Question 15 Marks
Find:
$\int\frac{x^{2}}{x^{4} + x^{2} - 2}dx$
Answer$\text{Let x}^{2} =\text{t} \therefore \frac{\text{x}^{2}}{\text{x}^{4} + \text{x}^{2} - 2} = \frac{\text{x}^{2}}{\text{(x}^{2} - 1)(\text{x}^{2} + 2)} = \frac{\text{t}}{(\text{t - 1)} (\text{t+2)}}=\frac{\text{A}}{\text{t - 1}} + \frac{\text{B}}{\text{t + 2}}$
Solving for A and B to get, $\text{A =}\frac{1}{3}, \text{B} = \frac{2}{3}$
$\int\frac{\text{x}^{2}}{\text{x}^{4} + \text{x}^{2} - 2}\text{dx} = \frac{1}{3}\int\frac{1}{\text{x}^{2} - 1} \text{dx} + \frac{2}{3}\int\frac{1}{\text{x}^{2} + 2}\text{dx} = \frac{1}{6}\log\bigg|\frac{\text{x - 1}}{\text{x + 1}}\bigg| + \frac{\sqrt{2}}{3}\tan^{-1}\frac{\text{x}}{\sqrt{2}} + \text{C}$
View full question & answer→Question 25 Marks
Evaluate: $\int\limits^{\frac{\pi}{2}}_{0}\frac{\sin^{2}x}{\sin x + \cos x}dx$
Answer$\text{Let I} = \int\limits^{\pi/2}_{0}\frac{\sin^{2}\text{x}}{\sin\text{x} + \cos \text{x}}\text{dx Also I} = \int\limits^{\pi/2}_{0}\frac{\sin^{2}\bigg(\frac{\pi}{2} - \text{x}\bigg)}{\sin\bigg(\frac{\pi}{2} - \text{x}\bigg) + \cos \bigg(\frac{\pi}{2} - \text{x}\bigg)} \text{dx} = \int\limits^{\pi/2}_{0}\frac{\cos^{2}\text{x}}{\cos \text{x} + \sin \text{x}}\text{dx}$
Adding to get, $\text{2I} = \int\limits^{\pi/2}_{0}\frac{1}{\sin\text{x} + \cos\text{x}}\text{dx}= \frac{1}{\sqrt{2}}\int\limits^{\pi/2}_{0}\frac{1}{\cos(\text{x} -\pi/4}\text{dx}$
$\Rightarrow\text{2I} = \frac{1}{\sqrt{2}}\int\limits^{\pi/2}_{0}\sec(\text{x} - \pi/4)\text{dx} = \frac{1}{\sqrt{2}}\log\Bigg|\sec\bigg(\text{x} - \frac{\pi}{4}\bigg) + \tan\bigg(\text{x} - \frac{\pi}{4}\bigg)\Bigg|^{\pi/2}_{0}$
$\Rightarrow \text{2I} = \frac{1}{\sqrt{2}}\left\{\log|\sqrt{2} + 1 | -\log|\sqrt{2} - 1|\right\}$
$\Rightarrow\text{I} = \frac{1}{2\sqrt{2}}\bigg\{\log|\sqrt{2} + 1 | -\log|\sqrt{2} - 1| \ \ \ \text{or}\ \ \ \frac{1}{2\sqrt{2}}\log \bigg|\frac{\sqrt{2}+1}{\sqrt{2} - 1}\bigg|$
View full question & answer→Question 35 Marks
Evaluate:
$\int\limits^{\pi/4}_{0}\bigg(\frac{\sin \text{x} +\cos \text{x}}{3 + \sin 2\text{x}}\bigg)\text{dx}$
Answer$\text{I} = -\int\limits^\frac{\pi}{4}_{0} \frac{\sin \text{x} + \cos \text{x}}{(\sin \text{x} - \cos \text{x})^{2} - 2^{2}}\text{dx}$
$\text{Put} \sin \text{x} - \cos \text{x} = \text{t} \Rightarrow \text{t} = \text{-1 to } 0$
$(\cos \text{x} + \sin {\text{x)}} \text{dx = dt} $
$\text{I} = - \int\limits^{0}_{-1}\frac{\text{dt}}{\text{t}^{2} - 2^{2}}$
$= -\frac{1}{4}\log\bigg|\frac{\text{t} - 2}{\text{t} + 2}\bigg|\bigg]^{0}_{-1}$
$= -\frac{1}{4}\big\{0 - \log 3\big\}$
$= \frac{1}{4}\log3$
View full question & answer→Question 45 Marks
Evaluate:
$\int\limits^{\pi}_{0} \frac{\text{x} \tan \text{x}}{\sec \text{x} + \tan \text{x}}\text{dx}$
Answer$\text{I} = \int\limits^{\pi}_{0} \frac{\text{x} \tan \text{x}}{\sec \text{x} + \tan \text{x}} \text{dx} = \int\limits^{\pi}_{0} \frac{(\pi - \text{x)} \tan \text{x}}{\sec \text{x} + \tan \text{x}} \text{dx}$
$\Rightarrow \text{2I} = \pi \int\limits^{\pi}_{0} \frac{{\tan \text{x}}}{\sec {\text{x} + \tan \text{x}}} \text{dx} = \pi \int\limits^{\pi}_{0} \tan \text{x} (\sec \text{x} - \tan \text{x}) \text{dx}$
$\text{I} = \frac{\pi}{2} \int\limits^{\pi}_{0} {(\sec \text{x} \tan \text{x} - \sec^{2} \text{x + 1) dx}}$
$= \frac{\pi}{2} [ \sec \text{x} - \tan \text{x + x}]^{\pi}_{0}$
$= \frac{\pi(\pi - 2)}{2}$
View full question & answer→Question 55 Marks
Evaluate: $\int\limits_{0}^{\pi}\frac{4\text{x}\sin\text{x}}{1 + \cos^{2}\text{x}}\text{dx}.$
AnswerLet $\text{I} = \int\limits_{0}^{ \pi}\frac{4\text{x}\sin\text{x}}{1 + \cos^{2}\text{x}}\text{dx}$
$\text{x}\rightarrow(\pi - \text{x})\text{gives }\text{ I} = \int\limits_{0}^{\pi}\frac{4(\pi - \text{x})\sin(\pi - \text{x})}{1 + \cos^{2}(\pi - \text{x})}\text{dx} = \int\limits_{0}^{\pi}\frac{4(\pi - \text{x})\sin\text{x}}{1 + \cos^{2}\text{x}}\text{dx}$
$\therefore2\text{I} =4\pi\int\limits_{0}^{\pi}\frac{\sin\text{x}}{1 + \cos^{2}\text{x}}\text{dx}$
Put cos x = t
$\therefore$ sin x dx = – dt
$\therefore\text{I} = 2 \pi\int\limits_{1}^{-1}\frac{-\text{dt}}{1 + \text{t}^{2}}\ \ \ \text{or}\ \ 2\pi\int\limits_{-1}^{1}\frac{\text{dt}}{1 + \text{t}^{2}}$
$ = 2\pi\bigg[\tan^{-1}\text{t}\bigg]_{-1}^{1} = 2\pi\bigg[\frac{\pi}{4} - \bigg( - \frac{\pi}{4}\bigg)\bigg] = \pi^{2}.$
View full question & answer→Question 65 Marks
Evaluate: $\int\frac{\text{x} + 2}{\sqrt{\text{x}^{2} + 5\text{x} + 6 }}\text{dx}.$
Answer$\text{I} = \int\frac{\text{x} + 2 }{\sqrt{\text{x}^{2} + 5\text{x} + 6 }}\text{dx} = \int\frac{\frac{1}{2}(2\text{x} + 5 ) - \frac{1}{2}}{\sqrt{\text{x}^{2} + 5 \text{x} + 6 }}\text{dx}$
$ =\frac{1}{2}\int\frac{2\text{x} + 5 }{\sqrt{\text{x}^{2} + 5\text{x} + 6 }}\text{dx}-\frac{1}{2}\int\frac{\text{dx}}{\sqrt{\big(\text{x} + 5/2\big)^{2} - \bigg(\frac{1}{2}\bigg)^{2}}}$
$ = \sqrt{\text{x}^{2} + 5\text{x} + 6 } -\frac{1}{2}\log\bigg|\bigg(\text{x} + \frac{5}{2}\bigg) + \sqrt{\text{x}^{2} + 5\text{x} + 6 }\bigg| + \text{c}.$
View full question & answer→Question 75 Marks
Evaluate:
$\int\frac{1}{\sin^{4}\text{x} +\sin^{2}\text{x}\cos^{2}\text{x}+\cos^{4}\text{x}}\text{dx}$
Answer$\text{I}=\int\frac{1}{\sin^{4}\text{x} +\sin^{2}\text{x}\cos^{2}\text{x}+\cos^{4}\text{x}}\text{dx}$
$ =\int\frac{(\tan^2\text{x}+1)\sec^{2}\text{x}}{\tan^{4}\text{x}+\tan^{2}\text{x}+1}\text{dx} , [$dividing $N \& D$ by $\cos^4 x]$
