Question 515 Marks
$\text{Find}: \int(x + 3) \sqrt{3 - 4x - x^{2}} dx.$
Answer$\text{Writing x} + 3 = -\frac{1}{2} (-4 - 2x) + 1$
$\therefore\int \text{(x + 3)} \sqrt{3 - 4\text{x} - \text{x}^{2}} \text{ dx} = -\frac{1}{2} \int(- 4 - 2\text{x}) \sqrt{3 - 4\text{x} - \text{x}^{2}} \text{ dx} + \int\sqrt{7 - (\text{x} + 2)^{2}} \text{dx}$
$= -\frac{1}{3} ( 3 - 4\text{x} - \text{x}^{2})^{3/2} + \frac{\text{x + 2}}{2} \sqrt{3 - 4\text{x}} - \text{x}^{2} + \frac{7}{2} \sin^{-1} \bigg(\frac{\text{x + 2}}{\sqrt{7}}\bigg) + \text{c}$
View full question & answer→Question 525 Marks
Evaluate: $\int\limits^{\pi/2}_{0}\frac{x\sin x\cos x}{\sin^4x+\cos^4x}\text{d}x$
Answer$\text{let I}=\int\limits^{\pi/2}_{0}\frac{x\sin x\cos x}{\sin^4x+\cos^4x}\text{d}x;$ $\therefore\ \ \text{I}=\int\limits^{\pi/2}_{0}\frac{\big(\pi/2-x\big)\cos x\sin x}{\cos^4x+\sin^4x}\text{d}x$
Adding we get, $\text{2I}=\frac{\pi}{2}\int\limits^{\pi/2}_{0}\frac{\sin x\cos x}{\sin^4x+\cos^4x}\text{d}x;$ $\ \ =\int\limits^{\pi/2}_{0}\frac{2\tan x \sec^2 x}{1+(\tan^2x)^2}\text{d}x$
$\ \ =\frac{\pi}{4}\tan^{-1}(\tan^2x)\bigg]^{\pi/2}_0=\frac{\pi^2}{8}$
$\therefore\ \ \text{I}=\frac{\pi^2}{16}$
View full question & answer→Question 535 Marks
$\text{Find}\int\frac{(3 - \sin\theta - 2)\cos\theta}{5 - \cos^{2}\theta - 4 \sin\theta} \text{d}\theta.$
Answer$\text{I} = \int\frac{(3\sin\theta - 2) \cos\theta}{5 - (1 -\sin^{2}\theta) - 4\sin\theta}\text{d}\theta$
$\sin\theta = \text{t} \Rightarrow \cos\theta\text{d}\theta = \text{dt}$
$\therefore\text{I} = \int\frac{3\text{t} - 2}{\text{t}^{2} - 4\text{t} + 4} = \int\frac{3\text{t} - 2}{(\text{t} - 2)^{2}}\text{dt}$
$= \int\frac{3(\text{t} - 2)}{(\text{t - 2)}^{2}}\text{dt} + 4 \int\frac{1}{(\text{t - 2)}^{2}}\text{dt}$
$= 3\log|\text{t} = 2| - \frac{4}{(\text{t - 2)}} + \text{C}$
$= 3\log|\sin\theta - 2| - \frac{4}{(\sin\theta - 2)} + \text{C}$
View full question & answer→Question 545 Marks
$\text{Find}:\int \frac{dx}{\sin x + \sin 2x}$
Answer$\int \frac{\text{dx}}{\sin \text{x} + \sin \text{2x}} = \int \frac{\text{dx}}{\sin \text{x} (1 + 2 \cos \text{x)}} = \int \frac{\sin\text{x}. \text{dx}}{(1 - \cos{\text{x})(1 + \cos\text{x}) (1 + 2\cos \text{x)}}}$
$=-\int \frac{\text{dt}}{(1 - \text{t})( 1 + \text{t}) ( 1 +\text{2t})} \text{where} \cos \text{x = t}$
$=\int \Bigg(\frac{\frac{-1}{6}}{1 - t}+\frac{\frac{1}{2}}{1 + t} - \frac{\frac{4}{3}}{1 + 2t}\Bigg)\text{dt}$
$= + \frac{1}{6} \log|1 -\text{t}| + \frac{1}{2}\log|1 + \text{t}| -\frac{2}{3}\log|1 + \text{2t}| + \text{c}$
$= \frac{1}{6}\log|1 - \cos\text{x}| + \frac{1}{2}\log| 1 + \cos \text{x}| - \frac{2}{3}\log|1 + 2\cos\text{x}| + \text{c}$
View full question & answer→Question 555 Marks
Evaluate: $\int(\text{x} - 3 )\sqrt{\text{x}^{2} + 3 \text{x}- 18 }\text{ dx}.$
Answer$\int(\text{x} - 3 ) \sqrt{\text{x}^{2} + 3 \text{x} - 18 }\text{ dx}$
$ = \frac{1}{2}\int(2 \text{x} + 3 ) \sqrt{\text{x}^{2} +3 \text{x} - 18}\text{ dx} - \frac{9}{2}\int\sqrt{\text{x}^{2} + 3 \text{x} - 18 |}\text{ dx}$
$ = \frac{1}{2}.\frac{2}{3}(\text{x}^{2} + 3 \text{x} - 18 )^{3/2} - \frac{9}{2}\int\sqrt{\bigg(\text{x} + 3/2\bigg)^{2} - \bigg(\frac{9}{2}\bigg)^{2}}\text{dx}$
$ =\frac{1}{3}(\text{x}^{2} + 3 \text{x} - 18 )^{3/2} -\frac{9}{2}$
$\frac{\bigg(\text{x} + \frac{3}{2}\bigg)}{2}\sqrt{\text{x}^{2} + 3\text{x}-18}-\frac{81}{8}\log\bigg|\text{x} + \frac{3}{2} + \sqrt{\text{x}^{2} + 3 \text{x}- 18 }\bigg| +\text{c}$
Or $ = \frac{1}{3}(\text{x}^{3} + 3\text{x} - 18)^{3/2} - \frac{9}{8}$
$(2 \text{x} + 3 ) \sqrt{\text{x}^{2} + 3\text{x} - 18 } -\frac{81}{2}\log\bigg|\text{x} + \frac{3}{2} + \sqrt{\text{x}^{2} + 3\text{x} - 18 }\bigg| + \text{c}.$
View full question & answer→Question 565 Marks
Evaluate: $\int\limits^\pi_0\frac{x\tan x}{\text{sec }x.\text{ cosec }x}\text{ d}x.$
Answer$\text{Let I}=\int\limits^\pi_0\frac{\text{x tan x}}{\text{sec x cosec x}}\text{ dx}$
$\therefore\text{ I}=\int\limits^\pi_0\frac{(\pi-\text{x})\tan(\pi-\text{x})}{ \sec (\pi-\text{x}) \text{ cosec} (\pi-\text{x})}\text{ dx}$
$\Rightarrow\text{ I}=\int\limits^\pi_0\frac{(\pi-\text{x})\tan\text{x}}{ \sec\text{x}.\text{cosec}\text{ x}}\text{ dx}$
$\text{Adding we ge t, 2I}=\pi\int\limits^\pi_0\frac{\tan\text{x}}{ \sec\text{x}.\text{cosec}\text{ x}}\text{ dx}$
$\text{2I}=\pi\int\limits^\pi_0\sin^2\text{x}\text{ dx}$
$=2\pi\int\limits^{\pi/2}_0\frac{1-\cos2\text{x}}{2}\text{ dx}$
$=\pi\bigg[\Big(\text{x}-\frac{\sin2\text{x}}{2}\Big)\bigg]^{\pi/2}_0$
$=\pi.\frac{\pi}{2}=\frac{\pi^2}{2}\ \ \therefore\ \ \text{I}=\frac{\pi^2}{4}$
View full question & answer→Question 575 Marks
Find $\int \frac{2x}{(x^{2} + 1)(x^{2} + 2)^{2}} \text{d}x.$
Answer$\int \frac{\text{2x}}{(\text{x}^{2} + 1)\text{(x}^{2} + 2)^{2}} = \int \frac{\text{dy}}{\text{(y + 1) (y + 2)}^{2}} [$by substituting $x^2 = y]$
$ = \int \frac{\text{dy}}{\text{y + 1}} - \int \frac{\text{dy}}{\text{y + 2}} - \int\frac{\text{dy}}{(\text{y + 2)}^{2}} ($using partial fraction$)$
$= \log \text{(y + 1)} - \log (\text{y + 2)} + \frac{1}{\text{y + 2}} + \text{C}$
$= \log (\text{x}^{2} + 1) - \log \text{(x}^{2} + 2) + \frac{1}{\text{x}^{2} + 2} + \text{C}$
View full question & answer→Question 585 Marks
$\text{Evaluate}\int\limits^{\pi}_{0} e^{2x} . \sin\big(\frac{\pi}{4} + x\big)\text{dx}$
Answer$\text{Let I } = \int^{\pi}_{0}\sin\bigg(\frac{\pi}{4} + \text{x}\bigg)e^{2x}\text{dx}$
$ = \sin\bigg(\frac{\pi}{4}+\text{x}\bigg)\frac{e^{2x}}{2}\Bigg]^{\pi}_{0} -\int^{\pi}_{0} \cos\bigg(\frac{\pi}{4} + \text{x}\bigg)\frac{e^{2x}}{2}\text{dx}$
$\text{I} = \Bigg[\sin\bigg(\frac{\pi}{4} + \text{x}\bigg)\frac{e^{2x}}{2}- \frac{1}{2}\Bigg(\cos\bigg(\frac{\pi}{4}+\text{x}\bigg)\frac{e^{2x}}{2}\Bigg)\Bigg]^{\pi}_{0} + \frac{1}{2}\int^{\pi}_{0}-\sin\bigg(\frac{\pi}{4}+\text{x}\bigg)\frac{e^{2x}}{x}\text{dx}$
