Question 12 Marks
Evaluate the integral $\int_{-1}^{1} \frac{d x}{x^{2}+2 x+5}$ using substitution.
AnswerGiven: $\int_{-1}^{1} \frac{d x}{x^{2}+2 x+5}$
= $\int_{-1}^{1} \frac{d x}{\left(x^{2}+2 x+1\right)+4}$
= $\int_{-1}^{1} \frac{d x}{(x+1)^{2}+(2)^{2}}$
Let x + 1 = t
⇒ dx = dt
When x = -1, t = 0 and when x = 1, t = 2
$\Rightarrow \int_{-1}^{1} \frac{d x}{(x+1)^{2}+(2)^{2}}=\int_{0}^{2} \frac{d t}{(t)^{2}+(2)^{2}}$
because, $\int \frac{d t}{x^{2}+a^{2}}=\frac{1}{a} \cdot \tan ^{-1} \frac{x}{a}+C$
$\Rightarrow \int_{0}^{2} \frac{d t}{(t)^{2}+(2)^{2}}=\left[\frac{1}{2} \tan ^{-1} \frac{t}{2}\right]_{0}^{2}$
= $\frac{1}{2} \tan ^{-1} 1-\frac{1}{2} \tan ^{-1} 0$
= $\frac{1}{2}\left(\frac{\pi}{4}\right)=\frac{\pi}{8}$
View full question & answer→Question 22 Marks
Evaluate the integral $\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} d x$ using substitution.
AnswerGiven: $\int_{0}^{2} \frac{d x}{x+4-x^{2}}$
$\int_{0}^{2} \frac{d x}{x+4-x^{2}}=\int_{0}^{2} \frac{d x}{-\left(x^{2}-x-4\right)}$
We can write it as, $\int_{0}^{2} \frac{d x}{-\left(x^{2}-x+\frac{1}{4}-\frac{1}{4}-4\right)}$
= $\int_{0}^{2} \frac{d x}{-\left[\left(x-\frac{1}{2}\right)^{2}-\frac{17}{4}\right]}$
= $\int\limits_0^2 {\frac{{dx}}{{\left[ {{{\left( {\frac{{\sqrt {17} }}{2}} \right)}^2} - {{\left( {x - \frac{1}{2}} \right)}^2}} \right]}}}$
Let $x-\frac{1}{2}=t \Rightarrow d x=d t$
when x = 0, $t=-\frac{1}{2}$ and when x = 2, $t=\frac{3}{2}$
$\Rightarrow \int_{0}^{2} \frac{d x}{\left[\left(\frac{\sqrt{17}}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}\right]}$ = $\int_{-\frac{1}{2}}^{\frac{3}{2}} \frac{d t}{\left(\frac{\sqrt{17}}{2}\right)^{2}-(t)^{2}}$
{because, $\int \frac{d x}{\left[(a)^{2}-(x)^{2}\right]}$ = $\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+C$ }
= $\Rightarrow \int_{-\frac{1}{2}}^{\frac{3}{2}} \frac{d t}{\left[\left(\frac{\sqrt{17}}{2}\right)^{2}-(t)^{2}\right]}$ = $\left[\frac{1}{2\left(\frac{\sqrt{17}}{2}\right)} \log \frac{\left(\frac{\sqrt{17}}{2}+t\right)}{\frac{\sqrt{17}}{2}-t}\right]_{-\frac{1}{2}}^{\frac{3}{2}}$
= $\frac{1}{\sqrt{17}}\left[\log \frac{\left(\frac{\sqrt{17}}{2}+\frac{3}{2}\right)}{\frac{\sqrt{17}}{2}-\frac{3}{2}}-\log \frac{\left(\frac{\sqrt{17}}{2}-\frac{1}{2}\right)}{\frac{\sqrt{17}}{2}+\frac{1}{2}}\right]$
= $\frac{1}{\sqrt{17}}\left[\log \frac{(\sqrt{17}+3)}{\sqrt{17}-3}-\log \frac{(\sqrt{17}-1)}{\sqrt{17}+1}\right]$
= $\frac{1}{\sqrt{17}}\left[\log \left\{\frac{(\sqrt{17}+3)}{\sqrt{17}-3} \times \frac{(\sqrt{17}+1)}{\sqrt{17}-1}\right\}\right]$
= $\frac{1}{\sqrt{17}}\left[\log \left\{\frac{(\sqrt{17}+3)(\sqrt{17}+1)}{(\sqrt{17}-3)(\sqrt{17}-1)}\right\}\right]$
= $\frac{1}{\sqrt{17}} \log \left[\frac{17+3+4 \sqrt{17}}{17+3-4 \sqrt{17}}\right]$
= $\frac{1}{\sqrt{17}} \log \left[\frac{20+4 \sqrt{17}}{20-4 \sqrt{17}}\right]$
= $\frac{1}{\sqrt{17}} \log \left[\frac{5+\sqrt{17}}{5-\sqrt{17}}\right]$
= $\frac{1}{\sqrt{17}} \log \left[\frac{(5+\sqrt{17})(5+\sqrt{17})}{(5-\sqrt{17})(5+\sqrt{17})}\right]$
= $\frac{1}{\sqrt{17}} \log \left[\frac{(25+17+10 \sqrt{17})}{25-17}\right]$
= $\frac{1}{\sqrt{17}} \log \left[\frac{(42+10 \sqrt{17})}{8}\right]$
= $\frac{1}{\sqrt{17}} \log \left[\frac{(21+5 \sqrt{17})}{4}\right]$
View full question & answer→Question 32 Marks
Evaluate the integral $\int_{0}^{1} \frac{x}{x^{2}+1} d x$ using substitution.
AnswerGiven integral is: $\int_{0}^{1} \frac{x}{x^{2}+1} d x$
Let $x^2 + 1 = t$
$\Rightarrow 2xdx = dt$
⇒ xdx = $\frac{1}{2}$ dt
When x = 0, t = 1 and when x = 1, t = 2
$\Rightarrow \int_{0}^{1} \frac{x}{x^{2}+1} d x=\int_{1}^{2} \frac{d t}{2 t}$
$=\frac{1}{2} \int_{1}^{2} \frac{d t}{t}$
$=\frac{1}{2}\left[\log |t| ]_{1}^{2}\right.$
= $\frac{1}{2}[\log 2-\log 1]$
= $\frac{1}{2} \log 2$
View full question & answer→Question 42 Marks
Evaluate the definite integral $\int_0^1 \frac{d x}{\sqrt{1-x^2}}$
AnswerAccording to the question , $I = \int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 1 - x ^ { 2 } } } d x$
= $\left[ \sin ^ { - 1 } x \right] _ { 0 } ^ { 1 } \quad \left[ \because \int_b^a \frac { 1 } { \sqrt { 1 - x ^ { 2 } } } d x = \sin ^ { - 1 } (a) - \sin ^ { - 1 } (b) \right]$
$= sin^{-1}(1)-sin^{-1}(0)$
$= \sin ^ { - 1 } \left( \sin \frac { \pi } { 2 } \right) - \sin ^ { - 1 } ( \sin 0 ) $ $= \frac { \pi } { 2 } - 0 = \frac { \pi } { 2 }$
View full question & answer→Question 52 Marks
Evaluate the definite integral $\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\cos ecxdx} $
Answer$\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\cos ecxdx} $ $= \left( {\log \left| {\cos ecx - \cot x} \right|} \right)_{\frac{\pi }{6}}^{\frac{\pi }{4}}$
$= \log \left| {\cos ec\frac{\pi }{4} - \cot \frac{\pi }{4}} \right| - \log \left| {\cos ec\frac{\pi }{6} - \cot \frac{\pi }{6}} \right|$
$= \log \left| {\sqrt 2 - 1} \right| - \log \left| {2 - \sqrt 3 } \right|$
$= \log \left( {\sqrt 2 - 1} \right) - \log \left( {2 - \sqrt 3 } \right)$
$= \log \left( {\frac{{\sqrt 2 - 1}}{{2 - \sqrt 3 }}} \right)$
View full question & answer→Question 62 Marks
Evaluate the definite integral $\int_0^{\frac{\pi}{4}} \tan x d x$.