$ = \int\frac{(\text{t}^{2} + 1)}{\text{t}^{4} +\text{t}^2+ 1 } \text{dx},$ where tan x = t
$ = \int\frac{1+\frac{1}{\text{t}^{2}}}{\text{t}^{2} +\frac{1}{\text{t}^2}+ 1 } \text{dx},[$dividing $N \& D$ by $t^2]$
Putting $\text{t}-\frac{1}{\text{t}}$ = z so that $\Big(\text{t}-\frac{1}{\text{t}^2}\Big)$ $dt = dz$
$\text{and}\ \text{t}^2-\frac{1}{\text{t}^2}=\text{z}^2+2$
$\therefore\ \text{I}=\int\frac{\text{dz}}{\text{z}^2+(\sqrt3)^2}$
$\frac{\text{1}}{\sqrt3}\tan^{-1}\frac{\text{z}}{\sqrt3}+\text{c}=\frac{\text{1}}{\sqrt3}\tan^{-1}\Big(\frac{\text{t}^2-1}{\sqrt3\ \text{t}}\Big)+\text{c}\frac{\text{1}}{\sqrt3}\tan^{-1}\Big(\frac{\tan^2\text{x}-1}{\sqrt3\tan\text{x}}\Big)+\text{c}$
View full question & answer→Question 85 Marks
Evaluate:
$\int\limits^{\pi/2}_{0} \frac{\cos^{2} \text{x dx}}{1 + 3\sin^{2}\text{x}}$
Answer$\text{I} = \int\limits^\frac{\pi}{2}_{0}\frac{\text{dx}}{1 + 4 \tan^{2}\text{x}} = \int\limits^\frac{\pi}{2}_0 \frac{\sec^{2}\text{x}}{( 1 + \tan^{2} \text{x})(1 + 4\tan^{2}\text{x})} \text{dx}$$\text{Put} \tan \text{x = t}$
$\text{I} = \int\limits^{\infty}_{0}\frac{\text{dt}}{(1 + \text{t}^{2)} (1 + 4 \text{t}^{2})} = -\frac{1}{3} \int\limits^\infty_{0} \frac{\text{dt}}{1 + \text{t}^{2}} + \frac{4}{3}\int\limits^\infty_{0} \frac{\text{dt}}{1 + (\text{2 t)}^{2}}$
$= - \frac{1}{3} \tan^{-1}\text{t}\bigg]^{\infty}_{0} + \frac{4}{3 \times 2} \tan^{-1}\text{(2 t})\bigg]^{\infty}_{0}$
$= -\frac{1}{3}\bigg(\frac{\pi}{2}\bigg) + \frac{2}{3}\bigg(\frac{\pi}{2}\bigg) = \frac{\pi}{6}$
View full question & answer→Question 95 Marks
$\text{Find}:\int \frac{\sin \theta\ \text{d}\theta}{(4 + \cos^{2} \theta) (2- \sin^{2} \theta)} \text{d} \theta$
Answer$\text{I} = \int \frac{\sin \theta\ \text{d}\theta}{(4 + \cos^{2} \theta) (2 - \sin^{2} \theta)} = \int\frac{\sin \theta\ \text{d}\theta}{(4 + \cos^{2} \theta) (1 + \cos^{2}\theta)}$$= -\int \frac{\text{dt}}{(4 + \text{t}^{2}) (1 + \text{t}^{2})}, \text{ where } \cos \theta = \text{t}$
$= \int \frac{\frac{1}{3}}{4 + \text{t}^{2}} \text{dt} - \int \frac{\frac{1}{3}}{1 + \text{t}^{2}} \text{dt}$
$= \frac{1}{6}\tan^{-1} \frac{\text{t}}{2} - \frac{1}{3} \tan^{-1} \text{t} + \text{c}$
$ = \frac{1}{6}\tan^{-1} \big(\frac{\cos \theta}{2}\big) - \frac{1}{3} \tan^{-1} \cos \theta + \text{c}$
View full question & answer→Question 105 Marks
Evaluate:
$\int\limits^{4}_{1} \left\{|\text{x - 1}| + |\text{x - 2}| + |\text{x - 4}| \right\} \text{dx}$
Answer$\text{I} = \int\limits^{4}_{1} \left\{|\text{x - 1}| + |\text{x - 2}| + |\text{x - 4}| \right\} \text{dx}$
$= \int\limits^{4}_{1} \text{(x - 1) dx} - \int\limits^{2}_{1} \text{(x - 2)} \text{dx} + \int\limits^{4}_{2} \text{(x - 2) dx} - \int\limits^{4}_{1} \text{x - 4) dx}$
$= \bigg[\frac{\text{(x - 1)}^{2}}{2}\bigg]^{4}_{1} - \bigg[\frac{\text{(x - 2)}^{2}}{2}\bigg]^{2}_{1} + \bigg[ \frac{\text{(x - 2)}^{2}}{2}\bigg]^{4}_{2} - \bigg[\frac{\text{(x - 4)}^{2}}{2}\bigg]^{4}_{1}$
$= \frac{9}{2} + \frac{1}{2} +2+ \frac{9}{2} = 11 \frac{1}{2} \text{ or } \frac{23}{2}$
View full question & answer→Question 115 Marks
If $\tan^{-1}\bigg(\frac{\text{x} - 2}{\text{x} - 4}\bigg) + \tan^{-1}\bigg(\frac{\text{x} + 2 }{\text{x} + 4 }\bigg) = \frac{\pi}{4} , $find the value of x.
AnswerGiven equation can be written as
$\tan^{-1}\bigg(\frac{\text{x} - 2}{\text{x} - 4}\bigg) =\tan^{-1} 1 - \tan^{-1}\bigg(\frac{\text{x} + 2}{\text{x} + 4}\bigg)$
$ = \tan^{-1}\bigg(\frac{1-\frac{\text{x} + 2}{\text{x} + 4 }}{1+ \frac{\text{x} + 2}{\text{x} + 4 }}\bigg) = \tan^{-1}\bigg(\frac{2}{2\text{x} + 6 }\bigg)$
$\therefore\frac{\text{x} - 2 }{\text{x} - 4} = \frac{1}{\text{x} + 3}$
$\Rightarrow\text{x}^{2} + \text{x} -6 =\text{x} - 4 \text{ or } \text{x}^{2} = 2 \therefore\text{x}= \pm\sqrt{2}.$
View full question & answer→Question 125 Marks
Evaluate :$\int\Big(\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}\Big)\text{dx}$
Answer$\text{I}=\int\Big(\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}\Big)\text{dx}=\int\frac{\cos\text{x}+\sin\text{x}}{\sqrt{\sin\text{x}\cos\text{x}}}\text{dx}$
Putting $\sin x – \cos x = t,$ so that $(\cos x + \sin x) dx = dt$
and $\sin x \cos x =\frac1 2(1 – t^2)$
$\therefore\ \text{I}=\sqrt2\int\frac{\text{dt}}{1-\text{t}^2}=\sqrt2\sin^{-1}\text{t}+\text{c}$
$=\sqrt2\sin^{-1}(\sin\text{x}-\cos\text{x})+\text{c}$
View full question & answer→Question 135 Marks
Find: $\int(3x + 1)\sqrt{4 - 3x - 2x^{2 } }\text{ dx}$
Answer$\int(3\text{x} + 1)\sqrt{4 - 3\text{x} - 2\text{x}^{2}}\text{ dx} = -\frac{3}{4}\int(-4\text{x} - 3)\sqrt{4 - \text{3x} - \text{2x}^{2}} \text{ dx} - \frac{5}{4}\int\sqrt{4 - 3\text{x} - 2\text{x}^{2}} \text{ dx}$
$= -\frac{1}{2}(4 - \text{3x} - \text{2x}^{2})^{3/2}-\frac{5}{4}\sqrt{2}\int\sqrt{\bigg(\frac{\sqrt{41}}{4}\bigg)^{2} - \bigg(\text{x} + \frac{3}{4}\bigg)^{2}} \text{dx}$
$= -\frac{1}{2}(4 - \text{3x} - \text{2x}^{2})^{3/2}-\frac{5}{4}\sqrt{2} \left\{\frac{4\text{x} + 3}{8} \sqrt{\frac{41}{16}- \bigg(\text{x} + \frac{3}{4}\bigg)^{2}} + \frac{41}{32}.\sin^{-1}\bigg(\frac{\text{4x + 3}}{\sqrt{41}}\bigg)\right\} + \text{C}$
$= -\frac{1}{2}(4 - \text{3x} - \text{2x}^{2})^{3/2}-\frac{5}{4}\left\{\frac{\text{4x + 3}}{8} \sqrt{4 - \text{3x - 2x}^{2}} + \frac{41\sqrt{2}}{32}.\sin^{-1}\bigg(\frac{\text{4x + 3}}{\sqrt{41}}\bigg)\right\} + \text{C}$
View full question & answer→Question 145 Marks