$\frac{5}{4}\text{I} =\bigg\{\frac{1}{2}\bigg[2\sin\bigg(\frac{\pi}{4} + \text{x}\bigg)-\cos\bigg(\frac{\pi}{4} + \text{x}\bigg)\bigg]e^{2x}\bigg\}^{\pi}_{0}$
$\text{I} = \frac{1}{5}\bigg\{2\bigg(-\frac{1}{\sqrt{2}}\bigg)+ \frac{1}{\sqrt{2}}\bigg\}e^{2\pi} - \bigg\{2\bigg(\frac{1}{\sqrt{2}}\bigg)-\frac{1}{\sqrt{2}}\bigg\}\Bigg] = \frac{-1}{5\sqrt{2}}(e^{2x + 1})$
View full question & answer→Question 595 Marks
$\text{Evaluate}: \int\limits^{\pi}_{-\pi} (\cos ax - \sin bx)^{2} dx$
Answer$\text{I} = \int\limits^{\pi}_{-\pi} (\cos \text{ax} - \sin \text{bx})^{2} \text{dx}=\int\limits^{\pi}_{-\pi} (\cos^{2} \text{ax} - \sin^{2} \text{bx}) \text{dx} - \int\limits^{\pi}_{-\pi} 2\cos \text{ax} \sin \text{bx dx}$
$\text{I}_{1} = 2 \int\limits^{\pi}_{0}(\cos^{2}\text{ax} + \sin^{2}\text{bx})\text{dx} \text{(being an even fun.)}$
$\text{I}_{2} = \text{0 (being an odd fun.)}$
$\therefore \text{I = I}_{1} = \int\limits^{\pi}_{0}( 1 + \cos \text{2ax} + 1 -\cos\text{2bx}) \text{dx}$
$= \bigg[2\text{x} + \frac{\sin \text{2ax}}{\text{2a}} - \frac{\sin \text{2bx}}{\text{2b}} \bigg]^{\pi}_{0}$
$= \bigg[2{\pi} + \frac{1}{\text{2a}} . \sin \text{2a}\pi - \frac{\sin 2\text{b}{\pi}}{2\text{b}}\bigg]\text{or 2} \pi$
View full question & answer→Question 605 Marks
$\text{Evaluate} \int\limits^{3}_{1} (e^{2} - 3x + x^{2} + 1)\text{dx as a limit of a sum.} $
Answer$\int\limits^{3}_{1}(e^{2 - 3x} + \text{x}^{2} + 1) \text{dx here h} = \frac{2}{\text{n}} $
$\lim\limits_{h\rightarrow{0}} \text{h}[\text{f(1) + f (1 + h) +f (1 + 2h) + }\dots\dots\dots\text{+ f (n - 1) h)] }$
$\lim\limits_{h\rightarrow{0}} \text{h}[(\text{e}^{-1} + 2) +\text{(e}^{-1 - 3\text{h}} + 2 + 2\text{h} +\text{h}^{2}) + (\text{e}^{-1-6\text{h}} + 2 + 4\text{h + 4h)}^{2}+ \dots\dots\dots$
$+ (\text{e}^{-1 - 3\text{(n - 1)h}} + 2 + 2 \text{(n - 1) h + (n - 1)}^{2}\text{h)}^{2}]$
$= \lim\limits_{h\rightarrow{0}} \text{h} \bigg[e^{-1}\bigg(1 + e^{-3\text{h}} + \text{e}^{-6\text{h}} + \dots\dots\text{+e}^{-3\text{(n- 1)h}}\bigg) + 2\text{n + 2h}\bigg(1 + 2 +\dots\dots\text{+ (n- 1)}\bigg) + \text{h}^{2}\bigg(1^{2} + 2^{2} +\dots\dots\text{+ (n - 1)}^{2}\bigg)\bigg]$
$= \lim\limits_{h\rightarrow{0}} \text{h} \bigg(e^{-1}.\frac{e^{\text-{3nh}} -1}{e^{-3} -1}. \text{h + 2nh + 2} \frac{\text{nh(nh -h)}}{2} + \frac{\text{nh (nh - h) (2nh -h)}}{6}\bigg)$
$= \text{e}^{-1}. \frac{(\text{e}^{-6} - 1)}{-3} + 4 + 4 + \frac{8}{3} = - \text{e}^{-1} \frac{(\text{e}^{-6} - 1)}{3} + \frac{32}{3}$
View full question & answer→Question 615 Marks
Evaluate: $\int\limits_{\pi/6}^{\pi/3}\frac{\text{dx}}{1 + \sqrt{\cot\text{x}}}.$
Answer$\text{I} = \int\limits_{\pi/6}^{\pi/3}\frac{\text{dx}}{1 + \sqrt{\cot\text{x}}} = \int\limits_{\pi/6}^{\pi/3}\frac{\sqrt{\sin\text{x}}}{\sqrt{\sin\text{x}} + \sqrt{\cos\text{x}}}\text{dx}$
$ = \int\limits_{\pi/6}^{\pi/3}\frac{\sqrt{\sin\bigg(\frac{\pi}{3} + \frac{\pi}{6} - \text{x}}\bigg)}{\sqrt{\sin\bigg(\frac{\pi}{3} + \frac{\pi}{6} - \text{x}\bigg) + \sqrt{\cos\bigg(\frac{\pi}{3} + \frac{\pi}{6} - \text{x}}\bigg)}}\text{dx}$
$\therefore\text{I} = \int\limits_{\pi/6}^{\pi/3}\frac{\sqrt{\cos\text{x}}}{\sqrt{\cos\text{x}} +\sqrt{\sin\text{x}}}\text{dx}$
Adding we get, $2 \text{I} = \int\limits_{\pi/6}^{\pi/3}\text{dx} = \big[\text{x}\big]_{\pi/6}^{\pi/3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$
$\therefore\text{I} = \frac{\pi}{12}.$
View full question & answer→Question 625 Marks
Evaluate: $\int\frac{\text{x}^{2}}{(\text{x}^{2} + 4)(\text{x}^{2} + 9 )}\text{dx}.$
AnswerLet $\text{I} = \int\frac{\text{x}^{2}}{(\text{x}^{2} + 4 )|(\text{x}^{2} + 9 )}\text{dx}$
Let $x^2 = t$
$\therefore\frac{\text{x}^{2}}{(\text{x}^{2} + 4 )(\text{x}^{2} + 9 )} =\frac{\text{t}}{(\text{t} + 4)(\text{t} + 9 )}$
Now $\frac{\text{t}}{(\text{t} + 4 )(\text{t} + 9 )} =\frac{\text{A}}{\text{t} + 4} + \frac{\text{B}}{\text{t} + 9 } = \frac{\text{A}(\text{t} + 9) + \text{B}(\text{t} + 4)}{(\text{t} + 4 )(\text{t} + 9 )}$
$\Rightarrow t =(A+ B) t +(9A+ 4B)$
Equating we get
$A + B=1, 9A + 4B= 0$
Solving above two equations, we get
$\text{A} = - \frac{4}{5},\text{B} = \frac{9}{5}$
$\therefore\frac{\text{x}^{2}}{(\text{x}^{2} + 4 )(\text{x}^{2} + 9 )} = -\frac{4}{5(\text{x}^{2} + 4 )} + \frac{9}{5(\text{x}^{2} + 9)}$
$\text{I} = -\frac{4}{5}\int\frac{\text{dx}}{\text{x}^{2} +2^{2}} +\frac{9}{5}\int\frac{\text{dx}}{\text{x}^{2} + 3^{2}}$
$ =-\frac{4}{5}\times\frac{1}{2}\tan^{-1} \frac{\text{x}}{2}+ \frac{9}{5}\times\frac{1}{3}\tan^{-1}\frac{\text{x}}{3} + \text{C}$
$ = - \frac{2}{5}\tan^{-1}\frac{\text{x}}{2} +\frac{3}{5}\tan^{-1}\frac{\text{x}}{3} + \text{C}.$
View full question & answer→Question 635 Marks
Evaluate: $\int\limits_{0}^{4}(|\text{x}|+|\text{x} - 2| + |\text{x} - 4 |)\text{dx}.$
AnswerLet $\text{I} =\int_{0}^{4}(\text{x}| + |\text{x} - 2 | + |\text{x} - 4 |)\text{dx|}$
$ = \int_{0}^{4}|\text{x}|\text{dx} + \int_{0}^{4}|\text{x} - 2|\text{dx} + \int_{0}^{4}|\text{x} - 4|\text{dx}$
$ =\int_{0}^{4}|\text{x}|\text{dx} + +\bigg[\int_{0}^{2}|\text{x} - 2|\text{dx} + \int_{2}^{4}|\text{x} - 2|\text{dx}\bigg] + \int_{0}^{4}|\text{x} - 4 |\text{dx}$ [By properties]
$ =\int_{0}^{4}\text{x dx} + \int_{0}^{2} - (\text{x} - 2 ) \text{dx} + \int_{2}^{4}(\text{x} - 2 )\text{dx} + \int_{0}^{4} - (\text{x} - 4)\text{dx}$
$ \begin{bmatrix}\because|\text{x}| =\text{x},\text{if }0\leq\text{x}\leq4 \$0.3em] |\text{x - 2}| = -(\text{x} - 2),\text{if }0\leq\text{x}\leq2 \$0.3em] |\text{x} - 2 | = (\text{x} - 2 ),\text{ if }2\leq\text{x}\leq4 & \\|\text{x} - 4| = -(\text{x} - 4),\text{ if }0\leq\text{x} \leq4 \end{bmatrix}$
$ =\bigg[\frac{\text{x}^{2}}{2}\bigg]^{4}_{0} - \bigg[\frac{(\text{x} - 2)^{2}}{2}\bigg]_{0}^{2} + \bigg[\frac{(\text{x} - 2 )^{2}}{2}\bigg]_{2}^{4} - \bigg[\frac{(\text{x} - 4)^{2}}{2}\bigg]_{0}^{4}$
$ = \frac{1}{2}\times16-\frac{1}{2}\times(0 - 4) + \frac{1}{2}(4- 0)-\frac{1}{2}\times(0- 16 )$
= 8 + 2 + 2 + 8 = 20.