Answer$\int _ { 0 } ^ { \pi / 4 } \tan x d x = [ \log | \sec x | ] _ { 0 } ^ { \pi / 4 }$
${ = \log \left| \sec \frac { \pi } { 4 } \right| - \log | \sec 0 | } \\ $
${ = \log | \sqrt { 2 } | - \log | 1 | } $
$\\ { = \frac { 1 } { 2 } \log 2 } $
View full question & answer→Question 72 Marks
Evaluate the definite integral $\int_{4}^{5} e^{x} d x$
AnswerLet $I=\int_{4}^{5} e^{x} d x$
$\Rightarrow I=\left[e^{x}\right]_{4}^{5}=e^{5}-e^{4}$
$\therefore \int_{4}^{5} \mathrm{e}^{\mathrm{x}} \mathrm{d} \mathrm{x}=\mathrm{e}^{4}(\mathrm{e}-1)$
View full question & answer→Question 82 Marks
Evaluate the definite integral$\int_{0}^{\frac{\pi}{2}} \cos 2 x d x$
AnswerLet $I=\int_{0}^{\frac{\pi}{2}} \cos 2 x d x$
$\Rightarrow \mathrm{I}=\left[\frac{\sin 2 \mathrm{x}}{2}\right]_{0}^{\pi / 2}$ [$\int \cos x d x=\sin x+c$ ]
$\Rightarrow I=\frac{1}{2}\left(\sin 2 \times \frac{\pi}{2}-\sin 2 \times 0\right)$
$\Rightarrow \mathrm{I}=\frac{1}{2}(\sin \pi-\sin 0)$
$\Rightarrow$ I = $\frac{1}{2}$(0 - 0) = 0
$\therefore \int_{0}^{\frac{\pi}{2}} \cos 2 x d x=0$
View full question & answer→Question 92 Marks
Evaluate the definite integral $\int\limits_0^{\frac{\pi }{4}} {\sin 2xdx}$
Answer$\int\limits_0^{\frac{\pi }{4}} {\sin 2xdx}$ $= \left( {\frac{{ - \cos 2x}}{2}} \right)_0^{\frac{\pi }{4}}$
$= \frac{{ - \cos \frac{\pi }{2}}}{2} - \left( {\frac{{ - \cos {0^o}}}{2}} \right)$
$ = \frac{0}{2} - \left( {\frac{{ - 1}}{2}} \right)$
$= 0 + \frac{1}{2}$
$= \frac{1}{2}$
View full question & answer→Question 102 Marks
Evaluate the definite integral $\int_{1}^{2}\left(4 x^{3}-5 x^{2}+6 x+9\right) d x$
AnswerLet $I=\int_{1}^{2}\left(4 x^{3}-5 x^{2}+6 x+9\right) d x$
$\Rightarrow \mathrm{I}=\int_{1}^{2}\left(4 \mathrm{x}^{3}-5 \mathrm{x}^{2}+6 \mathrm{x}+9\right) \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}=\int_{1}^{2} 4 \mathrm{x}^{3} \mathrm{d} \mathrm{x}-\int_{1}^{2} 5 \mathrm{x}^{2} \mathrm{dx}+\int_{1}^{2} 6 \mathrm{xdx}+\int_{1}^{2} 9 \mathrm{dx}$
$\Rightarrow \mathrm{I}=4 \int_{1}^{2} \mathrm{x}^{3} \mathrm{dx}-5 \int_{1}^{2} \mathrm{x}^{2} \mathrm{dx}+6 \int_{1}^{2} \mathrm{xdx}+9 \int_{1}^{2} \mathrm{d} \mathrm{x}$
$\Rightarrow$ I = $4 \times\left[\frac{x^{3+1}}{3+1}\right]_{1}^{2}-5 \times\left[\frac{x^{2+1}}{2+1}\right]_{1}^{2}+6 \times\left[\frac{x^{1+1}}{1+1}\right]_{1}^{2}+9 \times\left[\frac{x^{0+1}}{0+1}\right]_{1}^{2}$ [$\int x^{n} d x=\frac{x^{n+1}}{n+1}$ ]
$\Rightarrow $ I = $4 \times\left[\frac{\mathrm{x}^{4}}{4}\right]_{1}^{2}-5 \times\left[\frac{\mathrm{x}^{3}}{3}\right]_{1}^{2}+6 \times\left[\frac{\mathrm{x}^{2}}{2}\right]_{1}^{2}+9 \times[\mathrm{x}]_{1}^{2}$
= $2^{4}-1^{4}-5\left[\frac{2^{3}}{3}-\frac{1^{3}}{3}\right]+6\left[\frac{2^{2}}{2}-\frac{1^{2}}{2}\right]+9(2-1)$
= $16-1-5\left[\frac{7}{3}\right]+3(3)+9$
= $33-\frac{35}{3}$
= $\frac{99-35}{3}=\frac{64}{3}$
View full question & answer→Question 112 Marks
Evaluate the definite integrals $\int\limits_0^1 {\left( {x{e^x} + \sin \frac{{\pi x}}{4}} \right)dx} $
Answer$\int\limits_0^1 {\left( {x{e^x} + \sin \frac{{\pi x}}{4}} \right)dx} =$ $\int\limits_0^1 {x{e^x}dx + \int\limits_0^1 {\sin \frac{{\pi x}}{4}dx} } $
[Applying Product Rule on first definite integral]
$= \left( {x{e^x}} \right)_0^1 - \int\limits_0^1 {1.{e^x}dx - \frac{{\left( {\cos \frac{{\pi x}}{4}} \right)_0^1}}{{{ }{} {\text{ }}\frac{\pi }{4}}}} $
$ = {e^1} - 0 - \int\limits_0^1 {{e^x}dx - \frac{4}{\pi }\left[ {\cos \frac{\pi }{4} - \cos {0^o}} \right]} $
$= e - \left( {{e^x}} \right)_0^1 - \frac{4}{\pi }\left( {\frac{1}{{\sqrt 2 }} - 1} \right)$
$ = e - \left( {e - {e^0}} \right) - \frac{4}{{\pi \sqrt 2 }} + \frac{4}{\pi }$
$ = e - e + 1 - \frac{{2.2}}{{\pi \sqrt 2 }} + \frac{4}{\pi }$
$ = 1 + \frac{4}{\pi } - \frac{{2\sqrt 2 }}{\pi }$
View full question & answer→Question 122 Marks
Evaluate the definite integral $\int _ { 2 } ^ { 3 } \frac { 1 } { x } d x.$
AnswerAccording to the question , $I =\int _ { 2 } ^ { 3 } \frac { 1 } { x } d x$
$ = [ \log | x | ] _ { 2 } ^ { 3 }$
$ = \log 3 - \log 2 $
$= \log \frac { 3 } { 2 }\left[ \because \log m - \log n = \log \frac { m } { n } \right]$
View full question & answer→Question 132 Marks
Evaluate the definite integral $\int_{0}^{2} \frac{6 x+3}{x^{2}+4} d x$
AnswerLet $I=\int_{0}^{2} \frac{6 x+3}{x^{2}+4} d x$
$I=3 \int_{0}^{2} \frac{2 x+1}{x^{2}+4}=3 \int_{0}^{2} \frac{2 x}{x^{2}+4} d x+3 \int_{0}^{2} \frac{1}{x^{2}+4} d x$
$\therefore I = I_1 + I_2$
$I_{1}=3 \int_{0}^{2} \frac{2 x}{x^{2}+4} d x$
Let $x^2 + 4 = t$
2x dx = dt
When x = 0; t = 4
When $x = 2; t = 2^2 + 4 = 8$
Substituting t and dt in $I_1$
$\left.\Rightarrow \mathrm{I}_{1}=3 \int_{4}^{8} \frac{\mathrm{dt}}{\mathrm{t}}=3[\log | \mathrm{t}]\right]_{4}^{8}$ [$\int \frac{1}{x} d x=\log x$]
$\Rightarrow I_1$ = 3 [log |8| - log |4|] = 3 log $\frac{8}{4}$
$\Rightarrow I_1$ = 3 log |2|
$\mathrm{I}_{2}=3 \int_{0}^{2} \frac{1}{\mathrm{x}^{2}+4} \mathrm{dx}=3 \int_{0}^{2} \frac{1}{\mathrm{x}^{2}+2^{2}} \mathrm{d} \mathrm{x}$ [$\frac{d x}{a^{2}+x^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c$]
$\Rightarrow I_{2}=3 \times \frac{1}{2}\left[\tan ^{-1} \frac{x}{2}\right]_{0}^{2}$ = $\frac{3}{2}\left[\tan ^{-1} \frac{2}{2}-\tan ^{-1} \frac{0}{2}\right]$ = $\frac{3}{2}\left[\tan ^{-1} 1-\tan ^{-1} 0\right]$
$\Rightarrow I_{2}=\frac{3}{2} \times \frac{\pi}{4}=\frac{3 \pi}{ 8}$
Now $I = I_1 + I_2$
I = 3 log 2 + $\frac{3\pi}{8}$
$\int_{0}^{2} \frac{6 x+3}{x^{2}+4} d x=3 \log 2+\frac{3 \pi}{8}$
View full question & answer→Question 142 Marks
Evaluate the definite integral $\int\limits_0^\pi {\left( {{{\sin }^2}\frac{x}{2} - {{\cos }^2}\frac{x}{2}} \right)dx} $
Answer$\int\limits_0^\pi {\left( {{{\sin }^2}\frac{x}{2} - {{\cos }^2}\frac{x}{2}} \right)dx} $ $= \int\limits_0^\pi {\left( {\frac{{1 - \cos x}}{2} - \frac{{1 + \cos x}}{2}} \right)dx} $
$= \int\limits_0^\pi {\left( {\frac{{1 - \cos x - 1 - \cos x}}{2}} \right)dx} $
$ = \int\limits_0^\pi {\left( {\frac{{ - 2\cos x}}{2}} \right)dx} $
$ = - \int\limits_0^\pi {\cos xdx} $
$ = - \left( {\sin x} \right)_0^\pi $
$= - \left( {\sin \pi - \sin {0^o}} \right)$
= -(0 - 0) = 0
View full question & answer→Question 152 Marks
Evaluate the definite integral $\int\limits_0^{\frac{\pi }{4}} {\left( {2{{\sec }^2}x + {x^3} + 2} \right)dx} $
Answer$\int\limits_0^{\frac{\pi }{4}} {\left( {2{{\sec }^2}x + {x^3} + 2} \right)dx} $
$= 2\int\limits_0^{\frac{\pi }{4}} {{{\sec }^2}xdx + \int\limits_0^{\frac{\pi }{4}} {{x^3}dx + 2\int\limits_0^{\frac{\pi }{4}} {1dx} } } $