Evaluate: $\int\limits^\frac{3}{2}_{0}|x \cos\pi x| dx$
Answer$\int\limits^\frac{3}{2}_{0}|\text{x} \cos\pi \text{ x}| \text{dx} = \int\limits^{1/2}_{0}\text{x}\cos\pi \text{x dx}{-}\int\limits^{3/2}_{1/2}\text{x}\cos\pi \text{x dx}$
$= \left\{\frac{\text{x}\sin\pi \text{x}}{\pi} + \frac{\cos\pi\text{x}}{\pi^{2}}\right\}^{1/2}_{0}= \left\{\frac{\text{x}\sin\pi \text{x}}{\pi} + \frac{\cos\pi\text{x}}{\pi^{2}}\right\}^{3/2}_{1/2}$
$=\frac{1}{2\pi} - \frac{1}{\pi^{2}} - \bigg(-\frac{3}{2\pi}-\frac{1}{2\pi}\bigg) = \frac{5}{2\pi}-\frac{1}{\pi^{2}}$
View full question & answer→Question 155 Marks
Evaluate:
$\int\frac{\sin \text{x} - \text{x}\cos \text{x}}{\text{x} ( \text{x}+ \sin \text{x})} \text{dx}$
Answer$\text{I} = \int\frac{\text{x}+ \sin \text{x - x }(1 + \cos \text{x})}{\text{x} (\text{x} + \sin \text{x})}\text{dx}$
$= \int\frac{1}{\text{x}} \text{dx} - \int\frac{1 + \cos \text{x}}{\text{x}+ \sin \text{x}} \text{dx put x} + \sin \text{x} = \text{t} $
$\Rightarrow ( 1 + \cos \text{x}) \text{dx = dt}$
$= \log|\text{x}| - \log|\text{x} + \sin \text{x}| + \text{c} $
View full question & answer→Question 165 Marks
Find:
$\int \frac{\cos \theta}{(4 + \sin^{2} \theta) (5 - 4 \cos^{2} \theta)} \text{d} \theta$
Answer$\text{I} = \int \frac{\cos \theta}{(4 + \sin^{2} \theta) (5 - 4 \cos^{2} \theta)} \text{d} \theta = \int\frac{\cos \theta}{(4 + \sin^{2} \theta) (1 + 4 \sin^{2}\theta)} \text{d} \theta$
$= \int \frac{\text{dt}}{(4 + \text{t}^{2}) (1 + 4 \text{t}^{2})}, \text{ where } \sin \theta = \text{t}$
$= \int \frac{-\frac{1}{15}}{4 + \text{t}^{2}} \text{dt} + \int \frac{\frac{4}{15}}{1 + 4 \text{t}^{2}} \text{dt}$
$= -\frac{1}{30}\tan^{-1} \big(\frac{\text{t}}{2}\big) + \frac{4}{30} \tan^{-1} \text{(2 t)} + \text{c}$
$ = -\frac{1}{30}\tan^{-1} \big(\frac{\sin \theta}{2}\big) + \frac{2}{15} \tan^{-1} (2 \sin \theta) + \text{c}$
View full question & answer→Question 175 Marks
Prove that $ \tan^{-1}\bigg[\frac{\sqrt{1 + \text{x}} - \sqrt{1 - \text{x}}}{\sqrt{1 + \text{x}} + \sqrt{1 - \text{x}}}\bigg] = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}\text{x} , \frac{1}{\sqrt{2}}\leq\text{x}\leq1$
AnswerPutting x = cos θ in LHS,We get
LHS $ = \tan^{-1}\bigg[\frac{\sqrt{1 + \cos\theta} - \sqrt{1 - \cos\theta}}{\sqrt{1 + \cos\theta} + \sqrt{1 - \cos\theta}}\bigg]$
$ =\tan^{-1}\bigg[\frac{\sqrt{2}\cos\theta / 2 - \sqrt{2}\sin\theta/2 }{\sqrt{2}\cos\theta/2 + \sqrt{2}\sin\theta/2}\bigg]$
$ = \tan^{-1}\bigg[\frac{1 - \tan\theta/2}{1 + \tan\theta/2}\bigg] = \tan^{-1}\bigg[\tan\bigg(\frac{\pi}{4} - \theta / 2\bigg)\bigg]$
$ = \frac{\pi}{4} - \frac{1}{2}\theta =\frac{\pi}{4} - \frac{1}{2}\cos^{-1}\text{x} = \text{R.H.S.}$
View full question & answer→Question 185 Marks
Evaluate:
$\int\frac{1}{\cos^{4}\text{x} +\sin^{4}\text{x}}\text{dx}$
Answer$\text{I} = \int\frac{1}{\cos^{4}\text{x} + \sin^{4}\text{x}}\text{dx}$
dividing numerator and denominator by $\cos^4 x$
$ =\int\frac{\sec^{4}\text{x}}{1 + \tan^{4}\text{x}}\text{dx} = \int\frac{(1 + \tan^{2}\text{x})\sec^{2}\text{x}}{1 + \tan^{4}\text{x}}\text{dx}$
Putting $\tan x = t$
$\therefore\sec^{2}\text{x}\text{ dx} =\text{dt}$
$ = \int\frac{(\text{t}^{2} + 1)\text{dt}}{\text{t}^{4} + 1 } = \int\frac{1 +\frac{1}{\text{t}^{2}}}{\text{t}^{2} + \frac{1}{\text{t}^{2}}}\text{dt}$ $\left\{\text{ dividing by t}^{2}\right\}$
$ = \int\frac{\text{dz}}{\text{z}^{2} + (\sqrt{2})^{2}}\text{ where }\text{t} - \frac{1}{\text{t}} =\text{z}$
$ = \frac{1}{\sqrt{2}}\tan^{-1}\bigg(\frac{\text{z}}{\sqrt{2}}\bigg) +\text{c}$
$ =\frac{1}{\sqrt{2}}\tan^{-1}\bigg(\frac{\text{t}^{2} - 1 }{\sqrt{2}\text{t}}\bigg) + \text{c} = \frac{1}{\sqrt{2}}\tan^{-1}\bigg(\frac{\tan^{2}\text{x} - 1}{\sqrt{2}\tan\text{x}}\bigg) +\text{c}.$
View full question & answer→Question 195 Marks
Evaluate: $\int\frac{\text{cos 2x}\text{ - cos }\alpha}{\text{cos x - cos }\alpha}\text{dx}.$
AnswerEvaluate: $\text{I}=\int\frac{\text{cos 2x}\text{ - cos }\alpha}{\text{cos x - cos }\alpha}\text{dx}$
$=\int\frac{\text{(2cos}^{2}\text{x - 1)}-\text{(2 cos}^{2}\alpha- 1)}{\text{cos x - cos }\alpha}\text{dx}$
$=\int\frac{\text{2(cos}^{2}\text{x - cos}^{2}\alpha)}{\text{cos x - cos }\alpha}\text{dx}$ $=\int\frac{\text{2(cos x + cos }\alpha).\text{(cos x - cos }\alpha)}{\text{cos x - cos }\alpha}\text{dx}$
$=\text{2}\int\text{(cos x + cos }\alpha)\text{dx = 2}\int\text{cos x dx +2}\int\text{cos}\alpha\text{ dx}$
= 2 sin x + 2x cos $\alpha$ + C.
View full question & answer→Question 205 Marks
Evaluate: $\int\limits_0^{2\pi}\frac{1}{1+e^{\text {sin x}}}\text{dx}$
AnswerLet I = $\int\limits_0^{2\pi}\frac{1}{1+e^{\text {sin x}}}\text{dx}\ ......\text{(i)}$ Applying properties $\int\limits_0^{\alpha}f\text{(x)dx}=\int\limits_0^af\text{(a-x)dx}\text{ we get}$$\text{I}=\int\limits_{0}^{2\pi}\frac{dx}{1+e^{sin(2\pi-x)}}\int\limits_0^{2\pi}\frac{dx}{1+e^{-sin x}}=\int\limits_0^{2\pi}\frac{dx}{1+\frac{1}{e^{sinx}}}$
$\text{I}=\int\limits_{0}^{2\pi}\frac{e^{\text{sin x}}dx}{e^{\text{-sin x}}+1}\ .......\text{(ii)}$
Adding (i) and (ii) we get$\text{2I}=\int\limits_{0}^{2\pi}\frac{dx}{1+e^{\text{sin x}}}+\int\limits_{0}^{2\pi}\frac{e^{\text{sin x}}dx}{1+e^{\text{sin x}}}=\int\limits_{0}^{2\pi}\frac{1+e^{\text{sin x}}}{1+e^{\text{sin x}}}\text{dx}$
$=\int\limits_{0}^{2\pi}\text{dx} =\big[ \text{x}\big]^{2\pi}_{0}$
$\Rightarrow$ 2I = 2$\pi $
$\Rightarrow$ I = $\pi $.