View full question & answer→Question 645 Marks
Evaluate: $\int\frac{\sin(\text{x} - \text{a})}{\sin(\text{x + a})}\text{ dx}.$
AnswerLet $\text{I} =\int\frac{\sin(\text{x - a )}}{\sin\text{(x +a )}}\text{dx}$
Let x + a =t $\Rightarrow$x =t – a
$\Rightarrow$dx = dt
$\therefore\text{I} = \int\frac{\sin(\text{t} - 2\text{a})}{\sin\text{t}}\text{dt}$
$ =\int\frac{\sin\text{t}.\cos2\text{a} - \cos\text{t}.\sin2\text{a}}{\sin\text{t}}\text{dt}$
$= \cos 2\text{a} \int \text{dt} – \int \sin 2\text{a}. \cot \text{t dt} = \cos 2\text{a.t} – \sin 2\text{a}. \log|\sin \text{t}|+ \text{C}$
$= \cos 2\text{a}.\text{(x + a)} – \sin 2\text{a}. \log|\sin \text{(x + a)|+ C}$
$= \text{x} \cos 2\text{a + a} \cos 2\text{a} – (\sin 2\text{a)} \log|\sin \text{(x + a)|+ C}.$
View full question & answer→Question 655 Marks
$\int\limits_0^{\pi/4}\Bigg(\sqrt{\text{tan x}}+\sqrt{{\text{cot x}}}\Bigg)\text{ dx}=\sqrt{2}\cdot\frac{\pi}{2}$
Answer$\int\limits_0^{\pi/4}\Bigg(\sqrt{\text{tan x}}+\sqrt{{\text{cot x}}}\Bigg)\text{ dx}$ = $\int\limits_0^{\pi/4}\frac{\text{sin x + cos x}}{\sqrt{\text{sin x cos x }}}\text{dx}$Putting sin x – cos x = t, to get (cos x + sin x) dx = dt
and sin x cos x = $\frac{\text{1 - t}^{2}}{2}$
$\therefore 1=\sqrt{2}\int\limits_{-1}^{0}\frac{\text{dt}}{\sqrt{\text{1 - t}^{2}}}$= $\sqrt{2}\cdot[\sin^{-1}\text{t}]^{0}_{-1}$
$=\sqrt{2}(\sin^{-1}\text{0}-\sin^{-1}(-1)=\sqrt{2}\cdot\frac{\pi}{2}$.
View full question & answer→Question 665 Marks
Evaluate $\int\limits_1^3(\text{2x}^{2}+\text{5x})$ dx as a limit of a sum.
Answer$\int\limits_1^3(\text{2x}^{2}+\text{5x})\text{dx}=\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\text{h}[\text{f(1) + f(1 + h) + f(1 + 2 h)+..........+ f(1 }+\overline{\text{n - 1}}\text{ h)}]$
where $f(x) = 2x^2 + 5x$ and $h = \frac{2}{\text{n}}$ or nh $2$
$f(1) = 7$
$f(1 + h) = 2 (1 + h)^2 + 5 (1 + h) = 7 + 9h + 2h^2$
$f(1 + 2h) = 2 (1 + 2h)^2 + 5 (1 + 2h) = 7 + 18h + 2.2^2h^2$
$f(1 + 3h) = 2 (1 + 3h)^2 + 5 (1 + 3h) = 7 + 27h + 2.3^2h^2$
$f(1 + (n – 1) h) = 7 + 9 (n – 1) h + 2.(n – 1)^2 h^{2.}$
$\text{I}=\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\Bigg[\text{h}[\text{7n + 9h}\frac{\text{n(n-1)}}{{2}}+\text{2h}^{2}\cdot\frac{\text{n(n-1)(2n-1)}}{6}\Bigg]$
$=\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\Bigg[\text{7nh}+\frac{9}{2}\text{nh (nh - h)}+\frac{1}{3}\text{nh (nh - h)(2nh - h)}\Bigg]$
$= 14+18+\frac{16}{3}=\frac{112}{3}$.
View full question & answer→Question 675 Marks
Evaluate: $\int\frac{2}{\text{(1-x)(1+x)}^{2}}\text{dx}$
Answer$\frac{2}{\text{(1-x)}\text{(1+x}^{2})}=\frac{\text{A}}{\text{1-x}}+\frac{\text{Bx+C}}{\text{1+x}^{2}}$
$2 = A(1 + x^2) + (Bx + C) (1 – x)$
$\Rightarrow 0 = A – B, B – C = 0, A + C = 2$
$\Rightarrow A = B = C = 1$
$\therefore\int\frac{2}{\text{(1-x)(1+x}^{2})}\text{dx}=\int\frac{1}{\text{1-x}}\text{dx}+\frac{\text{x+1}}{\text{x}^{2}+{1}}\text{dx}$
$= – \log |1 – x| + \frac{1}{2} \log (x^2 + 1) + \tan^{-1} x + c.$
View full question & answer→Question 685 Marks
Evaluate: $\int$sin x sin 2x sin 3x dx.
Answer$\text{I}=\int\text{sin x sin 2x sin 3x dx}=\frac{1}{2}\int\text{2sin 3x sin x sin 2x dx}$$=\frac{1}{2}\int\text{(cos 2x - cos 4x) sin 2x dx}=\frac{1}{2}\int\text{(sin 2x cos 2x -cos 4x sin 2x)dx}$
$=\frac{1}{4}\int\text{sin 4x dx}-\frac{1}{4}\int\text{2 cos 4x sin 2x dx}$
$=-\frac{1}{16}\int\text{cos 4x }-\frac{1}{4}\int\text{(sin 6x - sin 2x)dx}$
$=-\frac{1}{16}\text{cos 4x}+\frac{1}{24}\text{cos 6x}-\frac{1}{8}\text{cos 2x + c}$
View full question & answer→Question 695 Marks
Evaluate: $\int\limits_0^{\pi/2} 2 \sin x \cos x \tan^{-1 }(\sin x) dx.$
Answer$\text{I}=\int\limits_{0}^{\pi/2} 2 \sin x \cos x \tan-1 (\sin x) dx = \int\limits_0^1\tan^{-1}\text{t}\cdot\text{(2t) dt}$
where $\sin x = t =\Bigg[\tan^{-1} t\times\text{t}^{2}\Bigg]_0^1-\int\limits_0^{1}\frac{{t}^{2}}{\text{t}^{2}+1}\text{dt}$
$=\frac{\pi}{4}-\int_0^1\Bigg(1-\frac{1}{\text{t}^{1}+1}\Bigg)\text{dt}$
$=\frac{\pi}{4}-[\text{t - tan}^{-1}\text{t}]^1_0$
$=\frac{\pi}{4}-\Bigg[1-\frac{\pi}{4}\Bigg]$
$=\Bigg(\frac{\pi}{2}-1\Bigg)$.