$= 2\left( {\tan x} \right)_0^{\frac{\pi }{4}} + \left( {\frac{{{x^4}}}{4}} \right)_0^{\frac{\pi }{4}} + 2\left( x \right)_0^{\frac{\pi }{4}}$
$= 2\left( {\tan \frac{\pi }{4} - \tan {0^o}} \right) + \frac{{{{\left( {\frac{\pi }{4}} \right)}^4}}}{4} - 0 $ $+ 2\left( {\frac{\pi }{4} - 0} \right)$
$= 2\left( {1 - 0} \right) + \frac{{\left( {\frac{{{\pi ^4}}}{{256}}} \right)}}{4} + \frac{{2\pi }}{4}$
$= 2 + \frac{{{\pi ^4}}}{{1024}} + \frac{\pi }{2}$
$= \frac{{{\pi ^4}}}{{1024}} + \frac{\pi }{2} + 2$
View full question & answer→Question 162 Marks
Evaluate the definite integral $ \int _ { 1 } ^ { 2 } \frac { 5 x ^ { 2 } } { x ^ { 2 } + 4 x + 3 }$
AnswerAccording to the question, $ I = \int _ { 1 } ^ { 2 } \frac { 5 x ^ { 2 } } { x ^ { 2 } + 4 x + 3 } d x$
$ I = 5 \int _ { 1 } ^ { 2 } \left( 1 + \frac { - 4 x - 3 } { x ^ { 2 } + 4 x + 3 } \right) d x$
$ = 5 \int _ { 1 } ^ { 2 } d x - 5 \int _ { 1 } ^ { 2 } \frac { 4 x + 3 } { x ^ { 2 } + 4 x + 3 } d x$
$ \Rightarrow \quad I = 5 [ x ] _ { 1 } ^ { 2 } - 5 \int _ { 1 } ^ { 2 } \frac { 4 x + 3 } { ( x + 3 ) ( x + 1 ) } d x$
Using partial fraction,
let $ \frac { 4 x + 3 } { ( x + 3 ) ( x + 1 ) } = \frac { A } { x + 3 } + \frac { B } { x + 1 }$
$ \Rightarrow \quad \frac { 4 x + 3 } { ( x + 3 ) ( x + 1 ) } = \frac { A ( x + 1 ) + B ( x + 3 ) } { ( x + 3 ) ( x + 1 ) }$
$ \Rightarrow$ $4x + 3 = A(x + 1) + B(x + 3)$
Comparing the coefficients of like from both sides,
$\Rightarrow A + B = 4 \Rightarrow A = 4 - B$
and A + 3B = 3$ \Rightarrow$ $4 - B + 3B = 3$
$ \Rightarrow \quad B = - \frac { 1 } { 2 },$then $ A = 4 + \frac { 1 } { 2 } = \frac { 9 } { 2 }$
Now, from Equation (i), we get
$ I = 5 ( 2 - 1 ) - 5 \int _ { 1 } ^ { 2 } \left( \frac { 9 / 2 } { x + 3 } + \frac { - 1 / 2 } { x + 1 } \right) d x$
$ = 5 - 5 \left[ \frac { 9 } { 2 } \log | x + 3 | - \frac { 1 } { 2 } \log | x + 1 | \right] _ { 1 } ^ { 2 }$
$ = 5 - 5\left. {\left[ {\left( {\frac{9}{2}\log 5 - \frac{1}{2}\log 3} \right)} \right. - \left( {\frac{9}{2}\log 4 - \frac{1}{2}\log 2} \right)} \right]$
$ = 5 - 5 \left[ \frac { 9 } { 2 } ( \log 5 - \log 4 ) - \frac { 1 } { 2 } ( \log 3 - \log 4 ]\right.$
$ = 5 - 5 \left[ \frac { 9 } { 2 } \log \frac { 5 } { 4 } - \frac { 1 } { 2 } \log \frac { 3 } { 2 } \right]$
$ = 5 - \frac { 45 } { 2 } \log \frac { 5 } { 4 } + \frac { 5 } { 2 } \log \frac { 3 } { 2 }$
View full question & answer→Question 172 Marks
Evaluate the definite integral $\int_{0}^{1} \frac{2 x+3}{5 x^{2}+1} d x$
AnswerLet $I=\int_{0}^{1} \frac{2 x+3}{5 x^{2}+1}$
Multiplying by 5 in numerator and denominator
$I=\frac{1}{5} \int_{0}^{1} \frac{5(2 x+3)}{5 x^{2}+1} d x=\frac{1}{5} \int_{0}^{1} \frac{10 x+15}{5 x^{2}+1} d x$
$\Rightarrow I=\frac{1}{5} \int_{0}^{1} \frac{10 x}{5 x^{2}+1} d x+3 \int_{0}^{1} \frac{5}{5 x^{2}+1}$
$\Rightarrow I = I_1 + I_2$
$I_{1}=\frac{1}{5} \int_{0}^{1} \frac{10 x}{5 x^{2}+1} d x$
Let $5x^2 + 1 = t …(i)$
10x dx = dt …(ii)
When $x = 0; t = 1$
When $x = 1; t = 6$
Substituting (i) and (ii) in $I_1$
$I_{1}=\frac{1}{5} \int_{1}^{6} \frac{d t}{t}=\frac{1}{5}[\log | t|]_{1}^{6}$
$I_{1}=\frac{1}{5}(\log |6|-\log |1|)=\frac{1}{5}(\log 6-0)$
$I_{1}=\frac{1}{5} \int_{0}^{1} \frac{10 x}{5 x^{2}+1} d x=\frac{\log 6}{5}$
$\mathrm{I}_{2}=3 \int {\frac{5}{5 \mathrm{x}^{2}+1}} \mathrm{dx}=\frac{3}{5} \int_{0}^{1} \frac{1}{\mathrm{x}^{2}+\frac{1}{5}} \mathrm{d} \mathrm{x}$ [$\int \frac{d x}{a^{2}+x^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c$]
$I_{2}=\frac{3}{5} \times \frac{1}{\frac{1}{\sqrt{5}}}\left[\tan ^{-1} \sqrt{5} x\right]_{0}^{1}=\frac{3}{5} \times \sqrt{5}\left(\tan ^{-1} \sqrt{5}-\tan ^{-1} 0\right)$
$I_2$ = $\frac{3}{\sqrt{5}}\tan ^{-1} \sqrt{5}$
$\because I = I_1 + I_2$
$\therefore \mathrm{I}=\frac{1}{5} \log 6+{\frac{3}{\sqrt{5}}} \tan ^{-1} \sqrt{5}$
$\int_{0}^{1} \frac{2 x+3}{5 x^{2}+1} d x$ = $\frac{1}{5} \log 6+{\frac{3}{\sqrt{5}}} \tan ^{-1} \sqrt{5}$
View full question & answer→Question 182 Marks
Evaluate the definite integral $\int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}x dx} $
Answer$\int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}xdx} $ $= \int\limits_0^{\frac{\pi }{2}} {\frac{{1 + \cos 2x}}{2}dx} $
$= \frac{1}{2}\int\limits_0^{\frac{\pi }{2}} {\left( {1 + \cos 2x} \right)dx} $
$= \frac{1}{2}\left( {x + \frac{{\sin 2x}}{2}} \right)_0^{\frac{\pi }{2}}$
$= \frac{1}{2}\left[ {\frac{\pi }{2} + \frac{1}{2}\sin \pi - \left( {0 + \frac{1}{2}\sin {0^o}} \right)} \right]$
$= \frac{1}{2}\left[ {\frac{\pi }{2} + 0 - 0} \right]$
$ = \frac{\pi }{4}$
View full question & answer→Question 192 Marks
Evaluate the definite integral $\int\limits_2^3 {\frac{{dx}}{{{x^2} - 1}}} $
Answer$\int\limits_2^3 {\frac{{dx}}{{{x^2} - 1}}} = \int\limits_2^3 {\frac{1}{{{x^2} - {1^2}}}dx} $ $= \left( {\frac{1}{{2\left( 1 \right)}}\log \left| {\frac{{x - 1}}{{x + 1}}} \right|} \right)_2^3$
$= \frac{1}{2}\log \left| {\frac{{3 - 1}}{{3 + 1}}} \right| - \frac{1}{2}\log \left| {\frac{{2 - 1}}{{2 + 1}}} \right|$
$= \frac{1}{2}\log \left| {\frac{1}{2}} \right| - \frac{1}{2}\log \left| {\frac{1}{3}} \right|$
$= \frac{1}{2}\left( {\log \frac{1}{2} - \log \frac{1}{3}} \right)$
$= \frac{1}{2}\log \frac{{\frac{1}{2}}}{{\frac{1}{3}}}$
$ = \frac{1}{2}\log \frac{3}{2}$
View full question & answer→Question 202 Marks
Evaluate the definite integral $\int_{0}^{1} \frac{d x}{1+x^{2}}$
AnswerLet $I=\int_{0}^{1} \frac{d x}{1+x^{2}}$
$I=\int_{0}^{1} \frac{d x}{1+x^{2}}$
We know that
$\int \frac{d x}{a^{2}+x^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c$
Therefore,
I = $\left[\tan ^{-1} x\right]_{0}^{1}$
= $\tan^{-1}(1) - \tan^{-1}(0)$ = $\frac{\pi}{4}$ - 0
= $\frac{\pi}{4}$
$\therefore \int_{0}^{1} \frac{d x}{1+x^{2}}=\frac{\pi}{ 4}$
View full question & answer→Question 212 Marks
Evaluate the definite integral $\int_{-1}^{1}(x+1) d x$
AnswerLet $I=\int_{-1}^{1}(x+1) d x$
$I=\int_{-1}^{1}(x+1) d x$
$\Rightarrow \mathrm{I}=\int_{-1}^{1} \mathrm{x} \mathrm{dx}+\int_{-1}^{1} 1 \cdot \mathrm{dx} ~~~~~\left[As \int \mathrm{x}^{\mathrm{n}} \mathrm{d} \mathrm{x}=\frac{\mathrm{x}^{\mathrm{n}+1}}{\mathrm{n}+1}\right]$
$\Rightarrow I=\left[\frac{x^{2}}{2}\right]_{-1}^{1}+[x]_{-1}^{1}$
$\Rightarrow \mathrm{I}=\left[\frac{1^{2}}{2}-\frac{(-1)^{2}}{2}\right]+[1-(-1)]$
$\Rightarrow I=\left[\frac{1}{2}-\frac{1}{2}\right]+[1+1]=0+2$
$\Rightarrow$ I = 2
$\int_{-1}^{1}(x+1) d x=2$
View full question & answer→Question 222 Marks
