Question 1011 Mark
Find $ \int \left[ \log ( \log x ) + \frac { 1 } { ( \log x ) ^ { 2 } } \right] d x$
AnswerAccording to the question , $ I = \int \left[ \log ( \log x ) + \frac { 1 } { ( \log x ) ^ { 2 } } \right] d x$
By Using integration by parts for first integral, we get
$ = \int {\log } (\mathop {\mathop{\rm l}\nolimits} \limits_Iogx) \cdot \mathop 1\limits_{II} dx + \int {\frac{1}{{{{(\log x)}^2}}}} dx$
$I = log (log x) \int 1 d x - \int \left[ \frac { d } { d x } \log ( \log x )\Big]\Big[ \int 1 d x \right] d x + \int \frac { 1 } { ( \log x ) ^ { 2 } } d x$ $= x log (log x) - \int \frac { 1 } { ( \log x ) } d x + \int \frac { 1 } { ( \log x ) ^ { 2 } } d x $
$= x log (log x)-\int (\mathop {\mathop{\rm\frac { 1}{logx}}\nolimits} \limits_I ) \cdot \mathop 1\limits_{II} dx + \int \frac { 1 } { ( \log x ) ^ { 2 } } d x $
By Using integration by parts for second integral, we get
$= x log (log x) -\frac{x}{logx}+ \int [\frac {d}{dx}(\frac {1}{logx})]. (\int 1dx)dx+ \int \frac { 1 } { ( \log x ) ^ { 2 } } d x $
$= x log (log x) -\frac{x}{logx}- \int \frac { 1 } { ( \log x ) ^ { 2 } } d x \frac {1}{x} . x+ \int \frac { 1 } { ( \log x ) ^ { 2 } } d x $
$= x log (log x) -\frac{x}{logx}- \int \frac { 1 } { ( \log x ) ^ { 2 } } d x + \int \frac { 1 } { ( \log x ) ^ { 2 } } d x $
$ = x \log ( \log x )-\frac{x}{logx}+ C$
$\therefore I = x \log ( \log x ) -\frac{x}{logx}+ C$
View full question & answer→Question 1021 Mark
Find $\int \frac{x^{4} d x}{(x-1)\left(x^{2}+1\right)}$
AnswerWe have $\frac{x^{4}}{(x-1)\left(x^{2}+1\right)}=(x+1)+\frac{1}{x^{3}-x^{2}+x-1}$
Using Partitial Fraction,
= $(x+1)+\frac{1}{(x-1)\left(x^{2}+1\right)}$ ......(i)
Now express $\frac{1}{(x-1)\left(x^{2}+1\right)}=\frac{\mathrm{A}}{(x-1)}+\frac{\mathrm{B} x+\mathrm{C}}{\left(x^{2}+1\right)}$ .....(ii)
So $1 = A (x^2 + 1) + (Bx + C) (x - 1)$
$= (A + B) x^2 + (C - B) x + A - C$
Equating coefficients on both sides, we get A + B = 0, C - B = 0 and A - C = 1, which gives
$A=\frac{1}{2}, B=C=-\frac{1}{2}$
Substituting values of A, B and C in (ii), we get
$\frac{1}{(x-1)\left(x^{2}+1\right)}=\frac{1}{2(x-1)}-\frac{1}{2} \frac{x}{\left(x^{2}+1\right)}-\frac{1}{2\left(x^{2}+1\right)}$ .....(iii)
Again, substituting (iii) in (i), we have
$\frac{x^{4}}{(x-1)\left(x^{2}+x+1\right)}$ = $(x+1)+\frac{1}{2(x-1)}-\frac{1}{2} \frac{x}{\left(x^{2}+1\right)}-\frac{1}{2\left(x^{2}+1\right)}$
Therefore
$\int \frac{x^{4}}{(x-1)\left(x^{2}+x+1\right)} d x$ = $\frac{x^{2}}{2}+x+\frac{1}{2} \log |x-1|-\frac{1}{4} \log \left(x^{2}+1\right)-\frac{1}{2} \tan ^{-1} x+C$