View full question & answer→Question 215 Marks
Evaluate: $\int\frac{\text{dx}}{\text{x}\text{(x}^{5}\text{+3)}}$
AnswerLet $I = \int\frac{\text{dx}}{\text{x}\text{(x}^{5}\text{+3)}}=\int\frac{\text{x}^{4}\text{dx}}{\text{x}^{5}\text{(x}^{5}\text{+3)}}=\frac{1}{5}\int\frac{\text{5x}^{4}\text{dx}}{\text{x}^{5}\text{(x}^{5}\text{+3)}}$ Let $x^5 = z$
$\Rightarrow 5x^4dx = dz$
$\therefore\text{I}=\frac{1}{5}\int\frac{\text{dz}}{\text{z(z+3)}}$
$=\frac{1}{5\times3}\int\frac{z+3-z}{z(z+3)}\text{dz}=\frac{1}{15}\int\frac{z+3}{z(z+3)}\text{dz}-\frac{1}{15}\frac{z}{z(z+3)}\text{dz}$
$=\frac{1}{15}\int.\frac{dz}{z}-\frac{1}{15}\int\frac{dz}{z+3}=\frac{1}{15}\left\{\text{log z}-\text{log }|z+3|\right\}+\text{C}$
$=\frac{1}{15}\log\Bigg|\frac{z}{z+3}\Bigg|+\text{C}=\frac{1}{15}\log\Bigg|\frac{x^5}{x^5+3}\Bigg|+\text{C}$
View full question & answer→Question 225 Marks
$\text{Evaluate:}\int\frac{\text{x + 2}}{\sqrt{\text{x}^{2}+\text{2x}+}\text{3}}\text{dx}$
Answer$\text{Let I}=\int\frac{\text{x + 2}}{\sqrt{\text{x}^{2}+\text{2x}+}\text{3}}\text{dx}$$=\frac{\text{1}}{\text{2}}\int\frac{\text{2x + 4}}{\sqrt{\text{x}^{2}+\text{2x}+}\text{3}}\text{dx}=\frac{\text{1}}{\text{2}}\int\frac{\text{(2x + 2) + 2 }}{\sqrt{\text{x}^{2}+\text{2x}+\text{3}}}\text{dx}$
$=\frac{\text{1}}{\text{2}}\int\frac{\text{(2x + 2)}\text{dx}}{\sqrt{\text{x}^{2}+\text{2x}+}\text{3}}+\frac{\text{1}}{\text{2}}\int\frac{\text{2dx }}{\sqrt{\text{x}^{2}+\text{2x}+\text{3}}}$
$\text{I}=\frac{\text{1}}{\text{2}}\text{I}_{1}+\text{I}_{2}-------(i)$
$\text{Where }\text{I}_{1}=\int\frac{\text{(2x+2)}\text{dx}}{\sqrt{\text{x}^{2}+\text{2x}+\text{3}}}\text{ and }\text{I}_{2}=\int\frac{\text{dx}}{\sqrt{\text{x}^{2}+\text{2x}+\text{3}}}$ $\text{I}_{1}=\int\frac{\text{2x+2}}{\sqrt{\text{x}^{2}+\text{2x}+\text{3}}}\text{dx}$ $x^2 + 2x + 3 = z^2$
(2x + 2)dx = 2z dz $\Rightarrow$ $\text{I}_{1}=\int\frac{\text{2z dz}}{\text{z}}$$=\text{2}\int{dz}=2z=2\sqrt{\text{x}^{2}+\text{2x}+\text{3}}+\text{C}_{1}$
$\Rightarrow$ $\text{I}_{1}=2\sqrt{\text{x}^{2}+\text{2x}+\text{3}}+\text{C}_{1}$ Again $\text{I}_{2}=\int\frac{\text{dx}}{\sqrt{\text{x}^{2}+\text{2x}+\text{3}}}=\int\frac{\text{dx}}{\sqrt{\text{(x+1)}^{2}+(\sqrt{\text{2}}})^{2}}$$=\log|\text{(x+1)}+\sqrt{\text{(x+1)}^{2}+(\sqrt{\text{2}})^{2}}|$
$=\log|\text{(x+1)}+\sqrt{\text{x}^{2}+\text{2x}+\text{3}|}+\text{C}_{2}$
Putting the value of $I_1$, and $I_2$ in $(i)$ we get$\text{I}=\text{2}\sqrt{\text{x}^{2}+\text{2x}+\text{3}}+\log|\text{(x+1)}+\sqrt{\text{x}^{2}+\text{2x}+\text{3}|}+\text{(C}_{1}+\text{C}_{2})$
$=\text{2}\sqrt{\text{x}^{2}+\text{2x}+\text{3}}+\log|\text{(x+1)}+\sqrt{\text{x}^{2}+\text{2x}+\text{3}|}+\text{C.}$
View full question & answer→Question 235 Marks
Evaluate: $\int\frac{\text{x sin}^{-1}\text{x}}{{\sqrt{\text{x-1}^{2}}}}\text{dx}.$
Answer$\text{I}=\int\frac{\text{x sin}^{-1}\text{x}}{{\sqrt{\text{1 - x}^{2}}}}\text{dx};$ Let x = sinθ $\Rightarrow$ dx = cosθ dθ $=\int\frac{\sin\theta\cdot\theta}{\cos\theta}\cos\theta\text{ d}\theta=\int\theta\sin\theta\text{ d}\theta$= –θ cosθ + $\int$cosθ dθ = –θ cosθ + sinθ + c
$=-\sin^{-1}\text{x}(\sqrt{\text{1-x}^{2}})+\text{x + c}$.
View full question & answer→Question 245 Marks
Evaluate:$\int\limits_0^{\pi/2}\frac{\text{x + sin x}}{\text{1 + cosx}}\text{dx}$.
Answer$\text{I}=\int\limits_{0}^{\pi/2}\frac{\text{x}} {\text{1 + cos x}}\text{dx}+\int\limits_0^{\pi/2}\frac{\text{sin x}}{\text{1 + cos x}}\text{dx}$
$=\int\limits_0^{\pi/2}\cdot\text{x}\cdot\frac{1}{2}\sec^{2}\frac{\text{x}}{2}\text{dx}+\int\limits_0^{\pi/2}\tan\frac{\text{x}}{2}\text{dx}$
$=\Bigg[\text{x tan}\frac{\text{x}}{2}\Bigg]_{0}^{\pi/2}-\int\limits_0^{\pi/2}\tan\frac{\text{x}}{2}\text{dx}+\int\limits_0^{\pi/2}\tan\frac{\text{x}}{2}\text{dx}$
$=\frac{\pi}{2}1-0=\frac{\pi}{2}.$
View full question & answer→Question 255 Marks
Evaluate:$\int\frac{\text{6x + 7}}{\sqrt{\text{(x - 5)(x - 4)}}}\text{dx}$.
Answer$\text{I}=\int\frac{\text{6x + 7}}{\sqrt{\text{(x - 5)(x - 4)}}}\text{dx}=\int\frac{\text{6x + 7}}{\sqrt{\text{x}^{2}-\text{9x + 20}}}\text{dx}$
$=\int\frac{\text{3(2x - 9)+ 34}}{\sqrt{\text{x}^{2}-\text{9x + 20}}}\text{dx}$
$=3\int\frac{\text{2x - 9}}{\sqrt{\text{x}^{2}-\text{9x + 20}}}\text{dx}\text{ } + \text{ }34\int\frac{\text{dx}}{\sqrt{\Big(\text{x}-\frac{9}{2}\Big)^{2}-\Big(\frac{1}{2}\Big)^{2}}}\text{dx}$
$=3.2\sqrt{\text{x}^{2}-\text{9x + 20 }}+34.\log\Bigg|\Bigg(\text{x}-\frac{9}{2}\Bigg)+\sqrt{\text{x}^{2}-\text{9x + 20 }}\Bigg|+\text{c}$
$=6\cdot\sqrt{\text{x}^{2}-\text{9x + 20 }}+34.\log\Bigg|\Bigg(\frac{\text{2x - 9}}{2}\Bigg)+\sqrt{\text{x}^{2}-\text{9x + 20 }}\Bigg|+\text{c}$.