Integrate the function $\sqrt {4 - {x^2}} $
Answer$\int {\sqrt {4 - {x^2}} dx} $ $ = \int {\sqrt {2^2 - {x^2}} dx} $
$= \frac{x}{2}\sqrt {{2^2} - {x^2}} + \frac{{{2^2}}}{2}{\sin ^{ - 1}}\frac{x}{2} + c$
$\left[ {\because \int {\sqrt {{a^2} - {x^2}} dx} } \right.$ $\left. { = \frac{x}{2}\sqrt {{a^2} - {x^2}} + \frac{{{a^2}}}{2}{{\sin }^{ - 1}}\frac{x}{a}} \right]$
$= \frac{x}{2}\sqrt {4 - {x^2}} + 2{\sin ^{ - 1}}\frac{x}{2} + c$
View full question & answer→Question 232 Marks
Integrate the rational function $\frac{1}{\left(e^{x}-1\right)}$ [Hint: $e^x = t$]
AnswerGiven function is, $\frac{1}{\left(e^{x}-1\right)}$
Let $e^x = t$
$e^x dx = dt$
$\int \frac{1}{\left(e^{x}-1\right)} d x=\int \frac{1}{t-1} \times \frac{d t}{t}=\int \frac{1}{t(t-1)} d t$
Let $\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}$
1 = A(t - 1) + Bt … (i)
Substituting t =1 and t = 0 in equation (i), we get,
A = -1 and B = 1
Therefore, $\frac{1}{t(t-1)}=\frac{-1}{t}+\frac{1}{t-1}$
$\int \frac{1}{t(t-1)} d t=\log \left|\frac{t-1}{t}\right|+C$
= $\log \left|\frac{e^{x}-1}{e^{x}}\right|+C$
View full question & answer→Question 242 Marks
Integrate the rational function $\frac{{2x}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 3} \right)}}$
AnswerLet $I = \int {\frac{{2x}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 3} \right)}}dx} $ ....(i) Putting ${x^2} = t \Rightarrow 2xdx = dt$
$\therefore$ From eq. (i),
$I = \int {\frac{{dt}}{{\left( {t + 1} \right)\left( {t + 3} \right)}}} $
$= \frac{1}{2}\int {\frac{2}{{\left( {t + 1} \right)\left( {t + 3} \right)}}dt} $
$= \frac{1}{2}\int {\frac{{\left( {t + 3} \right) - \left( {t + 1} \right)}}{{\left( {t + 1} \right)\left( {t + 3} \right)}}dt} $
$= \frac{1}{2}\int {\left( {\frac{{\left( {t + 3} \right)}}{{\left( {t + 1} \right)\left( {t + 3} \right)}} - \frac{{\left( {t + 1} \right)}}{{\left( {t + 1} \right)\left( {t + 3} \right)}}} \right)dt} $
$= \frac{1}{2}\int {\left( {\frac{1}{{\left( {t + 1} \right)}} - \frac{1}{{\left( {t + 3} \right)}}} \right)dt} $
$= \frac{1}{2}\left[ {\log \left| {t + 1} \right| - \log \left| {t + 3} \right|} \right] + c$
$= \frac{1}{2}\left[ {\log \left| {\frac{{t + 1}}{{t + 3}}} \right|} \right] + c$
$= \frac{1}{2}\log \left| {\frac{{{x^2} + 1}}{{{x^2} + 3}}} \right| + c$
$= \frac{1}{2}\log \left( {\frac{{{x^2} + 1}}{{{x^2} + 3}}} \right) + c$.
Which is the required solution.
View full question & answer→Question 252 Marks
Integrate the rational function $\frac{\cos x}{(1-\sin x)(2-\sin x)}$ [Hint: Put sin x = t]
AnswerGiven function is $\frac{\cos x}{(1-\sin x)(2-\sin x)}$
Let sin x = t
Therefore,
$\int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x=\int \frac{d t}{(1-t)(2-t)}$
Let $\frac{1}{(1-t)(2-t)}=\frac{A}{(1-t)}+\frac{B}{(2-t)}$
$\Rightarrow$ 1 = A(2 - t) + B (1 - t) …(i)
Substituting t = 2 and then t = 1 in equation (i), we get,
Therefore,
$\frac{1}{(1-t)(2-t)}=\frac{1}{(1-t)}-\frac{1}{(2-t)}$
$\Rightarrow$$\int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x=\int\left\{\frac{1}{(1-t)}-\frac{1}{(2-t)}\right\} d t$
= $-log |1-t|+\log |2-t|+C$
= $\log \left|\frac{2-t}{1-t}\right|+C$
= $\log \left|\frac{2-\sin x}{1-\sin x}\right|+C$
View full question & answer→Question 262 Marks
Integrate the rational function $\frac{3 x-1}{(x+2)^{2}}$
AnswerLet $\frac{3 x-1}{(x+2)^{2}}=\frac{A}{(x+2)}+\frac{B}{(x+2)^{2}}$
$\Rightarrow$ 3x -1 = A(x + 2) + B
Equating the coefficients of x and constant term, we get,
A = 3
2A + B = -1
B = -7
Thus,
$\frac{3 x-1}{(x+2)^{2}}=\frac{3}{(x+2)}-\frac{7}{(x+2)^{2}}$
$\Rightarrow$$\int \frac{3 x-1}{(x+2)^{2}}=\int\left\{\frac{3}{(x+2)}-\frac{7}{(x+2)^{2}}\right\} d x$
= $3 \log |x+2|-7\left(\frac{-1}{(x+2)}\right)+C$
= $3 \log |x+2|+\frac{7}{(x+2)}+C$
View full question & answer→Question 272 Marks
Integrate the rational function: $\frac{x}{(x+1)(x+2)}$
AnswerLet $\frac{x}{(x+1)(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}$
$\Rightarrow$ x = A(x + 2) + B(x + 1)
On comparing the coefficients of x and constant term, we get,
A + B = 1
2A + B = 0
On solving above two equations, we get,
A = -1 and B = 2
Thus,
$\frac{x}{(x+1)(x+2)}=\frac{-1}{(x+1)}+\frac{2}{(x+2)}$
$\Rightarrow$ $\int \frac{x}{(x+1)(x+2)}=\int\left\{\frac{-1}{(x+1)}+\frac{2}{(x+2)}\right\} d x$
= -log|x + 1| + 2 log|x + 2| + C
= $\log (x + 2)^2 - \log |x + 1| + C$
= $\log \frac{(x+2)^{2}}{(x+1)}+C$
View full question & answer→Question 282 Marks
Integrate the function $\frac{{{{\sec }^2}x}}{{\sqrt {{{\tan }^2}x + 4} }}$
AnswerLet $I = \int {\frac{{{{\sec }^2}x}}{{\sqrt {{{\tan }^2}x + 4} }}dx} $ ...(i)
Putting tan x = t
$ \Rightarrow {\sec ^2}x = \frac{{dt}}{{dx}}$
$\Rightarrow {\sec ^2}xdx = dt$
$\therefore $ From eq. (i),
$I = \int {\frac{{dt}}{{\sqrt {{t^2} + 4} }}} $
$= \int {\frac{1}{{\sqrt {{t^2} + {{\left( 2 \right)}^2}} }}dt} $
$ = \log \left| {t + \sqrt {{t^2} + {{\left( 2 \right)}^2}} } \right| + c$
$= \log \left| {\tan x + \sqrt {{{\tan }^2}x + 4} } \right| + c$
View full question & answer→Question 292 Marks
Integrate the function $\frac{4 x+1}{\sqrt{2 x^{2}+x-3}}$
AnswerThe given Integrand is:
Let $2x^2 + x – 3 =t$
$\Rightarrow (4x + 1) dx = dt$
$\Rightarrow \int \frac{4 x+1}{\sqrt{2 x^{2}+x-3}} d x=\int \frac{1}{\sqrt{t}} d t$
$=2 \sqrt{t}+c$
= $2 \sqrt{2 x^{2}+x-3}+C$
View full question & answer→Question 302 Marks
Integrate the function $\frac{3 x^{2}}{x^{6}+1}$
AnswerLet $x^3 = t$
$\Rightarrow 3x^2 dx = dt$
$\Rightarrow \int \frac{3 x^{2}}{x^{6}+1} d x=\int \frac{d t}{t^{2}+1}$
$= \tan^{-1}t + C$
$= \tan^{-1}(x^3) + C$
View full question & answer→Question 312 Marks
Find the integral of the function $\frac{{\cos x}}{{1 + \cos x}}$
Answer$\int {\frac{{\cos x}}{{1 + \cos x}}} dx$ $= \int {\frac{{1 + \cos x - 1}}{{1 + \cos x}}} dx$
$= \int {\frac{{1 + \cos x}}{{1 + \cos x}}} - \frac{1}{{1 + \cos x}}dx$
$= \int {\left( {1 - \frac{1}{{2{{\cos }^2}\frac{x}{2}}}} \right)} dx$
As $2{\cos ^2}\frac{\theta }{2} = 1 + \cos \theta$
$= \int 1 dx - \frac{1}{2}\int {{{\sec }^2}\frac{x}{2}dx} $
$= x - \frac{1}{2}\frac{{\tan \frac{x}{2}}}{{\frac{1}{2}}} + c$ $[\because\int {{{\sec }^2}\left( {ax + b} \right)} dx = \frac{{\tan \left( {ax + b} \right)}}{a} + c]$
$= x - \tan \frac{x}{2} + c$
View full question & answer→Question 322 Marks
Find the integral of the function $\frac{1-\cos x}{1+\cos x}$
AnswerWe have $\frac{1-\cos x}{1+\cos x}=\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}=2 \tan ^{2} \frac{x}{2}=\left(\sec ^{2} \frac{x}{2}-1\right)$