View full question & answer→Question 1031 Mark
Find $\int \frac{\left(x^{4}-x\right)^{\frac{1}{4}}}{x^{5}} d x$
AnswerWe have $\int \frac{\left(x^{4}-x\right)^{\frac{1}{4}}}{x^{5}} d x$ = $\int \frac{\left(1-\frac{1}{x^{3}}\right)^{\frac{1}{4}}}{x^{4}} d x$
Put, $1-\frac{1}{x^{3}}=1-x^{-3}=t$ $\Rightarrow \frac{3}{x^{4}} d x=d t$
Therefore $\int \frac{\left(x^{4}-x\right)^{\frac{1}{4}}}{x^{5}} d x$ = $\frac{1}{3} \int t^{\frac{1}{4}} d t=\frac{1}{3} \times \frac{4}{5} t^{\frac{5}{4}}+C=\frac{4}{15}\left(1-\frac{1}{x^{3}}\right)^{\frac{5}{4}}+C$
View full question & answer→Question 1041 Mark
Find $\int \cos 6 x \sqrt{1+\sin 6 x} d x$
AnswerPut t = 1 + sin 6x, so that dt = 6 cos 6x dx
Therefore $\int \cos 6 x \sqrt{1+\sin 6 x} d x$ = $\frac{1}{6} \int t^{\frac{1}{2}} d t$
= $\frac{1}{6} \times \frac{2}{3}(t)^{\frac{3}{2}}+\mathrm{C}=\frac{1}{9}(1+\sin 6 x)^{\frac{3}{2}}+\mathrm{C}$
View full question & answer→Question 1051 Mark
Evaluate $\int_{0}^{\frac{\pi}{2}} \log \sin x d x$
AnswerLet $I=\int_{0}^{\frac{\pi}{2}} \log \sin x d x$
$I=\int_{0}^{\frac{\pi}{2}} \log \sin \left(\frac{\pi}{2}-x\right) d x=\int_{0}^{\frac{\pi}{2}} \log \cos x d x$
Adding the two values of I, we get
$2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}}(\log \sin x+\log \cos x) d x$
= $\int_{0}^{\frac{\pi}{2}}(\log \sin x \cos x+\log 2-\log 2) d x$ (by adding and subtracting log 2)
= $\int_{0}^{\frac{\pi}{2}} \log \sin 2 x d x-\int_{0}^{\frac{\pi}{2}} \log 2 d x$
Put 2x = t in the first integral. Then 2 dx = dt, when x = 0, t = 0 and when $x=\frac{\pi}{2}$, t = $\pi$
Therefore, 2I = $\frac{1}{2} \int_{0}^{\pi} \log \sin t d t-\frac{\pi}{2} \log 2$
= $\frac{2}{2} \int_{0}^{\frac{\pi}{2}} \log \sin t d t-\frac{\pi}{2} \log 2$
= $\int_{0}^{\frac{\pi}{2}} \log \sin x d x-\frac{\pi}{2} \log 2$ ((by changing variable t to x)
= $I-\frac{\pi}{2} \log 2$
Hence, $\int_{0}^{\frac{\pi}{2}} \log \sin x d x=\frac{-\pi}{2} \log 2$
View full question & answer→Question 1061 Mark
Evaluate $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d x}{1+\sqrt{\tan x}}$
Answer$I=\int_{\pi /6}^{\pi /3} {\frac{{dx}}{{1 + \sqrt {\tan x} }}} $
$I = \int_{\pi /6}^{\pi /3} {\frac{{dx}}{{1 + \sqrt {\frac{{\sin x}}{{\cos x}}} }} = \int_{\pi /6}^{\pi /3} {\frac{{\sqrt {\cos x} }}{{\sqrt {\cos x} + \sqrt {\sin x} }}dx} } $...(1)
$I = \int_{\pi /6}^{\pi /3} {\frac{{\sqrt {\cos \left( {\frac{\pi }{3} + \frac{\pi }{6} - x} \right)} }}{{\sqrt {\cos \left( {\frac{\pi }{3} + \frac{\pi }{6} - x} \right)} + \sqrt {\sin \left( {\frac{\pi }{3} + \frac{\pi }{6} - x} \right)} }}dx} $
