3\int\text{x}^\frac{1}{2}\text{ dx}+4\int\text{x}^\frac{-1}{2}\text{ dx} $
$=\frac{\text{x}^{\frac{5}{2}+1}}{\frac{5}{2}+1}+3\frac{\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+4 \frac{\text{x}^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+\text{c} $
$=\frac{\text{x}^{\frac{7}{2}}}{\frac{7}{2}}+3 \frac{\text{x}^{\frac{3}{2}}}{\frac{3}{2}}+4 \frac{\text{x}^{\frac{1}{2}}}{\frac{1}{2}}+\text{c} $
$=\frac{2}{7}\text{x}^{\frac{7}{2}}+2\text{x}^{\frac{3}{2}}+8\text{x}^{\frac{1}{2}}+\text{c}$
View full question & answer→Question 751 Mark
Find the following integrals in Exercises. $\int\bigg(\sqrt{\text{x}}-\frac{1}{\sqrt{\text{x}}}\bigg)^2$
Answer$\int\bigg(\sqrt{\text{x}}-\frac{1}{\sqrt{\text{x}}}\bigg)^2\text{ dx}$
$=\int \bigg\{\big(\sqrt{\text{x } } \big)^2+\bigg(\frac{1}{\sqrt{\text{x}}}\bigg)^2-2\sqrt{\text{x}}\frac{1}{\sqrt{\text{x}}}\bigg\}\text{ dx}$
$ =\int\bigg(\text{x}+\frac{1}{\text{x}}-2\bigg)\text{ dx}$
$=\int\text{x}\text{ dx}+\int\frac{1}{\text{x}}\text{ dx}-\int 2\text{ dx}=\frac{\text{x}^2}{2}+\log\mid\text{x}\mid-2\text{x}+\text{c}$
View full question & answer→Question 761 Mark
Integrate the functions in Exercises:
$\frac{\text{x}}{\text{e}^{\text{x}^2}}$
Answer$\text{Let I}=\int\frac{\text{x}}{\text{e}^{\text{x}^{2}}} \text{ dx}=\frac{1}{2}\int\frac{\text{2x}}{\text{e}^{\text{x}^{2}}}\text{ dx}\ \ \ \ ....\text{(i)} $
$\text{Putting}\text{ x}^2=\text{t}\ \ \ \ \Rightarrow \ \ \ \ 2\text{x}=\frac{\text{dt}}{\text{dx}}\ \ \ \ \Rightarrow \ \ \ \ \ \text{2x}{\text{ dx}}=\text{ dt} $
$\therefore \ \ \ \ $From eq. (i), $\text{I}=\frac{1}{2}\int\frac{\text{dt}}{\text{e}^{\text{t}}}=\frac{1}{2}\int\text{e}^{\text{-t}}\text{ dt}$
$=\frac{1}{2}\cdot\frac{\text{e}^{-\text{t}}}{-1\rightarrow \text{Coeff. of t}}+\text{c}$
$=\frac{-1}{2(\text{e}^{\text{t}})}+\text{c}=\frac{-1}{2(\text{e}^{\text{x}^{2}})}+\text{c} $
View full question & answer→Question 771 Mark
Integrate the functions in Exercises:
$\sqrt{\sin2\text{x}}\cos2\text{x}$
Answer$\text{Let I}=\int\sqrt{\sin2\text{x }}\cos 2\text{x dx}=\frac{1}{2}\int\sqrt{\sin2\text{x }}(2\cos\text{2x dx}) \ \ \ \ \ \ \ ...\text{(i)} $
Putting $\sin2\text{x}=\text{t}\ \ \ \Rightarrow\ \ \ \ \cos2 \text{x}\frac{\text{d}}{\text{dx}}(2\text{x})=\frac{\text{dt}}{\text{dx}}\ \ \ \ \Rightarrow\ \ \ 2\cos 2\text{x dx = dt} $
$\therefore\ \ \ \ \ \ $From eq. (i), $\text{I}=\frac{1}{2}\int\sqrt{\text{t}}\text{ dt}=\frac{1}{2}\int\text{t}^{\frac{1}{2}}\text{ dt}=\frac{1}{2}\frac{\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\text{c}$
$=\frac{1}{2}\dot\ \frac{\text{t}^{^3/_2}}{^3/_2}+\text{c}=\frac{1}{3}(\sin 2\text{x})^{\frac{3}{2}}+\text{c} $
View full question & answer→Question 781 Mark
Find the following integrals in Exercises:
$\int\text{x}^2 \bigg(1-\frac{1}{\text{x}^2}\bigg)\text{dx}$
Answer$\int\text{x}^2 \bigg(1-\frac{1}{\text{x}^2}\bigg)\text{dx}=\int\bigg(\text{x}^2-\frac{\text{x}^2}{\text{x}^2}\bigg)\text {dx}$
$=\int(\text{x}^2-1)\text{dx}=\int\text{x}^2\text{ dx}-\int1\text{ dx}$
$\frac{\text{x}^3}{3}-\text{x+c}\ \ \ \ \ \ \ \ \ \ \ \bigg[\because\int\text{x}^\text{n} \text{ dx}=\frac{\text {x}^{\text{n}+1}}{\text{n+1}}\text{ if}\text{ n}\neq-1\bigg]$