$\Rightarrow \int \frac{1-\cos x}{1+\cos x} d x=\int\left(\sec ^{2} \frac{x}{2}-1\right) d x = \int \sec ^{2} \frac{x}{2} d x-\int 1 d x$
$=\frac{\tan \frac{x}{2}}{\frac{1}{2}}-x+C~~ \begin{equation} \left(\because \int \sec ^{2}(a x+b) d x=\frac{\tan (a x+b)}{a}+C\right) \end{equation}$
$= 2 \tan \frac{\mathrm{x}}{2}-\mathrm{x}+\mathrm{C}$
View full question & answer→Question 332 Marks
Find the integrals of the function sin 4x sin 8x
Answer$\int \sin 4 x \sin 8 x d x$ = $\int \frac{1}{2}\left\{\ cos (4 x-8 x)-\cos (4 x+8 x)\right\} d x$ , {As Sin(A)Sin(B)=Cos(A-B)-Cos(A+B)}
$= \frac{1}{2} \int(\cos (-4 x)-\cos 12 x d x$
$= \frac{1}{2} \int\{\cos 4 x-\cos 12 x\} d x$
$= \frac{1}{2}\left[\frac{\sin 4 x}{4}-\frac{\sin 12 x}{12}\right]+C$
View full question & answer→Question 342 Marks
Find the integral of the function $\sin^3 x \cos^3 x$
AnswerLet $I=\int \sin ^{3} x \cos ^{3} x \cdot d x$
$\Rightarrow \int \cos ^{3} x \cdot \sin ^{2} x \cdot \sin x \cdot d x$
$\Rightarrow \int \cos ^{3} x\left(1-\cos ^{2} x\right) \sin x \cdot d x$
Let cos x = t
$\Rightarrow$ -sinx.dx = dt
$\Rightarrow \mathrm{I}=-\int \mathrm{t}^{3}\left(1-\mathrm{t}^{2}\right) \mathrm{dt}$
$=-\int\left(t^{3}-t^{5}\right) d t$
$=-\left\{\frac{t^{4}}{4}-\frac{t^{6}}{6}\right\}+C$
$=-\left\{\frac{\cos ^{4} x}{4}-\frac{\cos ^{6} x}{6}\right\}+C$
$=\frac{\cos ^{6} x}{6}-\frac{\cos ^{4} x}{4}+C$
View full question & answer→Question 352 Marks
Find the integrals of the function $\frac{{\cos 2x + 2{{\sin }^2}x}}{{{{\cos }^2}x}}$
Answer$\int {\frac{{\cos 2x + 2{{\sin }^2}x}}{{{{\cos }^2}x}}dx} $ $= \int {\frac{{\left( {1 - 2{{\sin }^2}x} \right) + 2{{\sin }^2}x}}{{{{\cos }^2}x}}dx} $
using $\cos 2\theta = 1 - 2{\sin ^2}\theta $
$ = \int {\frac{1}{{{{\cos }^2}x}}dx} $
$= \int {{{\sec }^2}xdx} $
= tan x + c
View full question & answer→Question 362 Marks
Find the integral of the function $\frac{{{{\sin }^3}x + {{\cos }^3}x}}{{{{\sin }^2}x{{\cos }^2}x}}$
Answer$\int {\frac{{{{\sin }^3}x + {{\cos }^3}x}}{{{{\sin }^2}x{{\cos }^2}x}}dx} $ $= \int {\frac{{{{\sin }^3}x}}{{{{\sin }^2}x{{\cos }^2}x}} + \frac{{{{\cos }^3}x}}{{{{\sin }^2}x{{\cos }^2}x}}dx}$
$= \int {\frac{{\sin x}}{{{{\cos }^2}x}} + \frac{{\cos x}}{{{{\sin }^2}x}}dx} $
$ = \int {\frac{{\sin x}}{{\cos x\cos x}} + \frac{{\cos x}}{{\sin x\sin x}}dx} $
$ = \int {\left( {\tan x\sec x + \cot x\cos ecx} \right)dx} $
= sec x - cosecx + c
$$
View full question & answer→Question 372 Marks
Find the integral of the function $\tan^3 2x \sec 2x$
AnswerClearly, $\tan^32x \sec 2x = \tan^22x \tan 2x \sec 2x$
$= (\sec^22x - 1) \tan 2x \sec 2x$
$= \sec^22x.\tan 2x \sec 2x - \tan 2x \sec 2x$
$\Rightarrow \int \tan ^{3} 2 x \sec 2 x d x=\int \sec ^{2} 2 x \tan 2 x \sec 2 x d x-\int \tan 2 x \sec 2 x d x$
$= \int \sec ^{2} 2 x \tan 2 x \sec 2 x d x-\frac{\sec 2 x}{2}+C$
Now, Let sec 2x = t
⇒ 2 sec 2x tan 2x dx = dt
Thus, $\int \tan ^{3} 2 x \sec 2 x d x=\frac{1}{2} \int t^{2} d t-\frac{\sec 2 x}{2}+C$
$= \frac{t^{3}}{6}-\frac{\sec 2 x}{2}+C$
$= \frac{(\sec 2 x)^{3}}{6}-\frac{\sec 2 x}{2}+C$
View full question & answer→Question 382 Marks
Find the integral of the function $\frac{\cos x-\sin x}{1+\sin 2 x}$
AnswerClearly,
$\frac{\cos x-\sin x}{1+\sin 2 x}$ $= \frac{\cos x-\sin x}{\left(\sin ^{2} x+\cos ^{2} x\right)+2 \sin x \cos x}$
$= \frac{\cos x-\sin x}{(\sin x+\cos x)^{2}}$
Let sinx + cosx = t
$\Rightarrow$ (cosx-sinx)dx = dt
$\Rightarrow \int \frac{\cos x-\sin x}{1+\sin 2 x} d x=\int \frac{\cos x-\sin x}{(\sin x+\cos x)^{2}} d x$
$= \int \frac{d t}{t^{2}}$
$= -t^{-1} + C$
$=-\frac{1}{t}+C$
$\Rightarrow \int \frac{\cos x-\sin x}{1+\sin 2 x} d x = \frac{-1}{\sin x+\cos x}+C$
View full question & answer→Question 392 Marks
Find the integral of the function $\frac{{{{\sin }^2}x}}{{1 + \cos x}}$
Answer$\int {\frac{{{{\sin }^2}x}}{{1 + \cos x}}} dx$ $= \int {\frac{{1 - {{\cos }^2}x}}{{1 + \cos x}}} dx$
$= \int {\frac{{\left( {1 - \cos x} \right)\left( {1 + \cos x} \right)}}{{1 + \cos x}}} dx$
$= \int {\left( {1 - \cos x} \right)} dx$
$= \int {1dx - \int {\cos x} } dx$
= x - sin x + c
View full question & answer→Question 402 Marks
Integrate the function: $(4 x+2) \sqrt{x^{2}+x+1}$
AnswerLet $x^2 + x + 1 = t$
Differentiating both sides, we get,
$\Rightarrow (2x + 1)dx = dt$
Therefore
$\Rightarrow \int(4 x+2) \sqrt{x^{2}+x+1} d x=\int 2 \sqrt{t} d t$
$=2 \int \sqrt{t} d t$
$= 2\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)+C$
$=\frac{4}{3}\left(x^{2}+x+1\right)^{\frac{3}{2}}+C$
View full question & answer→Question 412 Marks
Integrate the function: $x \sqrt{1+2 x^{2}}$
AnswerLet $1 + 2x^2 = t$
$\Rightarrow 4xdx = dt$
$\Rightarrow\int x \sqrt{1+2 x^{2}} d x=\int \frac{\sqrt{t} d t}{4}$
$\Rightarrow\frac{1}{4} \int t^{\frac{1}{2}} d t$
$\Rightarrow \frac{1}{4}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)+C$
$\Rightarrow \frac{1}{6}\left(1+2 x^{2}\right)^{\frac{3}{2}}+C$
View full question & answer→Question 422 Marks
Integrate the function: $x \sqrt{x+2}$
AnswerLet I = $ \int x \sqrt{x+2} d x$
Put (x + 2) = t $\Rightarrow$ dx = dt
$\Rightarrow I = \int x \sqrt{x+2} d x=\int(t-2) \sqrt{t} d t$
$=\int\left(t^{\frac{3}{2}}-2 t^{\frac{t}{2}}\right) d t$
$=\int t^{\frac{3}{2}} d t-2 \int t^{\frac{1}{2}} d t$
$=\frac{t^{\frac{5}{2}}}{\frac{5}{2}}-2\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)+C$
$=\frac{2}{5} t^{\frac{5}{2}}-\frac{4}{3} t^{\frac{3}{2}}+C$
$=\frac{2}{5}(x+2)^{\frac{5}{2}}-\frac{4}{3}(x+2)^{\frac{3}{2}}+C$
View full question & answer→Question 432 Marks
Integrate the function: $\sqrt{a x+b}$
AnswerLet ax + b = t
$\Rightarrow$ adx = dt
$\Rightarrow$ $d x=\frac{1}{a} d t$
$\Rightarrow \int(a x+b)^{\frac{1}{2}}=\frac{1}{a} \int t^{\frac{1}{2}} d t$
= $\frac{1}{a}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)+C$
$\Rightarrow\frac{2}{3 a}(a x+b)^{\frac{3}{2}}+C$
View full question & answer→Question 442 Marks
Integrate the function: sin x sin(cos x)
AnswerPutting cos x = t
$\Rightarrow - \sin x = \frac{{dt}}{{dx}}$
$\Rightarrow - \sin xdx = dt$
$\therefore \int {\sin x\sin \left( {\cos x} \right)dx} $
$= \int {\sin \left( {\cos x} \right)\left( { - \sin xdx} \right)} $
$= - \int {\sin t} dt$
= - (-cos t) + c
= cos t + c = cos (cos x) + c
View full question & answer→Question 452 Marks
Integrate the function: $\frac{{{x^3}\sin \left( {{{\tan }^{ - 1}}{x^4}} \right)}}{{1 + {x^8}}}$
AnswerLet $I = \int {\frac{{{x^3}\sin \left( {{{\tan }^{ - 1}}{x^4}} \right)}}{{1 + {x^8}}}dx} $