View full question & answer→Question 705 Marks
Evaluate: $\int\frac{\text{2x}}{\text{(x}^{2}+1)\text{(x}^{2}+3)}\text{dx}$.
Answer$\text{I}=\int\frac{\text{2x}}{\text{(x}^{2}+1){\text{(x}^{2}+3)}}\text{dx}=\int\frac{\text{dt}}{\text{(t + 1)(t + 3)}}\text{where t = x}^{2}$
$=\frac{1}{2}\int\Bigg[\frac{1}{\text{(t + 1)}}-\frac{1}{{\text{(t + 3)}}}\Bigg]\text{dt}$
$=\frac{1}{2}[\log|\text{t + 1}|-\log|\text{t + 3}|]+\text{c}$
$=\frac{1}{2}[\log\text{(x}^{2}+1)-\log\text{(x}^{2}+3)]+\text{c}$.
View full question & answer→Question 715 Marks
Evaluate: $\int\frac{\text{5x + 3}}{\sqrt{\text{x}^{2}+\text{4x + 10}}}$.
Answer$\text{I}=\int\frac{\text{5x + 3}}{\sqrt{\text{x}^{2}+\text{4x + 10}}}\text{dx}=\int\frac{\frac{5}{2}\text{(2x + 4)}-7}{\sqrt{\text{x}^{2}+\text{4x}+10}}$$=\frac{5}{2} \int\frac{\text{2x + 4}}{\sqrt{\text{x}^{2}+\text{4x + 10}}}\text{dx}-7 \int\frac{1}{\sqrt{\text{(x+2)}^{2}+(\sqrt{6})^{2}}}\text{dx}$
$ =5\cdot\sqrt{\text{x}^{2}+\text{4x + 10}}-7\log|\text{(x + 2)}+\sqrt{\text{x}^{2}+\text{4x}+10}|+\text{c}$.
View full question & answer→Question 725 Marks
Evaluate: $\int\limits_0^{\pi/2}\frac{\text{x sin x cos x}}{\text{sin}^{4}\text{x + cos}^{4}\text{x}}\text{dx}$.
Answer$\text{I}=\int\limits_{0}^{\pi/2}\frac{\text{x sin x cos x}}{\text{sin}^{4}\text{x}+\text{cos}^{4}\text{x}}\text{dx}\ .....\text{(i)}$
$\text{I}=\int\limits_{0}^{\frac{\pi}{2}}{\pi}\frac{(\frac{\pi}{2}-x)\sin(\frac{\pi}{2}-x)\cos(\frac{\pi}{2}-x)}{\sin^{4}(\frac{\pi}{2}-x)+\cos^{4}(\frac{\pi}{2}-x)}\text{dx}=\int\limits_0^{\frac{\pi}{2}}\frac{(\frac{\pi}{2}-x)\text{cos x sin x dx}}{{\sin x}^4+\cos x^{4}}\ ......\text{(ii)}$
$2I = \text{I}=\frac{\pi}{2}\int\limits_{0}^{\frac{\pi}{2}}\frac{\text{sin x cos x}}{\text{sin}^{4}+\text{cos}^{4}\text{x}}\text{dx}=\frac{\pi}{2}\int\limits_0^{\frac{\pi}{2}}\frac{\text{tan x}\cdot\text{sec}^{2}\text{x dx}}{\text{1+tan}^{4}\text{x}}$
$\text{I}=\frac{\pi}{4}\frac{1}{2}\int\limits_0^{\infty}\frac{\text{dt}}{\text{1+t}^{2}}$ where $t = \tan^2 x.$
$=\frac{\pi}{8}[\tan ^{-1}\text{t}]_{0}^\infty=\frac{\pi}{8}\cdot\frac{\pi}{2}=\frac{\pi^{2}}{16}$.
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Evaluate: $ \int\limits_0^\pi\frac{\text{x tan x}}{\text{sec x + tan x}}\text{dx}$.
Answer$=\pi^{2}-2\pi$
$\text{I } = \int\limits^\pi_{0}\frac{\text{x tan x dx }}{\text{sec x + tan x }}\Rightarrow\text{I }=\int\limits_{0}^{\pi}\frac{(\pi-\text{x) tan x}}{\text{sec x + tan x}}\text{dx}$
$\text{2I } = \int\limits^\pi_{0}\frac{\pi\text{ tan x }}{\text{sec x + tan x }}=\pi\int\limits_{0}^{\pi}\frac{\text{sin x}}{\text{1+sin x}}\text{dx}$
$=\pi\int\limits_0^{\pi}\Bigg(1-\frac{1}{\text{1+sin x}}\Bigg)\text{dx}=\pi\int\limits_0^{\pi}\Bigg(1-\frac{\text{1-sin x}}{\text{cos}^{2}\text{x}}\text{dx}\Bigg)$
$=\pi\int\limits_0^\pi(\text{1-sec}^{2}\text{x}+\text{sec x tan x )}\text{dx}$
$=\pi[\text{x - tan x + sec x }]^{\pi}_{0}=\pi[\pi-1-1]$
$\therefore\text{I}=\frac{{\pi}^{2}}{2}-\pi=\frac{\pi}{2}[\pi-2]$.
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Evaluate: $\int\text{e}^{x}\Bigg(\frac{\sin 4x - 4}{1-\cos 4x}\Bigg)\text{dx}$.
Answer$\text{I}=\int\text{e}^{x}\Bigg(\frac{\sin 4x - 4}{1-\cos 4x}\Bigg)\text{dx}$
$=\int\text{e}^{x}\Bigg[\frac{\sin 4x}{1-\cos 4x}-\frac{4}{1 - \cos 4x}\Bigg]\text{dx}$
$=\int\text{e}^{x}\Bigg[\frac{2\sin 2x \cos 2x}{2\cdot\sin^{2} 2x}-\frac{4}{2\sin^{2}2x}\Bigg]\text{dx}$
$=\int\text{e}^{x}[\cot2x-2\text{cosec}^{2}2x]\text{dx}$
This is of the form $=\int\text{e}^{x}[\text{f(x)+f'(x)}]\text{dx}$
$\therefore \text{I}=\text{e}^{x}\cot\text{2x + c}.$
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Evaluate:$\int\limits_{\pi/6}^{\pi/3}\frac{\sin\text{x}+\cos{\text{x}}}{\sqrt{\sin\text{2x}}}\text{dx}$ .
AnswerLet $\sin x – \cos x = t $
$\Rightarrow (\cos x + \sin x) dx = dt,$
Also, $ \sin 2x = 1 – t^2$
When $\text{x}\frac{\pi}{3},\text{t}=\frac{\sqrt{3}-1}{2},\text{when x}=\frac{\pi}{6},\text{t}=\frac{1-\sqrt{3}}{2}$
$\therefore$ Given integral becomes $I = \int^{\frac{\sqrt{3}-1}{2}}_{\frac{1-\sqrt{3}}{2}}\frac{\text{dt}}{\sqrt{1-\text{t}^{2}}}$
$[\sin^{-1}\text{t}]^{\frac{\sqrt{3}-1}{2}}_{\frac{1-\sqrt{3}}{2}}=\sin^{-1}\bigg(\frac{\sqrt{3}-1}{2}\bigg)-\sin^{-1}\bigg(\frac{1-\sqrt{3}}{2}\bigg)$
OR $2\sin^{-1}\Bigg(\frac{\sqrt{3}-1}{2}\Bigg).$
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Evaluate:$\int\frac{1 - x^{2}}{x(1 - 2x)}\text{dx}$.