$\left[ {\because \int_a^b {f\left( x \right)dx = \int_a^b {f\left( {a + b - x} \right)dx} } } \right]$
$I = \int_{\pi /6}^{\pi /3} {\frac{{\sqrt {\cos \left( {\frac{\pi }{2} - x} \right)} }}{{\sqrt {\cos \left( {\frac{\pi }{2} - x} \right)} + \sqrt {\sin \left( {\frac{\pi }{2} - x} \right)} }}dx} $
$I = \int_{\pi /6}^{\pi /3} {\frac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}dx} $ ...(2)
Adding (1) and (2), we get
$2I = \int_{\pi /6}^{\pi /3} {1dx} $
$= \left[ x \right]_{\pi /6}^{\pi /3}$
$= \frac{\pi }{3} - \frac{\pi }{6} = \frac{\pi }{6}$
$I = \frac{\pi }{{12}}$
View full question & answer→Question 1071 Mark
Find the integral: $\int \frac{1-\sin x}{\cos ^{2} x} d x$
AnswerWe have
$\int \frac{1-\sin x}{\cos ^{2} x} d x=\int \frac{1}{\cos ^{2} x} d x-\int \frac{\sin x}{\cos ^{2} x} d x$
= $\int \sec ^{2} x d x-\int \tan x \sec x d x$
= tan x - sec x + C
View full question & answer→Question 1081 Mark
Evaluate $\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x$
AnswerLet $I=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x$ ...(i)
Then, by using, $\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x$, we get
$I=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{4}\left(\frac{\pi}{2}-x\right)}{\sin ^{4}\left(\frac{\pi}{2}-x\right)+\cos ^{4}\left(\frac{\pi}{2}-x\right)} d x$ = $\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{4} x}{\cos ^{4} x+\sin ^{4} x} d x$ ...(ii)
Adding (i) and (ii), we get
$2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{4} x+\cos ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x=\int_{0}^{\frac{\pi}{2}} 1 \cdot d x=[x]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}$
Hence, $I=\frac{\pi}{4}$
View full question & answer→Question 1091 Mark
Find the integral: $\int cosec x(cosec x+\cot x) d x$
AnswerWe have
$\int\left(cosec x(cosec x+\cot x) d x=\int cosec ^{2} x d x+\int cosec x \cot x d x\right.$
= -cot x - cosec x + C
View full question & answer→Question 1101 Mark
Evaluate $\int\limits_{ - 1}^1 {{{\sin }^5} \ x{{\cos }^4} \ xdx} $
AnswerLet $f(x) = \sin^5x.\cos^4x$
$f(-x) = \sin^5(-x).\cos^4(-x)$
$= -\sin^5x.\cos^4x$
$= -f(x)$
$\therefore$ f is odd function
$\therefore \int\limits_{ - 1}^1 {{{\sin }^5}x.{{\cos }^4}xdx = 0} $
View full question & answer→Question 1111 Mark
Find the integral: $\int(\sin x+\cos x) d x$
AnswerWe have
$\int(\sin x+\cos x) d x=\int \sin x d x+\int \cos x d x$
= -cos x + sin x + C
View full question & answer→Question 1121 Mark
Evaluate $\int _ { 0 } ^ { \pi } \frac { x \sin x } { 1 + \cos ^ { 2 } x } d x$.