$ = \frac{1}{4}\int {\sin \left( {{{\tan }^{ - 1}}{x^4}} \right).\frac{{4{x^3}}}{{1 + {x^8}}}dx} $…(i)
Putting $\tan^{-1}x^4 = t$
$\Rightarrow \frac{1}{{1 + {{\left( {{x^4}} \right)}^2}}}\frac{d}{{dx}}{x^4} = \frac{{dt}}{{dx}}$
$\Rightarrow \frac{{4{x^3}}}{{1 + {x^8}}}dx = dt$
From eq. (i), $I = \frac{1}{4}\int {\sin tdt} $
$= \frac{{ - 1}}{4}\cos t + c$
$ = \frac{{ - 1}}{4}\cos \left( {{{\tan }^{ - 1}}{x^4}} \right) + c$ View full question & answer→Question 462 Marks
Integrate the function: $\frac{{{{\left( {1 + \log x} \right)}^2}}}{x}$
AnswerLet $I = \int {\frac{{{{\left( {1 + \log x} \right)}^2}}}{x}dx}$ ...(i) Putting $1 + \log x = t$
$ \Rightarrow \frac{1}{x} = \frac{{dt}}{{dx}}$
$\Rightarrow \frac{{dx}}{x} = dt$
$\therefore$ From eq. (i), $I = \int {{t^2}dt} $
$= \frac{{{t^3}}}{3} + c$
$= \frac{1}{3}{\left( {1 + \log x} \right)^3} + c$
View full question & answer→Question 472 Marks
Integrate the function: $\frac{{\sin x}}{{{{\left( {1 + \cos x} \right)}^2}}}$
AnswerLet $I = \int {\frac{{\sin x}}{{{{\left( {1 + \cos x} \right)}^2}}}dx} $
$= - \int {\frac{{ - \sin x}}{{{{\left( {1 + \cos x} \right)}^2}}}dx} $…(i)
Putting 1 + cos x = t
$ \Rightarrow - \sin x = \frac{{dt}}{{dx}}$
$\Rightarrow $ -sin xdx = dt
$\therefore$ From eq. (i), $I = - \int {\frac{{dt}}{{{t^2}}}} $
$ = - \int {{t^{-2}}dt} $
$= \frac{{ - {t^{ - 1}}}}{{ - 1}} + c$
$= \frac{1}{t} + c$
$= \frac{1}{{1 + \cos x}} + c$
View full question & answer→Question 482 Marks
Integrate the function: $\frac{{\sin x}}{{1 + \cos x}}$
AnswerLet $I = \int {\frac{{\sin x}}{{1 + \cos x}}dx} $ $= - \int {\frac{{ - \sin x}}{{1 + \cos x}}dx} $ ...(i)
Putting 1 + cos x = t
$\Rightarrow - \sin x = \frac{{dt}}{{dx}}$
$\Rightarrow - \sin xdx = dt$
$\therefore$ From eq. (i), $I = - \int {\frac{{dt}}{t}}$
= - log |t| + c
= -log |1 + cos x| + c
View full question & answer→Question 492 Marks
Integrate the function: $\frac{1}{{x + x\log x}}$
AnswerPutting 1 + log x = t
$ \Rightarrow \frac{1}{x} = {dt}$
$ \Rightarrow \frac{{dx}}{x} = dt$
$\therefore \int {\frac{1}{{x + x\log x}}dx} $
$= \int {\frac{1}{{1 + \log x}}\frac{{dx}}{x}} $
$= \int {\frac{1}{t}dt = \log \left| t \right| + c} $
= log |1 + log x| + c
View full question & answer→Question 502 Marks
Integrate the function: cot x log sin x
AnswerLet $I = \int {\cot x\log \sin xdx} $ ...(i) Putting log sin x = t
$ \Rightarrow \frac{1}{{\sin x}}\frac{d}{{dx}}\left( {\sin x} \right) = \frac{{dt}}{{dx}}$
$ \Rightarrow \frac{1}{{\sin x}}\cos x = \frac{{dt}}{{dx}}$
$ \Rightarrow \cot xdx = dt$
$\therefore$ From eq. (i), $I = \int {tdt} $
$ = \frac{{{t^2}}}{2} + c$
$= \frac{1}{2}{\left( {\log \sin x} \right)^2} + c$
View full question & answer→Question 512 Marks
Integrate the function: $\frac{\cos x}{\sqrt{1+\sin x}}$
AnswerLet 1 + sinx = t
$\Rightarrow$ cosx dx = dt
$\Rightarrow \int \frac{\cos x}{\sqrt{1+\sin x}} d x=\int \frac{d t}{\sqrt{t}}$
$\Rightarrow \frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C$
$\Rightarrow 2 \sqrt{t}+C$
$\Rightarrow 2 \sqrt{1+\sin x}+C$
View full question & answer→Question 522 Marks
Integrate the function: $\sqrt{\sin 2 x} \cos 2 x$
AnswerLet sin2x = t
$\Rightarrow$ 2cos2xdx = dt
$=\int \sqrt{\sin 2 x} \cos 2 x d x=\frac{1}{2} \int \sqrt{t} d t$
$\Rightarrow \frac{1}{2}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)+C$
$\Rightarrow \frac{1}{3}(\sin 2 x)^{\frac{3}{2}}+C$
View full question & answer→Question 532 Marks
Integrate the function: $\frac{{\cos \sqrt x }}{{\sqrt x }}$
AnswerLet $I = \int {\frac{{\cos \sqrt x }}{{\sqrt x }}dx} $ ...(i)Putting $\sqrt x = t$
$\Rightarrow x = t^2$
$ \Rightarrow \frac{{dx}}{{dt}} = 2t$
$ \Rightarrow $ dx = 2tdt
$\therefore$ From eq. (i), $I = \int {\frac{{\cos t}}{t}2tdt} $
$ = 2\int {\cos t\,dt} $
= 2 sin t + c
$ = 2\sin \sqrt x + c$
View full question & answer→Question 542 Marks
Integrate the function: $\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}$
AnswerLet $I = \int {\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}} dx$ ...(i)
Putting ${\sin ^{ - 1}}x = t$
$ \Rightarrow \frac{1}{{\sqrt {1 - {x^2}} }}=\frac{{dt}}{{dx}}$
$ \Rightarrow \frac{{dx}}{{\sqrt {1 - {x^2}} }} = dt$
$\therefore$ From eq. (i), $I = \int {tdt} $
$= \frac{{{t^2}}}{2} + c$
$= \frac{1}{2}{\left( {{{\sin }^{ - 1}}x} \right)^2} + c$
View full question & answer→Question 552 Marks
Integrate the function: $\sec^2 (7 – 4x)$
AnswerLet $7 – 4x = t$
$\Rightarrow$ -4dx = dt
$\Rightarrow \int \sec ^{2}(7-4 x) d x=\frac{-1}{4} \int \sec ^{2} t d t$
$\Rightarrow \frac{-1}{4}(\tan t)+C$
$\Rightarrow \frac{-1}{4} \tan (7-4 x)+C$
View full question & answer→Question 562 Marks
Integrate the function: $\frac{{{{\left( {\log x} \right)}^2}}}{x}$
AnswerPutting log x = t
$ \Rightarrow \frac{dx}{x} = {{dt}}{{}}$
$\therefore \int {\frac{{{{\left( {\log x} \right)}^2}}}{x}dx} $ $ = \int {{t^2}dt} $
$= \frac{{{t^3}}}{3} + c$
$= \frac{1}{3}{\left( {\log x} \right)^3} + c$
View full question & answer→Question 572 Marks
Integrate the function: $\frac{e^{\tan ^{-1} x}}{1+x^{2}}$
Answerlet $\tan^{-1} x = t$
$\Rightarrow \frac{1}{1+x^{2}} d x=d t$
$\Rightarrow \int \frac{e^{t an^{-1} x}}{1+x^{2}} d x=\int e^{t} d t$
$= e^t + c$
= $e^{\tan ^{-1} x}+C$
View full question & answer→Question 582 Marks
Integrate the function: $e^{2 x+3}$
AnswerLet 2x + 3 = t
$\Rightarrow$ 2dx = dt
$\Rightarrow \int e^{2 x+3} d x=\frac{1}{2} \int e^{t} d t$
$\Rightarrow \frac{1}{2} e^{t}+C$
$\Rightarrow \frac{1}{2} e^{2 x+3}+C$
View full question & answer→Question 592 Marks
Integrate the function: $\frac{1}{{x{{\left( {\log x} \right)}^m}}},x > 0$, m $\neq$ 1
AnswerLet $I = \int {\frac{1}{{x{{\left( {\log x} \right)}^m}}}} dx$
$= \int {\frac{{\frac{1}{x}dx}}{{{{\left( {\log x} \right)}^m}}}} dx$…(i)
Putting $\log x = t$
$ \Rightarrow \frac{1}{x} = \frac{{dt}}{{dx}}$
$\Rightarrow \frac{{dx}}{x} = dt$
$\therefore $ From eq. (i), $I = \int {\frac{{dt}}{{{t^m}}} = \int {{t^{ - m}}dt} } $
$ = \frac{{{t^{ - m + 1}}}}{{ - m + 1}} + c$
$= \frac{{{{\left( {\log x} \right)}^{1 - m}}}}{{1 - m}} + c$
View full question & answer→Question 602 Marks
Integrate the function: $\frac{1}{{x - \sqrt x }}$
AnswerLet $I = \int\frac{1}{{x - \sqrt x }}dx$...(i) Putting $\sqrt x = t$
$ \Rightarrow x = {t^2}$
$\Rightarrow \frac{{dx}}{{dt}} = 2t$
$ \Rightarrow dx = 2tdt$
$\therefore $ From eq. (i),
$I = \int {\frac{1}{{{t^2} - t}}2t} dt$
$= 2\int {\frac{t}{{t\left( {t - 1} \right)}}} dt$