View full question & answer→Question 265 Marks
Evaluate:$\int\limits_{\pi/6}^{\pi/3}\frac{\text{dx}}{1+\sqrt{\tan\text{x}}}$.
Answer$\text{I}=\int\limits_{\pi/6}^{\pi/3}\frac{\text{dx}}{1+\sqrt{\text{tan x}}}=\int\limits_{\pi/6}^{\pi/3}\frac{\sqrt{\text{cos x}}}{\sqrt{\text{cos x}}+\sqrt{\text{sin x}}}\text{dx}.....................\text{(i)}$
$\text{x}\rightarrow\Big(\pi/3+\pi/6-\text{x}\Big)$
$=\int\limits^{\pi/3}_{\pi/6}\frac{\sqrt{\cos(\pi/2-\text{x})}}{\sqrt{\cos(\pi/2-\text{x)}}+\sqrt{\sin(\pi/2-\text{x})}}\text{dx}=\int\limits^{\pi/3}_{\pi/6}\frac{\sqrt{\sin\text{x}}}{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}\text{dx}............\text{(ii)}$
Adding (i) and (ii) to get
2I $=\int\limits^{\pi/3}_{\pi/6}1.\text{ dx}=[\text{x}]^{\pi/3}_{\pi/6}=\pi/3-\pi/6=\pi/6$
$\Rightarrow\text{I}=\pi/12$.
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Evaluate the following :
$\int\limits_1^2\frac{5\text{x}^2}{\text{x}^2+4\text{x}+3}\ \text{dx}$
Answer$\text{I}=\int\limits_1^2\frac{5\text{x}^2}{\text{x}^2+4\text{x}+3}\ \text{dx}=5\int\limits_1^21-\frac{4\text{x}+3}{\text{x}^2+4\text{x}+3}\text{ }\text{dx} $
$=5\big[\text{x}\big]_1^2-10\int\limits_1^2\frac{2\text{x}+4-\frac{5}{2}}{\text{x}^2+4\text{x}+3}\ \text{dx}$
$=5-10\big[\log|\text{x}^2+4\text{x}+3|\big]_1^2+25\int\limits_1^2\frac{1}{(\text{x}+2)^2-(1)^2}\ \text{dx}$
$=5-10\ \log\ \frac{15}{8}+25\ .\ \frac{1}{2}\bigg[\log\begin{vmatrix}\frac{\text{x}+2-1}{\text{x}+2+1}\end{vmatrix}\bigg]_1^2$
$=5-10\ \log\frac{15}{8}+\frac{25}{2}\ \log\frac{6}{5}$
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Evaluate the following :
$\int\frac{\text{x+2}}{\sqrt{(\text{(x-2)(x-3)}}}\text{dx}$
Answer$\text{Here I}=\int\frac{\text{x+2}}{\text{x}^2-5\text{x}+6}\text{ dx}=\frac{1}{2}\int\frac{2\text{x}-5+9}{\sqrt{{\text{x}^2}-5\text{x}+6}}\text{dx} $
$=\frac{1}{2}\int\frac{2\text{x}-5}{\sqrt{\text{x}^2-5\text{x}+6}}\ \text{dx}+\frac{9}{2}\int\frac{1}{\sqrt{\bigg(\text{x}-{5}/{2}\bigg)^2-\bigg(\frac{1}{2}\bigg)^2}}\ \text{dx}$
$\sqrt{\text{x}^2+4\text{x}+3}+\frac{9}{2}\log\begin{vmatrix}\bigg(\text{x}-\frac{5}{2}\bigg)+\sqrt{\text{x}^2-5\text{x}+6}\end{vmatrix}+\text{c}$
View full question & answer→Question 295 Marks
Evaluate: $\int\text{x sin}^{-1}\text{x dx}.$
Answer$\int\text{x}\sin^{-1}\text{x dx}=\sin^{-1}\text{x}\cdot\frac{\text{x}^{2}}{2}-\int\frac{\text{x}^{2}}{2}\cdot\frac{1}{\sqrt{1 - \text{x}^{2}}}\text{dx}$
$=\frac{\text{x}^{2}}{2}\sin^{-1}\text{x}+\frac{1}{2}\int\frac{\text{1 - x}^{2}-1}{\sqrt{\text{1 - x}^{2}}}\text{dx}$
$=\frac{\text{x}^{2}}{2}\sin^{-1}\text{x}+\frac{1}{2}\int\sqrt{\text{1 - x}^{2}}\text{dx}-\frac{1}{2}\int\frac{\text{dx}}{\sqrt{\text{1 - x}^{2}}}$
$=\frac{\text{x}^{2}}{2}\sin^{-1}\text{x}+\frac{1}{2}\Bigg[\frac{\text{x}}{2}\sqrt{\text{1 - x}^{2}}+\frac{1}{2}\sin^{-1}\text{x}\Bigg]-\frac{1}{2}\sin^{-1}\text{x}+\text{c}$
$=\sin^{-1}\cdot\text{ x }\cdot\Bigg[\frac{\text{x}^{2}}{2}+\frac{1}{4}-\frac{1}{2}\Bigg]+\frac{1}{4}\text{ x }\sqrt{\text{1 - x}^{2}}+\text{c}$
$\text{ Or, }\frac{\text{2x}^{2}-1}{4}\sin^{-1}\text{x}+\frac{1}{4}\text{ x }\sqrt{\text{1 - x}^{2}}+\text{c}.$
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Evaluate: $\int\frac{\text{dx}}{\sqrt{5-4\text{x - 2x}^{2}}}.$
Answer$\int\frac{\text{dx}}{\sqrt{5-4\text{x}-\text{2x}^{2}}}=\frac{1}{\sqrt{2}}\int\frac{\text{dx}}{\sqrt{\frac{5}{2}-\text{2x - x}^{2}}}=\frac{1}{\sqrt{2}}\int\frac{\text{dx}}{\sqrt{\Bigg(\sqrt\frac{7}{2}\Bigg)^{2}-(\text{x + 1})^{2}}}$
$=\frac{1}{\sqrt{2}}\cdot\sin^{-1}\Bigg(\frac{\text{x + 1}}{\frac{\sqrt{7}}{\sqrt{2}}}\Bigg)+\text{c}$
$\text{Or}\frac{1}{\sqrt{2}}\sin^{-1}\Bigg(\frac{\sqrt{2}}{7}\cdot\text{(x + 1)}\Bigg)+\text{c}.$
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Evaluate: $\int\limits^{\pi}_{0}\frac{\text{x dx}}{\text{a}^{2}\cos^{2}\text{x + b}^{2}\sin^{2}\text{x}}.$
Answer$\text{I}=\int\limits^{\pi}_{0}\frac{\text{x dx}}{\text{a}^{2}\cos^{2}\text{x + b}^{2}\sin^{2}\text{x}}=\int\limits_{0}^{\pi}\frac{(\pi-\text{x) dx}}{\text{a}^{2}\cos^{2}\text{x + b}^{2}\sin^{2}\text{x}}$
$\therefore2\text{I}=\pi\int\limits^{\pi}_{0}\frac{\text{dx}}{\text{a}^{2}\cos^{2}\text{x + b}^{2}\sin^{2}\text{x}}=2\pi\int\limits_{0}^{\pi/2}\frac{\sec^{2}\text{x}}{\text{a}^{2}+\text{b}^{2}\text{ }\tan^{2}\text{x}}\text{dx}$
$ =2\pi\int\limits^{\infty}_{0}\frac{\text{dt}}{\text{a}^{2}+\text{b}^{2}\text{ t}^{2}}=\frac{2\pi}{\text{ab}}\Bigg[\tan^{-1}\frac{\text{bt}}{\text{a}}\Bigg]^{\infty}_{0}\text{where tan x = t}$
$\text{2I}=\frac{2\pi}{\text{ab}}\cdot\frac{\pi}{2}=\frac{\pi^{2}}{\text{ab}}$
$\Rightarrow\text{I}=\frac{\pi^{2}}{\text{2ab}}.$
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Evaluate:$\int\limits_0^{\frac{\pi}{2}}$ log sin x dx.