AnswerAccording to the question , $I =\int _ { 0 } ^ { \pi } \frac { x \sin x } { 1 + \cos ^ { 2 } x } d x$ .......(i)
$\Rightarrow I = \int _ { 0 } ^ { \pi } \frac { ( \pi - x ) \sin ( \pi - x ) } { 1 + \cos ^ { 2 } ( \pi - x ) } d x$ $[\because \int _0^a f(x)= \int_0^a f(a-x)]$
$\Rightarrow I = \int _ { 0 } ^ { \pi } \frac { ( \pi - x ) \sin x } { 1 + \cos ^ { 2 } x } d x$.....(ii)
On Adding eqs.(i) and (ii),we get
$2 I = \int _ { 0 } ^ { \pi } \frac { \pi \sin x } { \left( 1 + \cos ^ { 2 } x \right) } d x$
Put , $cos x = t$
$-sin x dx = dt$
$\Rightarrow sinx dx = -dt$
Lower limit , when x = 0, then t = 1
Upper limit ,when $x= \pi$, then t = -1
$= - \pi \int _ { 1 } ^ { - 1 } \frac { d t } { 1 + t ^ { 2 } }$
$\Rightarrow 2 I = \pi \int _ { - 1 } ^ { 1 } \frac { d t } { \left( 1 + t ^ { 2 } \right) } $
$= \pi \left[ \tan ^ { - 1 } t \right] _ { - 1 } ^ { 1 }$
$= \pi \left[ \frac { \pi } { 4 } - \left( - \frac { \pi } { 4 } \right) \right] $
$= \frac { \pi ^ { 2 } } { 4 }$
$ \therefore I = \frac { \pi ^ { 2 } } { 4 }$
View full question & answer→Question 1131 Mark
Evaluate $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \sin ^2 x d x$
AnswerLet $f(x) = \sin^2x$
$f(-x) = \sin^2(-x) = \sin^2x = f(x)$
$\therefore$ function is even
$\therefore \int_{ - \pi /4}^{\pi /4} {{{\sin }^2}xdx = 2\int_0^{\pi /4} {{{\sin }^2}xdx} }$
$= \int_\limits0^{π/4} {2\left( {\frac{{1 - \cos 2x}}{2}} \right)dx} $
$= \int_\limits0^{π/4} {\left( {{{1 - \cos 2x}}{}} \right)dx} $
$= \left[ {x - \frac{{\sin 2x}}{2}} \right]_0^{\pi /4}$
$ = \frac{\pi }{4} - \frac{1}{2}$
View full question & answer→Question 1141 Mark
Evaluate $\int _ { 0 } ^ { 1 } \frac { \tan ^ { - 1 } x } { 1 + x ^ { 2 } } dx$
AnswerLet $I = \int _ { 0 } ^ { 1 } \frac { \tan ^ { - 1 } x } { 1 + x ^ { 2 } } d x$
Put $\tan ^ { - 1 } x = t \quad \Rightarrow \frac { 1 } { 1 + x ^ { 2 } } d x = d t$
Lower limit when x = 0, then t = 0
Upper limit when x = 1, then t = $\pi / 4.$
$\therefore I = \int _ { 0 } ^ { \pi / 4 } t d t = \left[ \frac { t ^ { 2 } } { 2 } \right] _ { 0 } ^ { \pi / 4 } = \frac { 1 } { 2 } \left[ \left( \frac { \pi } { 4 } \right) ^ { 2 } - ( 0 ) ^ { 2 } \right] = \frac { \pi ^ { 2 } } { 32 }$
View full question & answer→Question 1151 Mark
Evaluate $\int_{-1}^{1} 5 x^{4} \sqrt{x^{5}+1} d x$
AnswerPut $t = x^5 + 1$, then $dt = 5x^4 dx$.
Therefore, $\int 5 x^{4} \sqrt{x^{5}+1} d x$ = $\int \sqrt{t} d t=\frac{2}{3} t^{\frac{3}{2}}=\frac{2}{3}\left(x^{5}+1\right)^{\frac{3}{2}}$
Hence, $\int_{-1}^{1} 5 x^{4} \sqrt{x^{5}+1} d x$ = $\frac{2}{3}\left[\left(x^{5}+1\right)^{\frac{3}{2}}\right]_{-1}^{1}$
= $\frac{2}{3}\left[\left(1^{5}+1\right)^{\frac{3}{2}}-\left((-1)^{5}+1\right)^{\frac{3}{2}}\right]$
= $\frac{2}{3}\left[2^{\frac{3}{2}}-0^{\frac{3}{2}}\right]=\frac{2}{3}(2 \sqrt{2})=\frac{4 \sqrt{2}}{3}$
View full question & answer→Question 1161 Mark
Evaluate $\int_{0}^{\frac{\pi}{4}} \sin ^{3} 2 t \cos 2 t d t$
AnswerLet $I=\int_{0}^{\frac{\pi}{4}} \sin ^{3} 2 t \cos 2 t d t$.