$ = 2\int {\frac{1}{{\left( {t - 1} \right)}}} dt$
= 2 log |t - 1| + c
$ = 2\log \left| {\sqrt x - 1} \right| + c$
View full question & answer→Question 612 Marks
Integrate the function: $\frac{2 x}{1+x^{2}}$
AnswerLet $1 + x^2 = t$
$\Rightarrow$ 2x dx = dt
Now, $\int \frac{2 x}{1+x^{2}} d x=\int \frac{1}{t} d t$
$= \log |t| + C$
$= \log |1 + x^2| + C$
$= \log (1 + x^2) + C$
View full question & answer→Question 622 Marks
Find the integral: $\int\left(2 x^{2}+e^{x}\right) d x$
Answer$\int\left(2 x^{2}+e^{x}\right) d x$
= $2 \int x^{2} d x+\int e^{x} d x$
= $2\left(\frac{\mathrm{x}^{3}}{3}\right)+\mathrm{e}^{\mathrm{x}}+\mathrm{C}$
= $\frac{2 x^{3}}{3}+e^{x}+C$
View full question & answer→Question 632 Marks
Find the integral: $\int\left(a x^{2}+b x+c\right) d x$
Answer$\int\left(a x^{2}+b x+c\right) d x$
= $a \int x^{2} d x+b \int x d x+c \int 1 d x$
= $a\left(\frac{x^{3}}{3}\right)+b\left(\frac{x^{2}}{2}\right)+c x+c$
= $\frac{a x^{3}}{3}+\frac{b x^{2}}{2}+c x+C$
View full question & answer→Question 642 Marks
Find the integral: $\int x^{2}\left(1-\frac{1}{x^{2}}\right) d x$
Answer$\int x^{2}\left(1-\frac{1}{x^{2}}\right) d x$
= $\int\left(x^{2}-1\right) d x$
= $\int x^{2} d x-\int 1 d x$
= $\frac{x^{3}}{3}-x+C$
View full question & answer→Question 652 Marks
Find the integral: $\int\left(4 e^{3 x}+1\right) d x$
Answer$\int\left(4 e^{3 x}+1\right) d x=4 \int e^{3 x} d x+\int 1 d x$
= $4\left(\frac{\mathrm{e}^{3 \mathrm{x}}}{3}\right)+\mathrm{x}+\mathrm{C}$
= $\frac{4}{3} e^{3 x}+x+C$
View full question & answer→Question 662 Marks
Find an anti derivative (or integral) of function by the method of inspection $(ax + b)^2$
AnswerWe know that
$\frac{d}{d x}(a x+b)^{3}=3 a(a x+b)^{2}$
$\Rightarrow(a x+b)^{2}=\frac{1}{3 a} \frac{d}{d x}(a x+b)^{3}$
$=\frac{d}{d x}\left(\frac{1}{3 a}(a x+b)^{3}\right)$
Therefore, the anti-derivative of $(ax + b)^2$ is $\frac{1}{3 a}(a x+b)^{3}$.
View full question & answer→Question 672 Marks
Find an anti derivative (or integral) of function by the method of inspection $e^{2x}$.
AnswerWe know that
$\frac{d}{d x}\left(e^{2 x}\right)=2 e^{2 x}$
$\Rightarrow \mathrm{e}^{2 \mathrm{x}}=\frac{1}{2} \frac{d}{d x}\left(e^{2 x}\right)$
$=\frac{d}{d x}\left(\frac{1}{2} e^{2 x}\right)$
Therefore, the anti-derivative of $e^{2x}$ is $\frac{1}{2} e^{2 x}$.
View full question & answer→Question 682 Marks
Find the integral: $\int \frac{2-3 \sin x}{\cos ^{2} x} d x$
Answer$\int \frac{2-3 \sin x}{\cos ^{2} x} d x$
= $\int\left(\frac{2}{\cos ^{2} x}-\frac{3 \sin x}{\cos ^{2} x}\right) d x$
= $\int 2 \sec ^{2} x d x-3 \int \tan x \sec x d x$
= 2tan x - 3sec x + C
View full question & answer→Question 692 Marks
Find an anti derivative (or integral) of function by the method of inspection cos 3x.
AnswerWe know that $\frac{d}{d x}(\sin 3 x)=3 \cos 3 x$
$\Rightarrow \cos 3 x=\frac{1}{3} \frac{d}{d x}(\sin 3 x)$
$=\frac{d}{d x}\left(\frac{1}{3} \sin 3 x\right)$
Therefore, the anti-derivative of cos3x is $\frac{1}{3} \sin 3 x$.
View full question & answer→Question 702 Marks
Find the integral: $ \int \frac { \sec ^ { 2 } x } { cosec ^ { 2 } x } d x$
AnswerLet I = $\int \frac { \sec ^ { 2 } x } { cosec ^ { 2 } x } d x = \int \frac { \left( \frac { 1 } { \cos ^ { 2 } x } \right) } { \left( \frac { 1 } { \sin ^ { 2 } x } \right) } d x$
$= \int \frac { \sin ^ { 2 } x } { \cos ^ { 2 } x } d x$
$ = \int \tan ^ { 2 } x d x = \int \left( \sec ^ { 2 } x - 1 \right) d x$ $ \left[ \because \tan ^ { 2 } x = \sec ^ { 2 } x - 1 \right]$
$ = \int \sec ^ { 2 } x d x - \int 1 d x = \tan x - x + c$
View full question & answer→Question 712 Marks
Find the integral: $\int \sec x ( \sec x + \tan x ) d x.$
AnswerLet I = $\int \sec x ( \sec x + \tan x ) d x$
$= \int \left( \sec ^ { 2 } x + \sec x \tan x \right) d x$
$= \int \sec ^ { 2 } x d x + \int \sec x \tan x d x$
= tan x + secx + C
View full question & answer→Question 722 Marks
Find the integral: $\int\left(2 x^{2}-3 \sin x+5 \sqrt{x}\right) d x$
Answer$\int\left(2 x^{2}-3 \sin x+5 \sqrt{x}\right) d x$
= $2 \int x^{2} d x-3 \int \sin x d x+5 \int x^{\frac{1}{2}} d x$
= $\frac{2 x^{3}}{3}+3 \cos x+5\left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right)+C$
= $\frac{2 x^{3}}{3}+3 \cos x+\frac{10}{3} x^{\frac{3}{2}}+C$
View full question & answer→Question 732 Marks
Find the integral: $\int\left(2 x-3 \cos x+e^{x}\right) d x$
Answer$\int\left(2 x-3 cos x+e^{x}\right) d x$
= $2 \int x d x-3 \int \cos x d x+\int e^{x} d x$
= $\frac{2 x^{2}}{2}-3(\sin x)+e^{x}+C$
= $x^{2}-3 \sin x+e^{x}+C$
View full question & answer→Question 742 Marks
Find the integral: $\int(1-x) \sqrt{x} d x$
Answer$\int(1-x) \sqrt{x} d x$
= $\int\left(\sqrt{x}-x^{\frac{3}{2}}\right) d x$
= $\int x^{\frac{1}{2}} d x-\int x^{\frac{3}{2}} d x$
= $\frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^{\frac{5}{2}}}{\frac{5}{2}}+C$
= $\frac{2}{3} x^{\frac{3}{2}}-\frac{2}{5} x^{\frac{5}{2}}+C$
View full question & answer→Question 752 Marks
Find the integral: $\int \frac{x^{3}-x^{2}+x-1}{x-1} d x$
AnswerI = $\int \frac{x^{3}-x^{2}+x-1}{x-1} d x$
Now the numerator can be factorized as,
$x^3 - x^2 + x - 1 = x^2(x - 1) + 1(x - 1)$
$x^3 - x^2 + x - 1 = (x^2 + 1)(x - 1)$
Now putting this in given integral we get,
I = $\frac{x^{3}-x^{2}+x-1}{x-1}=\frac{\left(x^{2}+1\right)(x-1)}{x-1}=x^{2}+1$
= $\int\left(x^{2}+1\right) d x$
= $\int x^{2} d x+\int 1 . d x$
= $\frac{x^{3}}{3}+x+C$
View full question & answer→Question 762 Marks
Find the integral: $\int \frac{x^{3}+5 x^{2}-4}{x^{2}} d x$
AnswerLet I = $\int \frac{x^{3}+5 x^{2}-4}{x^{2}} d x$
Separating the terms we get,
I = $\int\left(x+5-4 x^{-2}\right) d x$
Applying the formula,
$\int x^{n} d x=\frac{x^{n+1}}{n+1}+c$ , gives
I = $\int x d x+5 \int 1 d x-4 \int x^{-2} d x$
= $\frac{x^{2}}{2}+5 x-4\left(\frac{x^{-1}}{-1}\right)+C$
= $\frac{x^{2}}{2}+5 x+\frac{4}{x}+C$
View full question & answer→Question 772 Marks
Find the integral: $\int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2} d x$
Answer$\int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2} d x$
= $\int\left(x+\frac{1}{x}-2\right) d x$
= $\int x d x+\int \frac{1}{x} d x-2 \int 1 d x$
Now we know that,
$^{\int x^{n} d x}=\frac{x^{n+1}}{n+1}+c$
Therefore,
$\int x d x+\int \frac{1}{x} d x-2 \int 1 d x$ = $\frac{x^{2}}{2}+\log |x|-2 x+C$
View full question & answer→Question 782 Marks
Find an anti derivative (or integral) of function sin 2x by the method of inspection.