Answer$\text{I}=\int\frac{1 - x^{2}}{x(1-2x)}\text{dx}=\frac{1}{2}\int\frac{2-2x^{2}}{2-2x^{2}}\text{dx}$
$=\frac{1}{2}\int\Bigg[1+\frac{2-x}{x(1-2x)}\Bigg]\text{dx}$
$=\frac{\text{x}}{2} +\frac{1}{2}\int\frac{\text{2-x}}{\text{x(1-2x)}}\text{dx}$
Let $\frac{\text{2 - x}}{\text{x(1 - 2x)}}=\frac{\text{A}}{\text{x}}+\frac{\text{B}}{\text{1 - 2x}}\cdot\text{Getting A = 2, B = 3}$
$\therefore\text{I}=\frac{\text{x}}{2}+\frac{1}{2}\int\Bigg(\frac{2}{\text{x}}+\frac{\text{3}}{\text{1 - 2x}}\Bigg)\text{dx}$
$=\frac{\text{x}}{2}+\log|\text{x}|-\frac{3}{4}\log|1-2\text{x}|+\text{c}.$
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$\text{Evaluate:} \int \frac{e^{x}}{\sqrt{5 - 4c^{X} - e^{2_{x}}}} \text{dx}$
Answer$\text{Getting I} = \int\frac{dt}{\sqrt{5 - 4t - t^{2}}} \text{where t = e}^{x}$$= \int\frac{dt}{\sqrt{(3)^{2}} - ( t + 2)^{2}}$
$= \sin^{-1} \frac{t + 2}{3} + c$
$= \sin^{-1} \frac{e^{x} + 2}{3} + c$
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$\text{Evaluate:} \int\limits_0^\frac{\pi}{2} (2\log \sin \text{x} - \log \sin 2\text{x}) \text{dx}$
Answer$\text{I} = \int\limits_0^\frac{\pi}{2} (2\log \sin \text{x} - \log \sin 2\text{x}) \text{dx} = \int\limits_0^\frac{\pi}{2}\log\bigg(\frac{\sin^{2}{x}}{2\sin x \cos x}\bigg)\text{dx}$
$= \int\limits_0^\frac{\pi}{2} \log \tan \text{x dx} - \int\limits_0^\frac{\pi}{2} \log 2 .\text{dx} = \text{I}_{1} - \text{I}_{2}$
$\text{I}_{1} = \int\limits_0^\frac{\pi}{2} \log \tan \text{x dx} = \int\limits_0^\frac{\pi}{2} \log \tan\bigg(\frac{\pi}{2} - \text{x}\bigg) \text{dx} = \int\limits_0^\frac{\pi}{2} \log \cot\text{x dx}$
$\Rightarrow \text{2 I}_{1} = 0 \Rightarrow \text{I}_{1} = 0$
$\text{I}_{2} = \log 2. \bigg[\text{x}\bigg]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} . \log 2$
$\therefore \text{I} = - \frac{\pi}{2} . \log 2$
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$\int\limits_0^{\pi} \frac{e^{\cos x}}{e^{\cos{x}}+ e^{-\cos x}} \text{dx}$
Answer$\text{I} = \int\limits_0^{\pi} \frac{e^{\cos x}}{e^{\cos{x}}+ e^{-\cos x}} \text{dx} = \int\limits_0^{\pi} \frac{e^{\cos}(\pi - x)}{e^{\cos{(\pi -x)}}+ e^{-\cos (\pi-x})} \text{dx}$
$= \int\limits_0^{\pi} \frac{e^{-\cos x}}{e^{-\cos{x}}+ e^{\cos x}}\text{dx}$
$\text{2I} = \int\limits_0^{\pi} 1. \text{dx} = \bigg[\text{x}\bigg]^{\pi}_{0}$
$= \pi$
$\text{I} = \frac{\pi}{2}$
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$\text{Evaluate:} \int\limits_{-a}^{a} \sqrt\frac{{a - x}}{a + x} {dx}$
Answer$\text{I} = \int\limits_{-a}^{a} \sqrt\frac{{a - x}}{a + x} {dx} = \int\limits_{-a}^{a}\frac{a - x}{\sqrt{a^{2} - x^{2}}} dx = \int\limits_{-a}^{a} \frac{x dx}{\sqrt{a^{2} - x^{2}}}$$I_{1}$ is even function and $I_{2}$ is odd function
$\therefore I_{2} = 0$
$i = 2a \int\limits_{-a}^{a} \frac{dx}{\sqrt{a^{2} - x^{2}}} = 2a. \frac{\pi}{2} = \pi\text{a}$
$\therefore I = \pi\text{a}$
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$\text{Evaluate :} \int\limits_0^{\pi} \frac{x \sin x}{1 + \cos^{2} x} dx$
Answer$\text{I} = \int\limits_0^{\pi} \frac{x \sin x}{1 + \cos^{2} x} \text{dx}$$= \int\limits_0^{\pi} \frac{(\pi - x) \sin (\pi - x)}{1 + \cos^{2} (\pi - x)} \text{dx} = \int\limits_0^{\pi} \frac{(\pi - x) \sin x}{1 + \cos^{2} x} \text{dx}$
$\Rightarrow \text{2I} = \int\limits_0^{\pi} \frac{\sin x}{1 + \cos^{2} x} \text{dx}$
$\text{I} = \frac{\pi}{2} . 2 \int\limits_0^{\pi} \frac{\sin x dx}{1 + \cos^{2} x} = -\pi \bigg[ \tan^{-1} (\cos x)\bigg]^\frac{\pi}{2}_{0}$
$= -\pi \bigg[ -\frac{\pi}{4}\bigg] = \frac{\pi^{2}}{4}$
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Evaluate: $\int \frac{\sin x}{(1 - \cos x) ( 2 - cos x)} \text{dx}$
Answer$\text{I} =\int \frac{\sin x}{(1 - \cos x) ( 2 - cos x)} \text{dx}$$\text{Let} - \cos \text{x} = \text{t} \Rightarrow \sin \text{x dx} = \text{dt}$
$\therefore \text{I}= \int \frac{dt}{(1+ t) ( 2 + t)} = \int\frac{dt}{1 + t} - \int \frac{dt}{2 + t}$
$= \log | 1 + t | - \log| 2 + t| + c$
$= \log | 1 - \cos x | - \log | 2 - \cos x | + c$
OR
$\log \bigg|\frac{1 - cos x}{2 - cos x}\bigg| + c$
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$\text{Evaluate} \int\limits_0^{2} (x^{2} + 2x + 1) \text{dx as limit of sums.}$
Answer$\text{Here h} = \frac{2 - 0}{n} =\frac{2}{n} , \text{f (x)} = x^{2} + 2x + 1$$\therefore \text{I} = \lim\limits_{ h\rightarrow 0} \bigg[ \text{f (0)} + \text{f (h)} + \text{f (2h)} +\dots\dots\dots\dots\text{+} \text{f} \overline{(\text{n} - 1)}\text{h}\bigg]$
$= \lim\limits_{ h\rightarrow 0} \text{h} \bigg[1 + (h^{2} + 2h + 1) + (2^{2}h^{2} + 2.2h + 1) + \dots\dots\dots(n- 1^{2}) h^{2} + 2 (n - 1) h + 1)\bigg]$
$= \lim\limits^{n \rightarrow \infty}_{h \rightarrow 0} \frac{2}{n} \Big[(1+1+1+1+1+\ ........)+\text{h}^2\big(1+2^2+3^2+\ ....\big)\Big]$
$= \lim\limits_{n \rightarrow \infty} \frac{2}{n} \bigg[n +\frac{4}{n^{2}} \frac{(n - 1)n(2n -1)}{6} + \frac{4}{n} .\frac{(n - 1)n}{n}\bigg]$
$= \lim\limits_{n \rightarrow \infty} \Big[2 + \frac{8}{6} \frac{n - 1}{n} \frac{2n - 1}{n} + 4\frac{n - 1}{n}\Big]$
$= 2 + 4 +\frac{8}{3} = \frac{26}{3}$
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Using properties of definite integrals, prove the following:$\int\limits_0^{\pi} \frac{x \tan x}{\sec x\text{ }cosec\text{ x}} dx = \frac{\pi^{2}}{4}$
Answer$\text{I} = \int\limits_0^{\pi} \text{x sin}^{2} \text{x dx} = \int\limits_0^{\pi} (\pi - x) . \sin^{2} ( \pi - x) \text{dx}$$\therefore \text{I} = \int\limits_0^{\pi} \pi \sin^{2} \text{x dx} - \int\limits_0^{\pi} x \sin^{2} \text{x dx}$
$\text{2 I} =\pi \int\limits_0^{\pi} \sin^{2} \text{x dx} = \frac{\pi}{2} = \int\limits_0^{\pi}(1 - \cos 2x) \text{dx}$
$= \frac{\pi}{2} \bigg[ x- \frac{\sin 2x}{2} \bigg]^{\pi}_{0}$
$= \frac{\pi}{2} [\pi] = \frac{\pi^2}{2}$
$\Rightarrow \text{I} = \frac{\pi^{2}}{4}$