Consider $\int \sin ^{3} 2 t \cos 2 t d t$
Put sin 2t = u so that 2 cos 2t dt = du or cos 2t dt = $\frac{1}{2}$du
So $\int \sin ^{3} 2 t \cos 2 t d t=\frac{1}{2} \int u^{3} d u$
= $\frac{1}{8}\left[u^{4}\right]=\frac{1}{8} \sin ^{4} 2 t=\mathrm{F}(t)$
Therefore, by the second fundamental theorem of integral calculus
$I=F\left(\frac{\pi}{4}\right)-F(0)=\frac{1}{8}\left[\sin ^{4} \frac{\pi}{2}-\sin ^{4} 0\right]=\frac{1}{8}$
View full question & answer→Question 1171 Mark
Evaluate $\int_{1}^{2} \frac{x d x}{(x+1)(x+2)}$
AnswerLet $I=\int_{1}^{2} \frac{x d x}{(x+1)(x+2)}$
Using partial fraction, we get $\frac{x}{(x+1)(x+2)}=\frac{-1}{x+1}+\frac{2}{x+2}$
So, $\int \frac{x d x}{(x+1)(x+2)}$ = - log |x + 1| + 2log |x + 2| = F(x)
Therefore, by the second fundamental theorem of calculus, we have
I = F(2) - F(1) = [- log 3 + 2 log 4] - [- log 2 + 2 log 3]
= -3 log 3 + log 2 + 2 log 4 = log$ \left(\frac{32}{27}\right)$
View full question & answer→Question 1181 Mark
Evaluate $\int_{4}^{9} \frac{\sqrt{x}}{\left(30-x^{\frac{3}{2}}\right)^{2}} d x$
AnswerLet $I=\int_{4}^{9} \frac{\sqrt{x}}{\left(30-x^{\frac{3}{2}}\right)^{2}} d x$.
We first find the antiderivative of the integrand.
Put $30-x^{\frac{3}{2}}=t$. Then $-\frac{3}{2} \sqrt{x} d x=d t$ or $\sqrt{x} d x=-\frac{2}{3} d t$
Thus, $\int \frac{\sqrt{x}}{\left(30-x^{\frac{3}{2}}\right)^{2}} d x$ = $-\frac{2}{3} \int \frac{d t}{t^{2}}=\frac{2}{3}\left[\frac{1}{t}\right]=\frac{2}{3}\left[\frac{1}{\left(30-x^{\frac{3}{2}}\right)}\right]=\mathrm{F}(x)$
Therefore, by the second fundamental theorem of calculus, we have
I = F(9) - F(4)
= $\frac{2}{3}\left[\frac{1}{(30-27)}-\frac{1}{30-8}\right]$
$=\frac{2}{3}\left[\frac{1}{3}-\frac{1}{22}\right]=\frac{19}{99}$
View full question & answer→Question 1191 Mark
Evaluate $\int_{2}^{3} x^{2} d x$
Answerf(x) is continuous in [2, 3]
$\int_{a}^{b} f(x) d x$ = $\mathop {\lim }\limits_{n \to \infty } h\sum\limits_{r = 0}^{n - 1} f (a + rh)$ where $\mathrm{h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}$
Here,
$\int_{2}^{3}\left(x^{2}\right) d x$ = $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n}} \right)\sum\limits_{r = 0}^{n - 1} f \left( {2 + \left( {\frac{r}{n}} \right)} \right)$
= $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n}} \right)\sum\limits_{r = 0}^{n - 1} {{{\left( {2 + \left( {\frac{r}{n}} \right)} \right)}^2}}$
= $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n}} \right)\sum\limits_{r = 0}^{n - 1} {\left( {\frac{{{r^2}}}{{{n^2}}} + 4 + \frac{{4r}}{n}} \right)}$
= $\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left( {\frac{{(n - 1)(n)(2n - 1)}}{{6{n^2}}} + 4n + \frac{{4(n - 1)(n)}}{{2n}}} \right)$