AnswerWe know that
$\frac{d}{d x}(\cos 2 x)=-2 \sin 2 x$
$\Rightarrow \sin 2 x=-\frac{1}{2} \frac{d}{d x}(\cos 2 x)$
$=\frac{d}{d x}\left(-\frac{1}{2} \cos 2 x\right)$
Therefore, the anti-derivative of sin2x is $-\frac{1}{2} \cos 2 x$
View full question & answer→Question 792 Marks
By using the properties of definite integrals, evaluate the integral $\int\limits_0^4 {\left| {x - 1} \right|dx} $
AnswerLet $I = \int\limits_0^4 {\left| {x - 1} \right|dx} $ ...(i)
Here x - 1 = 0
$ \Rightarrow x = 1 \in \left( {0,4} \right)$
$\therefore$ From eq. (i),
$I = \int\limits_0^1 {\left| {x - 1} \right|dx + \int\limits_1^4 {\left| {x - 1} \right|} dx} $
$= - \int\limits_0^1 {\left( {x - 1} \right)dx + \int\limits_1^4 {\left( {x - 1} \right)} dx} $
$ \Rightarrow I = - \left( {\frac{{{x^2}}}{2} - x} \right)_0^1 + \left( {\frac{{{x^2}}}{2} - x} \right)_1^4$
$= - \left\{ {\left( {\frac{1}{2} - 1} \right) - 0} \right\} + \left\{ {\frac{{16}}{2} - 4 - \left( {\frac{1}{2} - 1} \right)} \right\}$ $= \frac{1}{2} + 8 - 4 + \frac{1}{2}$
= 9 - 4
= 5
View full question & answer→Question 802 Marks
By using the properties of definite integrals, evaluate the integral $\int\limits_0^{2\pi } {{{\cos }^5}x} dx$
Answer$\int\limits_0^{2\pi } {{{\cos }^5}x} dx$ $= 2\int\limits_0^{\pi } {{{\cos }^5}x} dx$
$\left[ {\because \int\limits_0^{2a} {f\left( x \right)dx = 2\int\limits_0^a {f\left( x \right)dx,if\,\,f\left( {2a - x} \right) = f\left( x \right)} } } \right]$
Here $f\left( x \right) = {\cos ^5}x$
$\therefore f\left( {\pi - x} \right) = {\cos ^5}\left( {\pi - x} \right) = -{\cos ^5}x$=-f(x)
$ \Rightarrow\int\limits_0^{\pi } {{{\cos }^5}x} dx=0$$ [\because \int _0^{2a}f(x)dx=0,if\ f(2a-x)=-f(x)]$
$\therefore \int\limits_0^{2\pi } {{{\cos }^5}x} dx=0$
View full question & answer→Question 812 Marks
By using the properties of definite integrals, evaluate the integral $\int\limits_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {{{\sin }^7}xdx} $
AnswerLet $I = \int\limits_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {{{\sin }^7}xdx} $Here $f(x) = \sin^7x$
$\therefore f\left( { - x} \right) = {\sin ^7}\left( { - x} \right)$
$(-\sin x)^7$
$= -\sin^7x = -f(x)$
$\therefore $ f(x) is an odd function of x.
$\therefore I = \int\limits_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {{{\sin }^7}xdx} = 0$
${\left[ {\because \int\limits_{ - a}^a {f\left( x \right)dx = 0} } \right.}$ when f(x) is an odd function]
View full question & answer→Question 822 Marks
By using the properties of definite integrals, evaluate the integral $\int\limits_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {{{\sin }^2}xdx}$
AnswerLet $I = \int\limits_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {{{\sin }^2}xdx} $
$= 2\int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}xdx} $ ...(i) ${\because \int\limits_{ - a}^a {f\left( x \right)dx = 2\int\limits_0^a {f\left( x \right)dx,} } }$ when f(x) is even function]
$\Rightarrow I = 2\int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}\left( {\frac{\pi }{2} - x} \right)dx} $
$\left[ {\because \int\limits_0^a {f\left( x \right)dx = \int\limits_0^a {f\left( {a - x} \right)dx = } } } \right]$
$\Rightarrow I = 2\int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}xdx} $ …(ii)
Adding eq. (i) and (ii),
$2I = 2\int\limits_0^{\frac{\pi }{2}} {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)dx} $
$= 2\int\limits_0^{\frac{\pi }{2}} {1dx} $
$= 2\left( x \right)_0^{\frac{\pi }{2}}$
$ = 2.\frac{\pi }{2} = \pi $
$\Rightarrow I = \frac{\pi }{2}$
View full question & answer→Question 832 Marks
Find $\int \frac { \left( x ^ { 2 } + 1 \right) e ^ { x } } { ( x + 1 ) ^ { 2 } } d x.$
AnswerAccording to the question, $I = \int e ^ { x } \frac { \left( x ^ { 2 } + 1 \right) } { ( x + 1 ) ^ { 2 } } d x$
$= \int e ^ { x } \frac { \left( x ^ { 2 } + 1 + 2 x - 2 x \right) } { ( x + 1 ) ^ { 2 } } d x$
$= \int e ^ { x } \left( \frac { ( x + 1 ) ^ { 2 } - 2 x } { ( x + 1 ) ^ { 2 } } \right) d x [\therefore (a+b)^2 = a^2 +b^2+2ab]$
$= \int e ^ { x } \left( 1 - \frac { 2 x } { ( x + 1 ) ^ { 2 } } \right) d x$
$= \int e ^ { x } d x - 2 \int e ^ { x } \cdot \frac { x } { ( x + 1 ) ^ { 2 } } d x$
$= e ^ { x } - 2 \int e ^ { x } \left( \frac { x + 1 - 1 } { ( x + 1 ) ^ { 2 } } \right) d x$ $= e ^ { x } - 2 \int e ^ { x } \left( \frac { x+1 } { ( x + 1 )^2 } + \frac { ( - 1 ) } { ( x + 1 ) ^ { 2 } } \right) d x$
$= e ^ { x } - 2 \int e ^ { x } \left( \frac { 1 } { ( x + 1 ) } + \frac { ( - 1 ) } { ( x + 1 ) ^ { 2 } } \right) d x$
Consider $f ( x ) = \frac { 1 } { x + 1 },$$\therefore$$f ^ { \prime } ( x ) = \frac { ( - 1 ) } { ( x + 1 ) ^ { 2 } }$
Thus, the above integral is of the form
$\because \int e ^ { x } f ( x ) + f ^ { \prime } ( x ) d x = e ^ { x } f ( x ) + C $
$\therefore \quad I = e ^ { x } - 2 e ^ { x }\Big[ \frac { 1 } { ( x + 1 ) } \Big]+ C$
$\Rightarrow I = e ^ { x } \left( \frac { x + 1 - 2 } { x + 1 } \right) + C $
$\Rightarrow I = e ^ { x } \left( \frac { x - 1 } { x + 1 } \right) + C$
$\therefore \int e ^ { x } \frac { \left( x ^ { 2 } + 1 \right) } { ( x + 1 ) ^ { 2 } } d x = e ^ { x } \left( \frac { x - 1 } { x + 1 } \right) + C$
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