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$\text{Evaluate} \int\limits_0^{2} (x^{2} +x + a) \text{dx as limit of a sum.}$
AnswerWe have to find $\text{I} = \int\limits_0^{2} (x^{2} +x + a) \text{dx as limit of a sum.}$$\text{Here h} = \frac{2 - 0 }{n}= \frac{2}{n}$
$\lim\limits_{X\rightarrow \infty} \text{h} [f(0 +(0 +h) +f(0 +2h) +{\dots}\text{ } { \dots}+ f(0 +\overline{n-1}\text{ h})]$
$= \lim\limits_{X\rightarrow \infty} \frac{2}{n}[1 + (h^{2} +h +1 )+ (4h^{2} + 2h +1) + {\dots\dots}+ (n-1) ^{2}h^{2} + (n-1)h + 1]$
$\lim\limits_{X\rightarrow \infty} \frac{2}{n} \bigg[n +h^{2} \frac{(n-1)n(2n-1)}{6} +\frac{(n-1)n}{2}\bigg]$
$\lim\limits_{X\rightarrow \infty} \frac{2}{n} \bigg[n\frac{4}{6n^{2}}(n-1)(n)(2n-1) +\frac{(n-1)n}{n}\bigg]$
$= 2\bigg[1 + \frac{4}{3} +1\bigg] = \frac{20}{3} $
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Evaluate:$\int\limits_0^\frac{\pi}{4} \sin 2x \sin 3\text{x dx}$
Answer$\frac{1}{2}\int\limits_0^\frac{\pi}{4} 2\sin 2x \sin 3\text{x dx}$$= \frac{1}{2} \int\limits_0^\frac{\pi}{4} (\cos x - \cos 5\text{x)dx}$
$= \frac{1}{2} \bigg [\sin x - \frac{\sin 5x}{5}\bigg]_{0}^{\frac{\pi}{4}}$
$= \frac{1}{2} \bigg[\sqrt\frac{1}{2} + \frac{1}{5} \sqrt\frac{1}{2}\bigg] = \frac{1}{2} \bigg[\frac{5 + 1}{5\sqrt{2}}\bigg] = \frac{3}{5\sqrt{2}} $
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Evaluate:$\int\limits_0^\frac{\pi}{4} \log (1 + \tan\text{ x)dx}$
Answer$\int\limits_0^\frac{\pi}{4} \log (1 + \tan\text{ x)dx}$$\text{Using} $ $\int\limits_0^{a} \text f (x) {dx} = \int\limits_0^{a} f (a - x) \text{dx, we get}$
$\text {I} = \int\limits_0^\frac{\pi}{4} \log \bigg[1 + tan \bigg(\frac{x}{4} - {x}\bigg)\bigg] \text{dx}$
$\int\limits_0^\frac{\pi}{4} \log \bigg[1 + \frac{1 - \tan x}{1 + tan x}\bigg] \text{dx} = \int\limits_0^\frac{\pi}{4}[\log 2 - \log(1 + tan x)] \text{dx}$
$\therefore 2\text{I} = \bigg[\int\limits_0^\frac{\pi}{4} \log 2.\text{dx}\bigg] \Rightarrow \text{I} = \frac{\pi}{8} \log 2$
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Evaluate:$\int\frac{3x + 1}{2x^{2} -2x + 3} dx$
Answer$\int\frac{3x + 1}{2x^{2} -2x + 3} = \int \frac{\frac{3}{4}\big(4x -2\big)+\frac{5}{2}}{2x^{2}-2x + 3 }\text{dx}$$= \frac{3}{4}\int\frac{(4x - 2) dx}{2x^{2} - 2x + 3} + \frac{5}{4} \int \frac{dx}{x^{2} - x +\frac{3}{2}}$
$\frac{3}{4} \log | 2x^{2} - 2x + 3| + \frac{5}{4} \int \frac{dx} {\bigg(x - \frac{1}{2}\bigg)^{2} + \bigg( \frac{\sqrt{5}}{2}\bigg)^{2}} +c$
$= \frac{3}{4} \log | 2x^{2} - 2x + 3| \frac{\sqrt{5}}{2}\tan ^{-1} \frac{2x - 1}{\sqrt{5}}$
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Evaluate: $\int\limits_1^3(\text{x}^2+3\text{x}+\text{e}^\text{x})\text{dx},$ as the limit of the sum.
Answer$\int\limits_1^3(\text{x}^2+3\text{x}+\text{e}^\text{x})\text{dx}$
$\lim\limits_{\text{h} \rightarrow 0}\ \text{h}\big[\text{f}(1)+\text{f}(1+\text{h})+\text{f}(1+2\text{h})+\ ....\ +\text{f}(1+2(\text{n}-1)\text{h})\big]$
$\lim\limits_{\text{h}\rightarrow0}\ \text{h}\big[\big(1+3+\text{e})+((1+\text{h})^2+3(1+\text{h})+\text{e}^{1+\text{h}}\big)\\+\big((1+2\text{h})^2+3(1+2\text{h})+\text{e}^{1+2\text{h}}\big)+\ ...\big]$
$\lim\limits_{\text{h}\rightarrow0}\ \text{h}\big[4+\text{e}+\big(1+\text{h}^2+2\text{h}+3+3\text{h}+\text{e}^{1+\text{h}}\big)\\+\big(1+4\text{h}^2+4\text{h}+3+6\text{h}+\text{e}^{1+2\text{h}}\big)+\ ...\big]$
$\lim\limits_{\text{h}\rightarrow0}\ \text{h}\big[4+\text{e}+\big(4+\text{h}^2+5\text{h}+\text{e}^{1+\text{h}}\big)\\+\big(4+4\text{h}^2+10\text{h}+\text{e}^{1+2\text{h}}\big)+\ ...\big]$
$\lim\limits_{\text{h}\rightarrow0}\ \text{h}\big[4\text{n}+\text{e}\big(1+\text{e}^\text{h}+\text{e}^{2\text{h}}+\ ...\big)\\+\text{h}^2\big[1^2+2^2+ ...\big]+5\text{h}(1+2+\ ...)\big]$
$\lim\limits_{\text{h}\rightarrow0}\ \text{h}\Bigg[4\text{n}+\text{e}\bigg(\frac{1(\text{e}^\text{hn}-1)}{\text{e}^\text{h}-1}\bigg)+\text{h}^2\bigg(\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}\bigg)+5\text{h}\bigg(\frac{\text{n}(\text{n}-1)}{2}\bigg)\Bigg]$
$\lim\limits_{\text{h}\rightarrow0}\ \text{h}\Bigg[4\text{n}+\text{e}\bigg(\frac{\text{e}^\text{nh}-1}{\text{e}^\text{h}-1}\bigg)+\frac{\text{h}^2\text{n}^3}{6}\bigg(1-\frac{1}{\text{n}}\bigg)\bigg(2-\frac{1}{\text{n}}\bigg)+\frac{5\text{hn}^2}{2}\bigg(1-\frac{1}{\text{n}}\bigg)\Bigg]$
$\lim\limits_{\text{n}\rightarrow\infty}\ \frac{2}{\text{n}}\Bigg[4\text{n }+\text{e}\Bigg(\frac{\text{e}^{\frac{\text{n}\times2}{\text{n}}}}{\text{e}^\frac{2}{\text{n}}-1}\Bigg)+\frac{4}{\text{n}^2}\frac{\text{n}^3}{6}\bigg(1-\frac{1}{\text{n}}\bigg)\bigg(2-\frac{1}{\text{n}}\bigg)+\frac{5}{2}\times\text{n}^2\times\frac{2}{\text{n}}\bigg(1-\frac{1}{\text{n}}\bigg)\Bigg]$
$\lim\limits_{\text{n}\rightarrow\infty}\ \frac{2}{\text{n}}\Bigg[4\text{n}+\text{e}\bigg(\frac{\text{e}^2-1}{\text{e}^\frac{2}{\text{n}}-1}\bigg)+\frac{4\text{n}}{6}\bigg(1-\frac{1}{\text{n}}\bigg)\bigg(2-\frac{1}{\text{n}}\bigg)+5\text{n}\bigg(1-\frac{1}{\text{n}}\bigg)\Bigg]$
$\lim\limits_{\text{n}\rightarrow\infty}\ 2\Bigg[4+\frac{\text{e}}{\text{n}}\frac{(\text{e}^2-1)}{\text{e}^\frac{2}{\text{n}}-1}\bigg)+\frac{2}{3}\bigg(1-\frac{1}{\text{n}}\bigg)\bigg(2-\frac{1}{\text{n}}\bigg)+5\bigg(1-\frac{1}{\text{n}}\bigg)\Bigg]$
$=8+\text{e}(\text{e}^2-1)+\frac{4}{3}+5$
$=\frac{24+4+15}{3}+\text{e}(\text{e}^2-1)$
$=\frac{43}{3}+\text{e}(\text{e}^2-1)$
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Evaluate: $\int\limits_0^{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{16+9\sin2\text{x}}$
Answer$\int\limits_0^\frac{\pi}{4}\frac{\sin\text{x}+\cos\text{x}}{16+9\sin2\text{x}}\text{dx}$
$\int\limits_0^{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{16+9\big[1-(\sin\text{x}-\cos\text{x})^2\big]}\text{dx}$
$\int\limits_0^{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{25-9(\sin\text{x}-\cos\text{x})^2}\text{dx}$