Answer$\text{I}=\int\limits_0^{\pi/2}\log\sin\text{x dx}=\int\limits_0^{\pi/2}\log \sin\Bigg(\frac{\pi}{2}-\text{x}\Bigg)\text{dx}=\int\limits_0^{\pi/2}\log\cos\text{x dx}$
$\therefore\text{2I}=\int\limits_0^{\pi/2}\log(\text{sin x cos x)dx}=\int\limits_0^{\pi/2}\log\frac{\sin\text{2x}}{2}\text{dx}$
$=\int\limits_0^{\pi/2}\log\sin\text{2x dx}-\int\limits_0^{\pi/2}\log\text{ 2dx}$
$=\frac{1}{2}\times2\int\limits_0^{\pi/2}\log\sin\text{t}\text{ dt}-\frac{\pi}{2}\log2;\Bigg[\text{t = 2x,}\frac{\text{dt}}{2}=\text{dx}\Bigg]$
$=\text{I}-\frac{\pi}{2}\log2$
$\text{I}=-\frac{\pi}{2}\log2$.
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Evaluate:$\int\frac{\text{x + 2}}{\text{2x}^{2}+\text{6x + 5}}\text{dx}$.
Answer$\text{For I}=\int\frac{\text{x + 2}}{\text{2x}^{2}+\text{6x + 5}}\text{dx}=\frac{1}{4}\int\frac{\text{4x + 6 + 2}}{\text{2x}^{2}+\text{6x + 5}}\text{dx}$$=\frac{1}{4}\int\frac{\text{4x + 6}}{\text{2x}^{2}+\text{6x}+\text{5}}\text{dx}+\frac{1}{4}\int\frac{\text{dx}}{\text{x}^{2}+\text{3x}+\frac{5}{2}}$
$=\frac{1}{4}\cdot\log|\text{2x}^{2}+\text{6x}+5|+\frac{1}{4}\int\frac{\text{dx}}{\Big(\text{x}+\frac{3}{2}\Big)^{2}+\Big(\frac{5}{2}-\frac{9}{4}\Big)}$
$=\frac{1}{4}\log|\text{2x}^{2}+\text{6x}+5|+\frac{1}{4}\int\frac{\text{dx}}{\Big(\text{x}+\frac{3}{2}\Big)^{2}+\Big(\frac{1}{2}\Big)^{2}}$
$=\frac{1}{4}\log|\text{2x}^{2}+\text{6x}+5|+\frac{1}{2}\tan^{-1}(\text{2x + 3})+\text{c}$.
View full question & answer→Question 345 Marks
Using properties of definite integrals, evaluate:$\int\limits^{\pi/4}_{0}\text{log (1 + tan x) dx}$.
Answer$\text{I}=\int_{0}^{\pi/4}\log\text{(1 + tan x ) dx}=\int_0^{\pi/4}\log\Bigg(1+\tan(\frac{\pi}{4}-\text{x})\Bigg)\text{dx}$$=\int_{0}^{\pi/4}\log\Bigg(1+\frac{1-\tan\text{x}}{\text{1 + tan x}}\Bigg)\text{dx}=\int_{0}^{\pi/4}\log\Bigg(\frac{2}{\text{1 + tan x}}\Bigg)\text{dx}$
$=\int_{0}^{\pi/4}\log2\text{ dx}-\int_0^{\pi/4}\log(1+\tan\text{ x})\text{ dx}$
$\Rightarrow\text{2I}=\log2\cdot\int_0^{\pi/4}1\cdot\text{dx}=\log2\cdot\pi/4$
$\Rightarrow\text{I}=\pi/8\cdot\log2.$
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Evaluate $\int\limits_0^{3} (2x^2 + 3x + 5)dx$ as limit of a sum.
AnswerHere h = $\frac{3}{\text{n}}$ and $f(x) = 2x^2 + 3x + 5$
$\therefore$ I = $\lim\limits_{\text{ h} \to 0}\text{h}\cdot[\text{f(0) + f(h) + f(2h) + f(3h) +.......+ f}\left\{\overline{\text{(n - 1)}}\text{h}\right\}]$
$=\lim _{\substack{h \rightarrow 0 \\ h \rightarrow \infty}} \frac{3}{n} \cdot\left[(5)+\left(2 h^2+3 h+5\right)+\left(2.2^2 h^2+3.2 h+5\right)+\ldots \ldots \ldots+\left\{2(n-1)^2 h^2+3(n-1) h+5\right\}\right]$
$=\lim\limits_{\text{n} \to \infty} [\frac{3}{\text{n}}\cdot [(5 + 5 + 5 +.........n \text{terms}) + 2h^{2 }{1^{2 }+ 2^{2 }+ 3^{2 }+ ........+ (n - 1)^2} + 3h {1 + 2 + 3 + ..... (n - 1)}]$
$= \lim\limits_{\text{n} \to \infty} \frac{3} {\text{n}}\cdot \Bigg[\text{5n+2}\cdot\frac{9}{\text{n}^{2}}\cdot\text{n}\frac{\text{(n - 1)(2n - 1)}}{6}+\frac{3.3}{\text{n}}\cdot\frac{\text{n(n - 1)}}{2}\Bigg]$
$= \lim\limits_{\text{n} \to \infty} 3\Bigg[5+3\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)+\frac{9}{2}\Big(1-\frac{1}{\text{n}}\Big)\Bigg]$
$= 3\Big[5+6+\frac{9}{2}\Big]=\frac{93}{2}.$
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Prove that $\int\limits_{0}^{\text{a}}\text{f(x)dx}=\int\limits_{0}^{\text{a}}\text{f (a - x) dx.}$
Hence, evaluate $\int\limits_{0}^{\pi/2}\frac{\text{dx}}{\text{1 + tan x}}$.
Answer$\text{RHS}=\int\limits^{a}_{0}\text{f (a - x)dx}=-\int\limits^{a}_{0}\text{f (y)dy }\text{ }\text{ where (a - x)}=\text{y} $
$=\int\limits_{0}^{a}\text{f (y) dy}=\int\limits_{0}^{a}\text{f (x) dx}=\text{LHS}$
$\text{I}=\int\limits_{0}^{\pi/2}\frac{\text{dx}}{\text{1 + tan x}}=\int\limits_{0}^{\pi/2}\frac{\cos\text{x}}{\cos\text{x + sin x}}\text{dx}...........\text{(i)}$
$=\int\limits_{0}^{\pi/2}\frac{\cos\text{ }(\pi/2-\text{x})}{\cos(\pi/2-\text{x})+\sin(\pi/2-\text{x})}\text{dx}=\int\limits_{0}^{\pi/2}\frac{\sin\text{ x}}{\text{sin x + cos x}}\text{dx}...........\text{(ii)}$
$\Rightarrow\text{2I}=\int_{0}^{\pi/2}\text{1 dx}=[\text{x}]_0^{\pi/2}=\pi/2$
$\Rightarrow\text{I}=\pi/4.$
View full question & answer→Question 375 Marks
Evaluate:$\int\sqrt{\tan\theta}\text{ d}\theta. $
Answer$\text{I}=\text{ }\text{ }\text{ put tan}\theta=\text{x}^{2}\text{ }\text{ }\therefore\text{ }\sec^{2}\theta\text{d}\theta=2\text{x dx}\Rightarrow\text{d}\theta=\frac{\text{2x}}{\text{x}^{4}+1}\text{dx}$
$=\int\frac{\text{2x}^{2}}{\text{x}^{4}+1}\text{dx}=\int\frac{\text{x}^{2}+1}{\text{x}^{4}+1}\text{dx}+\int\frac{\text{x}^{2}-1}{\text{x}^{4}+1}\text{dx }=\int\frac{1+1/\text{x}^{2}}{\text{x}^{2}+1/\text{x}^{2}}\text{dx}+\int\frac{1-1/\text{x}^{2}}{\text{x}^{2}+1/\text{x}^{2}}\text{dx}=\text{I}_{1}+\text{I}_{2} $
$\text{I}_{1}=\int\frac{1+1/\text{x}^{2}}{\text{x}^{2}+1/\text{x}^{2}}\text{dx}=\int\frac{1+1/\text{x}^{2}}{\text{(x - 1/x)}^{2}+2}\text{dx}=\int\frac{\text{dt}}{\text{t}^{2}+2}\text{ }\text{ }\text{ }\text{where }\text{x}-\frac{1}{\text{x}}=\text{t}$
$=\frac{1}{\sqrt{2}}\tan^{-1}\frac{\text{t}}{\sqrt{2}}+\text{c}_{1}=\frac{1}{\sqrt{2}}\tan^{-1}\Bigg(\frac{\text{x}^{2}-1}{\sqrt{2}\text{x}}\Bigg)+\text{c}_{1}=\frac{1}{\sqrt{2}}\tan^{-1}\Bigg(\frac{\tan\theta-1}{\sqrt{2\tan\theta}}\Bigg)+\text{c}_{1}$
$\text{I}_{2}=\int\frac{1-1/\text{x}^{2}}{\text{x}^{2}+1/\text{x}^{2}}\text{dx}=\int\frac{1-1/\text{x}^{2}}{\text{(x + 1/x)}^{2}-2}\text{dx}=\frac{\text{dz}}{\text{z}^{2}-2}\text{ }\text{ }\text{ }\text{ where x}+\frac{1}{\text{x}}=\text{z}$
$=\frac{1}{\sqrt{2}}\log\Bigg(\frac{\text{z}-\sqrt{2}}{\text{z}+\sqrt{2}}\Bigg)+\text{c}_{2}=\frac{1}{2\sqrt{2}}\log\Bigg(\frac{\text{x}^{2}+1-\sqrt{\text{2x}}}{\text{x}^{2}+1-\sqrt{\text{2x}}}\Bigg)+\text{c}_{2}=\frac{1}{2\sqrt{2}}\log\Bigg(\frac{\tan\theta+1-\sqrt{2\tan\theta}}{\tan\theta+1-\sqrt{2\tan\theta}}\Bigg)+\text{c}_{2}$
$\therefore\text{ }\text{ }\text{ I}=\frac{1}{\sqrt{2}}\tan^{-1}\Bigg(\frac{\tan\theta-1}{\sqrt{2\tan\theta}}\Bigg)+\frac{1}{2\sqrt{2}}\log\Bigg(\frac{\tan\theta+1-\sqrt{2\tan\theta}}{\tan\theta+1-\sqrt{2\tan\theta}}\Bigg)2+\text{c}.$
View full question & answer→Question 385 Marks
Evaluate $\int\limits_{0}^{2}\text{(x}^{2}+\text{x}+2)\text{dx}$ as limit of sums.