View full question & answer→Question 791 Mark
Integrate the functions in Exercises:
$\frac{\text{e}^{\tan^{-1}\text {x}}}{1+\text{x}^2}$
Answer$\text{Let I}=\int\frac{\text{e}^{\tan^{-1}\text{x}}}{1+\text{x}^2}\text{ dx} \ \ \ \ \ \ \ ...\text{(i)} $
$\text{Putting }\tan^{-1}\text{x}=\text{t}\ \ \ \Rightarrow \ \ \ \frac{1}{1+\text{x}^2}=\frac{\text{dt}}{\text{dx}}\ \ \ \ \Rightarrow \ \ \ \ \frac{\text{dx}}{1+\text{x}^2}=\text{ dt} $
$\therefore \ \ \ \ $From eq. (i), $\text{I}=\int\text{e}^{\text{t}}\text{ dt}=\text{e}^{\text{t}}+\text{c}=\text{e}^{{\tan}^{-1}\text{x}}+\text{c}$
View full question & answer→Question 801 Mark
Integrate the functions in Exercises:
$\frac{(1+\log\text{x})^2}{\text{x}}$
Answer$\text{Let I}=\int\frac{(1+\log\text{x})^2}{\text{x}}\text{ dx} \ \ \ \ ... \text{(i)}$
Putting $1+\log\text{x}=\text{t} \ \ \ \ \Rightarrow \ \ \ \frac{1}{\text{x}}=\frac{\text{dt}{}}{\text{dx}}\ \ \Rightarrow \ \ \ \frac{\text{dx}{}}{\text{x}}=\text{dt} $
$\therefore \ \ \ \ $From eq. (i), $\text{I}=\int\text{t}^2\text{ dt}=\frac{\text{t}^3}{3}+\text{c}= \frac{1}{3}(1+\log\text{x})^3+\text{c}$
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By Using properties of definite integral, evaluate the following integral in Exercise:
$\int^{2\pi}_{0}\cos^{5}\text{x}\ \text{dx}$
Answer$\int^{2\pi}\limits_{0}\cos^{5}\ \text{x}\ \text{dx}=2\int^{\pi}\limits_{0}\cos^{5}\text{x}\ \bigg[\because\int^{2\text{a}}\limits_{0}\text{f}\text{(x)}\text{dx}=2\int^{\text{a}}\limits_{0}\text{f}\text{(x)}\text{dx},\text{if}\ \text{f}(2\text{a}-\text{x})=\text{f}\text{(x)}\bigg]$
$\text{Here}\ \text{f}\text{(x)}=\cos^{5}\text{x}\ \therefore\ \text{f}(2\pi-\text{x}=\cos^{5}(2\pi-\text{x})=\cos^{5}\text{x}$
$\Rightarrow\ \ \text{f}\ \text{(x)}=2(0)=0$
View full question & answer→Question 821 Mark
Evaluate $\int2^{\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int2^{\text{x}}\text{ dx}$
$\text{I}=\frac{2^{\text{x}}}{\log_\text{e}2}+\text{C}$
View full question & answer→Question 831 Mark
Integrate the functions in Exercises:
$\frac{\sin\text{x}}{(1+\cos\text{x})^2}$
Answer$ \text{Let I}=\int\frac{\sin\text{x}}{(1 +\cos\text{x})^2}\text{ dx}=-\int\frac{-\sin\text{x}}{(1+\cos\text{x})^2}\text{ dx} \ \ \ \ \ \ \ \ ...\text{(i)} $
Putting $1+\cos\text{x}=\text{t}\ \ \ \Rightarrow\ \ \ -\sin\text{x}=\frac{\text{dt}}{\text{dx}}\ \ \ \Rightarrow\ \ \ -\sin\text{x dx}=\text{dt} $
$\therefore \ \ \ \ $From eq. (i), $\text{I}=-\int\frac{\text{dt}}{\text{t}^2}=-\int\text{t}^{-2}\text{ dt}=\frac{-\text{t}^{-1}}{-1}+\text{c}$
$=\frac{1}{1+\cos\text{x}}+\text{c}$
View full question & answer→Question 841 Mark
Find the following integrals in Exercises:
$\int(4\text{ e}^{3\text{x}} + 1)\text{ dx}$
Answer$\int\bigg(4\text{e}^{3\text{x}}+1\bigg)\text{dx}$
$=\int4\text{e}^{3\text{x}}\text{ dx}+\int1 \text{ dx}=4\int{e}^{3\text {x}}\text{ dx}+\text{x}$
$=4 \frac{\text{e}^{3\text{x}}}{3}+\text{x}+\text{c} \ \ \ \ \ \ \ \ \ \ \bigg[\because \text{e}^ \text{ax}\text{ dx}=\frac{\text{e}^\text{ax}}{\text{x}}\text{and}\int1\text{ dx}=\text{x}\bigg]$
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