$\sin\text{x}=\cos\text{x}=\text{t}$
$(\cos\text{x}+\sin\text{x})\text{dx}=\text{dt}$
$\int\limits_{-1}^0\frac{\text{dx}}{25-9\text{t}^2}$
$\int\limits_{-1}^0\frac{1}{9\Big(\frac{25}{9}-\text{t}^2\Big)}\text{dt}$
$\frac{1}{9}\int\limits_{-1}^0\frac{1}{\Big(\frac{5}{3}\Big)^2-\text{t}^2}$
$\frac{1}{9}\Bigg[\frac{1}{2\times\frac{5}{3}}\log\begin{vmatrix}\frac{\frac{5}{3}+\text{t}}{\frac{5}{3}-\text{t}}\end{vmatrix}\Bigg]_{-1}^0$
$\frac{1}{9}\times\frac{1}{\frac{10}{3}}\Bigg(\log\begin{vmatrix}\frac{\frac{5}{3}}{\frac{5}{3}}\end{vmatrix}\Bigg)-\log\begin{vmatrix}\frac{\frac{2}{3}}{\frac{8}{3}} \end{vmatrix}$
$\frac{1}{30}\Big[\log1-\log\frac{1}{4}\Big]$
$\frac{1}{30}\log4=\frac{1}{15}\log2$
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Find: $\int\frac{2\cos\text{x}}{(1-\sin\text{x})(1+\sin^2\text{x})}\text{dx}$
Answer$\int\frac{2\cos\text{x}}{(1-\sin\text{x})(1+\sin^2\text{x})}\text{dx}$
$\text{put}\sin\text{x}=\text{t}$
$\cos\text{x dx}=\text{dt}$
$\int\frac{2\text{dt}}{(1-\text{t})(1+\text{t}^2)}$
$\frac{2}{(1-\text{t})(1+\text{t}^2)}=\frac{\text{A}}{1-\text{t}}+\frac{\text{Bt}+\text{C}}{1+\text{t}^2}$
$2=\text{A}(1+\text{t}^2)+(\text{Bt}+\text{C})(1-\text{t})$
$\text{put}\ 1-\text{t}=0\ \ |\ \ 2=\text{A}(2)\\\text{t}=1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \text{A}=1$
Comparing coefficients of $t^2 \& t$
$t^2 \rightarrow A + (–B) = 0$
$B = A$
$B = 1$
$t \rightarrow B – C = 0$
$B = C = 1$
$\int\Big(\frac{1}{1-\text{t}}+\frac{\text{t}+1}{\text{t}^2+1}\Big)\text{dt}$
$\frac{\log(1-\text{t})}{-1}+\int\frac{\text{t}}{\text{t}^2+1}\text{dt}+\int\frac{1}{\text{t}^2+1}\text{dt}$
$-\log(1-\sin\text{x})+\frac{1}{2}\log(\text{t}^2+1)+\tan^{-1}\text{t}+\text{C}$
$-\log(1-\sin\text{x})+\frac{1}{2}\log(\sin^2\text{x}+1)+\tan^{-1}(\sin\text{x})+\text{C}$
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If $=\text{y}=(\sin^{-1}\text{x})^2,$ prove that $(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}\frac{\text{dy}}{\text{dx}}-2=0.$
AnswerHere $=\text{y}=(\sin^{-1}\text{x})^2$
$\Rightarrow\text{y}'=2(\sin^{-1}\text{x})\times\frac{1}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\text{y}'\sqrt{1-\text{x}^2}=2(\sin^{-1}\text{x})$
$\Rightarrow\text{y}''\sqrt{1-\text{x}^2}+\text{y}'\times\frac{-2\text{x}}{2\sqrt{1-\text{x}^2}}=\frac{2}{\sqrt{1-\text{x}^2}}$
$\Rightarrow(1-\text{x}^2)\text{y}''-\text{xy}'=2$
$\therefore(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}\frac{\text{dy}}{\text{dx}}-2=0.$
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If $\text{x}=\cos\text{t}+\log\tan\Big(\frac{\text{t}}{2}\Big),\ \text{y}=\sin\text{t},$ then find the values of $\frac{\text{d}^2\text{y}}{\text{dt}^2}$ and $\frac{\text{d}^2\text{y}}{\text{dx}^2}$ at $\text{t}=\frac{\pi}{4}.$
AnswerGiven $\text{y}=\sin\text{t}$
Differentiating w.r.t. t
$\frac{\text{dy}}{\text{dt}}=\cos\text{t}\ \dots(\text{i})$
Differentiating again w.r.t. t
$\frac{\text{d}^2\text{y}}{\text{dt}^2}=-\sin\text{t}$
$\frac{\text{d}^2\text{y}}{\text{dt}^2}$ at $\Big(\text{t}=\frac{\pi}{4}\Big)$ $=-\sin\frac{\pi}{4}=-\frac{1}{\sqrt{2}}$
Given $\text{x}=\cos\text{t}+\log\Big(\tan\frac{\text{t}}{2}\Big)$
Differentiating w.r.t. t
$\frac{\text{dx}}{\text{dt}}=-\sin\text{t}+\frac{\frac{1}{2}\sec^2\frac{1}{2}}{\tan\frac{\text{t}}{2}}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=-\sin\text{t}+\frac{1}{\sin\text{t}}\ \dots(\text{ii})$ $\big(\therefore2\sin\frac{\text{t}}{2}.\cos\frac{\text{t}}{2}=\sin\text{t}\big)$
From (i) and (ii)
$\frac{\text{dy}}{\text{dt}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\cos\text{t}}{-\sin\text{t}+\frac{1}{\sin\text{t}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin\text{t}.\cos\text{t}}{1-\sin^2\text{t}}=\frac{\sin\text{t}.\cos\text{t}}{\cos^2\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\tan\text{t}$
Differentiating w.r.t. x
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}(\tan\text{t})$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\sec^2\text{t}.\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\sec^2\text{t}}{\Big(-\sin\text{t}+\frac{1}{\sin\text{t}}\Big)}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\sec^2\text{t}.\sin\text{t}}{\cos^2\text{t}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\sin\text{t}}{\cos^4\text{t}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}$ at $\Big(\text{t}=\frac{\pi}{4}\Big)$ $=\frac{\frac{1}{\sqrt{2}}}{\Big(\frac{1}{\sqrt{2}}\Big)^4}=2\sqrt{2}$
View full question & answer→Question 945 Marks
Prove that $\int\limits_{0}^{\text{a}}\text{f(x)}\text{dx}=\int\limits_{0}^{\text{a}}\text{f}(\text{a}-\text{x})\text{dx},$ hence evaluate $\int\limits_{0}^\pi\frac{\text{x}\sin\text{x}}{1+\cos^2\text{x}}\text{dx}.$
AnswerTo prove: $\int\limits_{0}^{\text{a}}\text{f(x)}\text{dx}=\int\limits_{0}^{\text{a}}\text{f}(\text{a}-\text{x})\text{dx}$
Proof: Let
$\text{Let} \ \text{a}-\text{x}=\text{t}$
$\Rightarrow\text{dx}=-\text{dt}$
When $\text{x}=0,\text{t}=\text{a}$
When $\text{x}=\text{a},\text{t}=0$
Putting the value of x in LHS
$\int\limits_\text{a}^{0}\text{f}(\text{a}-\text{t})(-\text{dt})$
$=-\int\limits_\text{a}^{0}\text{f}(\text{a}-\text{t})(\text{dt})$
$=\int\limits_\text{a}^{0}\text{f}(\text{a}-\text{t})(\text{dt})$
$\Big(\therefore\int\limits_\text{a}^\text{b}\text{f(t)}\text{d}\text{t}=\int\limits_\text{a}^\text{b}\text{f(x)dx}\Big)$
$\text{LHS}=\text{RHS}$
Using this we can solve the given question as follows:
$=\int\limits_0^\pi\frac{\text{x}\sin\text{x}}{1+\cos^2\text{x}}\text{dx}=\text{I}\dots(1)$
On Putting $\text{x}=\pi-\text{x}$
$\int\limits_0^\pi\frac{(\pi-\text{x})\sin(\pi-\text{x})}{1+\cos^2(\pi-\text{x})}=\text{I}\dots(2)$