Answer$\int\limits_0^2\text{f(x) dx}=\lim_{\substack{ \text{h}\rightarrow0\\ \text{or n}\rightarrow\infty }} \text{h[f (0) + f(0 + h) + f (0 + 2h) + ...........+ f (0 + (n - 1)h]} $
where $f(x) = x^2 + x + 2, \text{h}=\frac{2}{\text{n}}$
$=\lim\limits_{\text{n}\rightarrow\infty}\text{h [ 2 + (h}^{2}\text{h + 2) + (2}^{2}\text{h}^{2}+\text{2h + 2) +......+((n - 1)}^{2}\text{h}^{2}+\text{(n - 1) h + 2)}]$
$=\lim\limits_{\text{n}\rightarrow\infty}\text{h [2n + h}^{2}\text{(1}^{2}+2^{2}+3^{3}+........(\text{n - 1))}^{2}+\text{h}(1+2+3+......+\text{(n - 1))}]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\Bigg[\text{2n}+\frac{4}{\text{n}^{2}}\cdot\frac{\text{n(n - 1)(2n - 1)}}{6}+\frac{2}{\text{n}}\cdot\frac{\text{n (n -1)}}{2}\Bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\Bigg[2+\frac{2}{3}\Bigg(1-\frac{1}{\text{n}}\Bigg)\Bigg(2-\frac{1}{\text{n}}\Bigg)+\Bigg(1-\frac{1}{\text{n}}\Bigg)\Bigg]$
$=2\Bigg[2+\frac{4}{3}+1\Bigg]=\frac{26}{3}.$
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Find:
$\int \frac{\text{e}^{\text{x}}}{(2 + \text{e}^{\text{x}}) (4 + \text{e}^{2\text{x}})} \text{dx}.$
Answer$\text{I} = \int \frac{\text{dt}}{\text{(2 + t) (4 +} \text{t}^{2})} \text{ } \text{where} \text{ e}^{\text{x}} = \text{t}$
$\text{Now}, \frac{1}{\text{(2 + t) (4 + t}^{2})} = \frac{1}{8 (2 + \text{t})} - \frac{1}{8} \bigg(\frac{\text{t - 2}}{\text{4 + t}^{2}}\bigg)$
$\Rightarrow \int \frac{\text{dt}}{\text{(2 + t) (4 + t}^{2})} = \frac{1}{8} \log | \text{2 + t|} = \frac{1}{16} \log |\text{4 + t}^{2}| + \frac{1}{8} \tan^{-1} \bigg(\frac{\text{t}}{2}\bigg) + \text{c}$
$\Rightarrow \int \frac{\text{e}^{\text{x}}\text{dx}}{(2 + \text{e}^{\text{x}}) (4 + \text{e}^{2\text{x}})} = \frac{1}{8} \log |\text{2 + e}^{\text{x}}| - \frac{1}{16} \log \text{|4 + e}^{\text{2x}}| + \frac{1}{8} \tan^{-1} \bigg(\frac{\text{e}^{\text{x}}}{2}\bigg) + \text{c}$
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Find: $\int (2x + 5) \sqrt{10 - 4x - 3x^{2}} \text{ dx}$
Answer$\text{I} = \int {(\text{2x + 5)}} \sqrt{10 - \text{4x - 3x}^{2}} \text{dx}$
$= -\frac{1}{3} \int(-4-6\text{x}) \sqrt{10 - \text{4x - 3x}^{2}} \text{dx} + \frac{11}{3} \int \sqrt{10 - \text{4x - 3x}^{2}} \text{ dx}$
$= -\frac{2}{9} (10 - \text{4x - 3x}^{2})^{3/2} + \frac{11\sqrt{3}}{3} \int \sqrt{\bigg(\frac{\sqrt{34}}{3}\bigg)^{2} - \bigg(\text{x} - \frac{2}{3}\bigg)^{2}} \text{dx}$
$= -\frac{2}{9} (10 - \text{4x - 3x}^{2})^{3/2} + \frac{11\sqrt{3}}{3}\Bigg[\frac{\big(\text{x} - \frac{2}{3}\big) - \sqrt{\big(\frac{\sqrt{34}}{3}\big)^{2} - \big(\text{x} - \frac{2}{3}\big)^{2}}}{2} + \frac{17}{9} \sin^{-1} \frac{3\text{x - 2}}{\sqrt{34}} \Bigg] + \text{C}$
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Evaluate:
$\int\limits^{1}_{-2} \big|\text{x}^{3} - \text{x}\big| \text{ dx}$
Answer$\int\limits^{1}_{-2} \big|\text{x}^{3} - \text{x}\big| \text{ dx} = \int\limits^{-1}_{-2} - \text{(x}^{3} - \text{x})\text{dx} + \int\limits^{0}_{-1} \text{(x}^{3} - \text{x}) \text{dx} - \int\limits^{1}_{0} \text{(x}^{3} - \text{x}) \text{dx}$
$= \Bigg| \frac{\text{x}^{2}}{2} - \frac{\text{x}^{4}}{4}\Bigg|^{1}_{-2} + \Bigg|\frac{\text{x}^{4}}{4} - \frac{\text{x}^{2}}{2}\Bigg|^{0}_{-1} - \Bigg|\frac{\text{x}^{4}}{4} - \frac{\text{x}^{2}}{2}\Bigg|^{1}_{0}$
$= \frac{11}{4}$
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$\text{Find:}\int \frac{\text{x}\ \text{dx}}{(2 + \text{x}^{\text{2}}) (4 + \text{e}^{4\text{}})}$
Answer$\text{I} =\frac{1}{2} \int \frac{\text{1}}{\text{(2 + t) (4 +} \text{t}^{2})}\text{dt} \text{ } \text{where} \ \text{t} = \text{x}^2$
$\text{Now},\Bigg[ \frac{1}{\text{(2 + t) (4 + t}^{2})}\Bigg] =\frac{1}{2}\Bigg[ \frac{1}{8 (2 + \text{t})} - \frac{1}{8} \bigg(\frac{\text{t - 2}}{\text{4 + t}^{2}}\bigg)\Bigg]$
$\Rightarrow \int \frac{\text{1}}{\text{(2 + t) (4 + t}^{2})} \text{dt}= \frac{1}{16} \log | \text{2 + t|} = -\frac{1}{32} \log |\text{4 + t}^{2}| + \frac{1}{16} \tan^{-1} \bigg(\frac{\text{t}}{2}\bigg) + \text{c}$
$\Rightarrow \int \frac{\text{x}}{(2 + \text{x}^{\text{2}}) (4 + \text{x}^{4\text{}})}\text{dx} = \frac{1}{16} \log |\text{2 + x}^{\text{2}}| - \frac{1}{32} \log \text{|4 + x}^{\text{4}}| + \frac{1}{16} \tan^{-1} \bigg(\frac{\text{x}^{\text{2}}}{2}\bigg) + \text{c}$
View full question & answer→Question 435 Marks
Find: $\int \frac{x \sin^{-1} x}{\sqrt{1 - x^{2}}} \text{d}x.$
Answer$\text{I} = \int \frac{\text{x} \sin^{-1} \text{x}}{\sqrt{1 - \text{x}^{2}}} \text{ dx}$
$\text{put} \sin^{-1} \text{x = t} \Rightarrow \frac{\text{dx}}{\sqrt{1 - \text{x}^{2}}} = \text{dt}$
$= \int \text{t.} \sin \text{t dt}$
$ = \text{ - t} \cos \text{t} + \sin \text{t + c}$
$= -\sqrt{1 - \text{x}^{2}} \sin^{-1} \text{x + x + c}$
View full question & answer→Question 445 Marks
Find:
$\int \text{e}^{2\text{x}} \sin \text{(3x + 1)} \text{ dx}$
Answer$\text{I} = \int \text{e}^{2\text{x}} \sin \text{(3x + 1)} \text{ dx}$
$= \sin \text{(3x + 1)}. \frac{\text{e}^{2x}}{2} - \int 3 \cos \text{(3x + 1)} . \frac{\text{e}^{2\text{x}}}{2} \text{dx}$
$= \frac{\text{e}^{\text{2x}}}{2}. \sin \text{(3x + 1)} - \frac{3}{2} \bigg[\cos \text{(3x + 1)} . \frac{\text{e}^{\text{2x}}}{2} - \int -3 \sin \text{(3x + 1)}. \frac{\text{e}^{\text{2x}}}{2} \text{dx}\bigg]$
$= \frac{\text{e}^{\text{2x}}}{2} \sin \text{(3x + 1)} - \frac{3}{4} \cos \text{(3x + 1)} . \text{e}^{\text{2x}} - \frac{9}{4} \text{I + c}$
$\Rightarrow \frac{13}{4} \text{I} = \frac{\text{e}^{\text{2x}}}{4} [2 \sin \text{(3x + 1)} - 3 \cos \text{(3x + 1)}] + \text{c}$
$\Rightarrow \text{I} = \frac{\text{e}^{\text{2x}}}{13} [ 2 \sin \text{(3x + 1)} - 3 \cos \text{(3x + 1)}] + \text{c}$
View full question & answer→Question 455 Marks
Evaluate: $\int\limits^{\pi}_{0} \frac{\text{x}}{1 + \sin \alpha \sin x} \text{dx}.$
Answer$\text{I} = \int\limits^{\pi}_{0} \frac{(\pi - x)}{1 + \sin \alpha \sin (\pi - x)} \text{dx}$
$\text{2I} = \pi \int\limits^{\pi}_{0} \frac{\text{dx}}{1 + \sin \alpha \sin \text{x}}$
$= 2\pi \int\limits^{\pi/2}_{0} \frac{\text{dx}}{1 + \sin \alpha \sin \text{x}}$
$\frac{\text{dx}}{1 + \sin \alpha \frac{2\tan\frac{\text{x}}{2}}{1 + \tan^{2} \frac{\text{x}}{2}}}$
$\text{I} = \pi \int\limits^{1}_{0}\frac{\text{2dt}}{1 + \text{t}^{2} + 2\text{t} \sin \alpha} \text{ }\text{ }\text{ }\text{ }\text{put} \tan \frac{\text{x}}{2} = \text{t}$
$\Rightarrow \text{I} = 2\pi \int\limits^{1}_{0} \frac{\text{dt}}{\text{(t} + \sin \alpha)^{2} + \cos^{2} \alpha}$
$= \frac{2\pi}{\cos \alpha} \bigg[\tan^{-1}\bigg(\frac{\text{t} + \sin \alpha}{\cos \alpha}\bigg) \bigg]^{1}_{0}$
$\Rightarrow \text{I} = \frac{\pi}{\cos \alpha} \bigg(\frac{\pi}{2} - \alpha\bigg)$
View full question & answer→Question 465 Marks
$\text{Find}:\int\frac{(3\sin x-2)\cos x}{13\ -\ \cos^2x \ - \ 7\sin x}\text{d}x$
Answer$\int\frac{(3\sin\text{x}-2)\cos\text{x}}{13\ -\ \cos^2\text{x}\ -\ 7\sin\text{x}}\text{dx}$
$\int\frac{(3\sin\text{x}-2)\cos\text{x}}{\sin^2\text{x}\ -\ 7\sin\text{x}\ +12}\text{dx}$
put sin x = y, cos x dx = dy
$=\int\frac{(3\text{y}-2)\text{dy}}{\text{y}^2-7\text{y}+12}$
$=\int\frac{(3\text{y}-2)\text{dy}}{(\text{y}-4)(\text{y}-3)}$
$=\int\Big(\frac{10}{\text{y}-4}-\frac{7}{\text{y}-3}\Big)\text{dy}$
= 10 log | y – 4 | –7 log | y – 3 | + C
= 10 log | sin x – 4 | – 7 log | sin x – 3 | + C
View full question & answer→Question 475 Marks
Evaluate: $\int\limits^{3/2}_{0} |x \sin \pi \text{ } x| \text{dx}.$
Answer$\text{I} = \int\limits^{3/2}_{0} | \text{x} \sin \pi \text{ x}| \text{dx}$
$= \int\limits^{1}_{0} \text{x} \sin \pi \text{ x} . \text{dx} - \int\limits^{3/2}_{1} \text{x} \sin \pi \text{x dx}$
$= \bigg[-\text{x} \frac{\cos \pi \text{x}}{\pi} + \frac{\sin \pi\text{x}}{\pi^{2}} \bigg]^{1}_{0} - \bigg[-\frac{\text{x}\cos \pi \text{x}}{\pi} + \frac{\sin \pi \text{x}}{\pi}^{2} \bigg]^{3/2}_{1}$
$= \frac{2}{\pi} + \frac{1}{\pi^{2}}$
View full question & answer→Question 485 Marks
Evaluate: $\int\limits^{\pi}_{0} \frac{x \sin x}{1 + \cos^{2} x} \text{d}x.$
Answer$\text{I} = \int\limits^{\pi}_{0} \frac{\text{x} \sin \text{x}}{1 + \cos^{2} \text{x}} \text{dx}$
$= \int\limits^{\pi}_{0} \frac{(\pi - \text{x}) \sin \text{x}}{1 + \cos^{2} \text{x}} \text{x}$
$\Rightarrow \text{2I} = \pi \int\limits^{\pi}_{0} \frac{\sin \text{x dx}}{1 + \cos^{2}\text{x}}$
$\text{Put} \cos \text{x = t and} -\sin \text{x dx = dt}$
$= -\pi \int\limits^{-1}_{1} \frac{\text{dt}}{1 + \text{t}^{2}}$
$= \pi [ \tan^{-1} \text{t}]^{1}_{-1} = \frac{\pi^{2}}{2}$
$\Rightarrow \text{I} = \frac{\pi^{2}}{4}$
View full question & answer→Question 495 Marks
$\text{Find} \int \frac{\sqrt{x}}{\sqrt{\text{a}^{3}} - \text{x}^{3}}\text{dx}.$
Answer$\text{I} = \int \frac{\sqrt{x}}{\sqrt{\text{a}^{3}} - \text{x}^{3}}\text{dx}$
$\text{Put x}^{3/2} = \text{t}\Rightarrow\frac{3}{2}.\text{x}^{1/2}\text{dx = dt or}\sqrt{\text{x }} dx= \frac{3}{2}\text{dt}$
$\text{I} = \frac{2}{3}\int\frac{\text{dt}}{\sqrt{\text{(a}^{3/2})^{2} - \text{t}^{2}}}$
$= \frac{2}{3}.\sin^{-1}\bigg(\frac{\text{t}}{\text{a}^{3/2}}\bigg)+\text{C}$
$= \frac{2}{3} \sin^{-1}\bigg(\frac{\text{x}^{3/2}}{\text{a}^{3/2}}\bigg)+ \text{C}$
View full question & answer→Question 505 Marks
Integrate the following w.r.t. x
$\frac{x^{2} - 3x + 1}{\sqrt{1 - x^{2}}}$
Answer$\int\frac{\text{x}^{2} - 3\text{x} + 1}{\sqrt{1 - \text{x}^{2}}} \text{dx} = \int\frac{2 - 3\text{x} - (1 - \text{x}^{2})}{\sqrt{1 - \text{x}^{2}}} \text{dx}$
$= 2\int\frac{1}{\sqrt{1 - \text{x}^{2}}} \text{dx} -3 \int\frac{\text{x}}{\sqrt{1 - \text{x}^{2}}} \text{dx} - \int\sqrt{1 - \text{x}^{2}} \text{dx}$
$= 2 \sin^{-1}\text{x} + 3 \sqrt{1 - \text{x}^{2}} - \frac{\text{x}}{2} \sqrt{1 - {\text{x}^{2}}} - \frac{1}{2} \sin^{-1} \text{x + c}$
$\text{or} = \frac{3}{2} \sin^{-1} \text{x} + \frac{1}{2} (6 - \text{x}) \sqrt{1 - \text{x}^{2}} + \text{c}$
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