$\text{I}+\text{I}=\int\limits_{0}^{\pi}\frac{\text{x}\sin\text{x}\text{ dx}}{1+\cos^{2}\text{x}}+\int\limits^{\pi}_0\frac{(\pi-\text{x})\sin(\pi-\text{x})}{1+\cos^2(\pi-\text{x})}$
$\Rightarrow2\text{I}=\int\limits_0^\pi\frac{\text{x}\sin\text{x}}{1+\cos^2\text{x}}\text{dx}+0$
$\Rightarrow2\text{I}=\int\limits_0^\pi\frac{\text{x}\sin\text{x}}{1+\cos^2\text{x}}\text{dx}$
$\text{Let}\cos\text{x}=0$
$-\sin\text{x}\text{ dx}=\text{dv}$
$\Rightarrow2\text{I}=-\text{x}\int\limits^{\pi}_{\text{0}}\frac{\text{dv}}{1+\text{v}^2}$
$\Rightarrow2\text{I}=-\pi\int\limits^{-1}_1\frac{1}{1+\text{v}^2}\text{dv}$ $[\text{Given}\text{ x}=\pi(\text{limit})]$
$\Rightarrow2\text{I}=-\pi[\tan^{-1}\text{v}]_1^{-1}$
$\Rightarrow2\text{I}=-\pi\Big[-\frac{\pi}{4}-\frac{\pi}{4}\Big]=\frac{\pi^{2}}{2}$
$\text{I}=\frac{\pi^{2}}{4}$
View full question & answer→Question 955 Marks
Find $\int\sec^3\text{xdx}.$
Answer$\int\sec^3\text{xdx}$
$=\int\sec\text{x}\cdot\sec^2\text{xdx}$
$=\int\sqrt{1+\tan^2\text{x}}\cdot\sec^2\text{xdx}$
$\big(\text{Put}\tan\text{x}=\text{t };\sec^2\text{x dx}=\text{dt}\big)$
$=\int\sqrt{1+\text{t}^2}\text{dt}$
$=\frac{\text{t}}{2}\sqrt{1+\text{t}^2}+\frac{1}{2}\log\Big|\text{t}+\sqrt{1+\text{t}^2}\Big|+\text{c}$
$=\frac{\sec\text{x}\cdot\tan\text{x}}{2}+\frac{1}{2}\log\big|\tan\text{x}+\sec\text{x}\big|+\text{c}$
View full question & answer→Question 965 Marks
If $\text{y}=\text{e}^{\text{x}^2\cos\text{x}}+(\cos\text{x})^{\text{x}}, $ then find $\frac{\text{dy}}{\text{dx}}.$
AnswerLet $\text{u}=(\cos\text{x})^{\text{x}}\Rightarrow\text{y}=\text{e}^{\text{x}^2\cos\text{x}}+\text{u}$
$\therefore\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}^2\cos\text{x}}(2\text{x}\cdot\cos\text{x}-\text{x}^2\cdot\sin\text{x})+\frac{\text{du}}{\text{dx}}$
$\log\text{u}=\log(\cos\text{x})^\text{x}\Rightarrow\log\text{u}=\text{x}\cdot\log(\cos\text{x})$
Differentiate w.r.t. “x”
$\frac{\text{1}}{\text{u}}\frac{\text{du}}{\text{dx}}=\log(\cos\text{x})-\text{x}\tan\text{x}$ $\Rightarrow\frac{\text{du}}{\text{dx}}=(\cos\text{x})^{\text{x}}\{\log(\cos\text{x})-\text{x}\tan\text{x}\}$
Therefore,
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}^2\cos\text{x}}(2\text{x}\cdot\cos\text{x}-\text{x}^2\cdot\sin\text{x})\\+(\cos\text{x})^{\text{x}}\{\log(\cos\text{x})-\text{x}\tan\text{x}\}$
View full question & answer→Question 975 Marks
Fine: $\int\frac{3\text{x}+5}{\text{x}^2+3\text{x}-18}\text{dx}.$
AnswerLet $\text{I}=\int\frac{(3\text{x}+5)}{\text{x}^2+3\text{x}-18}$
$\text{I}=\int\frac{(3\text{x}+5)\text{dx}}{(\text{x}+6)(\text{x}-3)}$
Let $\int\frac{(3\text{x}+5)}{(\text{x}+6)(\text{x}-3)}=\frac{\text{A}}{\text{x}+6}+\frac{\text{B}}{\text{x}-3}$
So, $3\text{x}+5=\text{A}(\text{x}-3)+\text{B}(\text{x}+6)$
On comparing,
$\text{A}+\text{B}=3\ \dots(\text{i})$
$-3\text{A}+6\text{B}=5\ \dots(\text{ii})$
$-3\text{A}+6(3-\text{A})=5$
$-3\text{A}+18-6\text{A}=5$
$\text{A}=\frac{-13}{9}=\frac{13}{9}$ and $\text{B}=3-\text{A}=3-\frac{13}{9}=\frac{14}{9}$
So, $\int\frac{(3\text{x}+5)\text{dx}}{(\text{x}+6)(\text{x}-3)}$
$=\int\frac{13\text{dx}}{9(\text{x}+6)}+\int\frac{14\text{dx}}{9(\text{x}-3)}$
$=\frac{13}{9}\text{ln}(\text{x}+6)+\frac{14}{9}\text{ln}(\text{x}-3)+\text{C}$
View full question & answer→Question 985 Marks
Evaluate the following integrals:
$\int\text{cosec x}\log({\text{cosec x}-\cot\text{x})}\text{dx}$
AnswerLet $\text{I}=\int\text{cosec x}\log(\text{cosec x}-\cot\text{x})\text{dx}\ ....(1)$
Let $\log(\text{cosec x}-\cot\text{x})=\text{t}$ then,
$\text{dx}[\log(\text{cosec x}-\cot\text{x})]=\text{dt}$
$\Rightarrow\text{cosec x}\text{ dx}=\text{dt}$ $\Big[\because\ \frac{\text{d}}{\text{dx}}(\log(\text{cosec x}-\cot\text{x}))=\text{cosec x}\Big]$
Putting $\log(\text{cosec x}-\cot\text{x})=\text{t}$ and $\text{cosec x}\text{ dx}=\text{dt}$ in equation (1), we get
$\text{I}=\int\text{t dt}$
$=\frac{\text{t}^2}{2}+\text{C}$
$\text{I}=\frac{1}{2}[\log(\text{cosec x}-\cot\text{x})]^2+\text{C}$
View full question & answer→Question 995 Marks
Evaluate the following integrals:
$\int\frac{\tan\text{x}}{\sqrt{\cos\text{x}}}\text{dx}$
Answer$\int\frac{\tan\text{x}}{\sqrt{\cos\text{x}}}\text{dx}$
$\Rightarrow\int\frac{\sin\text{x}}{\cos\text{x}\sqrt{\cos\text{x}}}\text{dx}$
$\Rightarrow\int\frac{\sin\text{x}}{\cos^\frac{3}{2}\text{x}}\text{dx}$
$\text{Let }\cos\text{x}=\text{t}$
$\Rightarrow-\sin\text{x}\text{ dx}=\text{dt}$
$\Rightarrow\sin\text{x}=-\frac{\text{dt}}{\text{dx}}$
$\text{Now,}\int\frac{\sin\text{x}}{\cos^\frac{3}{2}\text{x}}\text{dx}$
$=\int-\frac{1}{\text{t}^\frac{3}{2}}\text{dt}$
$=-\int\text{t}^{-\frac{3}{2}}\text{dt}$
$=-\bigg[\frac{\text{t}^{-\frac{3}{2}+1}}{\frac{-3}{2}+1}\bigg]+\text{C}$
$=\frac{2}{\sqrt{\text{t}}}+\text{C}$
$=\frac{2}{\sqrt{\cos\text{x}}}+\text{C}$
View full question & answer→Question 1005 Marks
Evalute the following integrals:
$\int\frac{\sin(\text{x}-\alpha)}{\sin(\text{x}+\alpha)}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sin(\text{x}-\alpha)}{\sin(\text{x}+\alpha)}\text{dx}$ then,
$\text{I}=\int\frac{\sin(\text{x}-\alpha+\alpha-\alpha)}{\sin(\text{x}+\alpha)}\text{dx}$
$=\int\frac{\sin(\text{x}+\alpha-2\alpha)}{\sin(\text{x}+\alpha)}\text{dx}$
$=\int\frac{\sin(\text{x}+\alpha)\cos2\alpha-\cos(\text{x}+\alpha)\sin2\alpha}{\sin(\text{x}+\alpha)}\text{dx}$
$=\int\Big[\frac{\sin(\text{x}+\alpha)\cos2\alpha}{\sin(\text{x}+\alpha)}-\frac{\cos(\text{x}+\alpha)\sin2\alpha}{\sin(\text{x}+\alpha)}\Big]\text{dx}$
$=\int\big(\cos2\alpha-\cot(\text{x}+\alpha)\sin2\alpha\big)\text{dx}$
$=\cos2\alpha\int\text{dx}-\sin2\alpha\int\cot(\text{x}+\alpha)\text{dx}$
$=\text{x}\cos2\alpha-\sin2\alpha\log|\sin(\text{x}+\alpha)|+\text{C}$
$\therefore\text{I}=\text{x}\cos2\alpha-\sin2\alpha\log|\sin(\text{x}+\alpha)|+\text{C}$
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