= $\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left( {\frac{{\left( {{n^2} - n} \right)(2n - 1)}}{{6{n^2}}} + 4n + \frac{{2\left( {{n^2} - n} \right)}}{n}} \right)$
= $\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left(\frac{\left(2 n^{3}-2 n^{2}-n^{2}+n\right)}{6 n^{2}}+4 n+\frac{2\left(n^{2}-n\right)}{n}\right)$
= $\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left( {\frac{{\left( {2{n^3} - 3{n^2} + n} \right) + \left( {24{n^3}} \right) + \left( {12{n^3} - 12{n^2}} \right)}}{{6{n^2}}}} \right)$
= $\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left( {\frac{{38{n^3} - 15{n^2} + n}}{{6{n^2}}}} \right)$
= $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{38{n^3} - 15{n^2} + n}}{{6{n^3}}}} \right)$
= $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{38}}{6}} \right) - \left( {\frac{{15}}{{6n}}} \right) + \left( {\frac{1}{{6{n^2}}}} \right)$
= $\frac{38}{6}$
= $\frac{19}{3}$
View full question & answer→Question 1201 Mark
Find $\int \sqrt{3-2 x-x^{2}} d x$
AnswerLet I = $\int \sqrt{3-2 x-x^{2}} \;d x$
= $\int \sqrt{-\left(x^{2}+2 x-3\right)} d x$
$=\int \sqrt{-\left(x^{2}+1+2 x-3-1\right)} d x$
= $\int \sqrt{-\left[(x+1)^{2}-4\right]} d x$
= $\int \sqrt{4-(x+1)^{2}} d x$
= $\int \sqrt{2^{2}-(x+1)^{2}} d x$ $\left[\because\int\sqrt{a^2-x^2}dx=\frac x2\sqrt{a^2-x^2}+\frac{a^2}2\sin^{-1}\left(\frac xa\right)+c\right]$
= $\frac{1}{2}(x+1) \sqrt{4-(x+1)^{2}} +\frac{4}{2} \sin ^{-1}\left(\frac{x+1}{2}\right)+c$
$\therefore \ \mathrm{I}=\int \sqrt{2^{2}-(x+1)^{2}} d x$
$=\frac{1}{2}(x+1) \sqrt{3-2 x-x^{2}}+2 \sin ^{-1} \frac{(x+1)}{2}+c$
View full question & answer→Question 1211 Mark
Find $\int \sqrt{x^{2}+2 x+5} d x$
AnswerNote that
$\int \sqrt{x^{2}+2 x+5} d x=\int \sqrt{(x+1)^{2}+4} d x$
Put x + 1 = y, so that dx = dy. Then
$\int \sqrt{x^{2}+2 x+5} d x=\int \sqrt{y^{2}+2^{2}} d y$
= $\frac{1}{2} y \sqrt{y^{2}+4}+\frac{4}{2} \log |y+\sqrt{y^{2}+4}|+C$
= $\frac{1}{2}(x+1) \sqrt{x^{2}+2 x+5}+2 \log |x+1+\sqrt{x^{2}+2 x+5}|+\mathrm{C}$
View full question & answer→Question 1221 Mark
Find the integral: $\int\left(x^{\frac{3}{2}}+2 e^{x}-\frac{1}{x}\right) d x$
AnswerWe have
$\int\left(x^{\frac{3}{2}}+2 e^{x}-\frac{1}{x}\right) d x=\int x^{\frac{3}{2}} d x+\int 2 e^{x} d x-\int \frac{1}{x} d x$
= $\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}+2 e^{x}-\log |x|+C$
= $\frac{2}{5} x^{\frac{5}{2}}+2 e^{x}-\log |x|+C$
View full question & answer→Question 1231 Mark
Find $\int e^{x}\left(\tan ^{-1} x+\frac{1}{1+x^{2}}\right) d x$
AnswerWe have $I=\int e^{x}\left(\tan ^{-1} x+\frac{1}{1+x^{2}}\right) d x$
Consider $f(x) = \tan^{-1} x$, then f′(x) = $\frac{1}{1+x^{2}}$
Thus, the given integrand is of the form $e^x [f(x) + f′(x)]$.
Therefore, $I=\int e^{x}\left(\tan ^{-1} x+\frac{1}{1+x^{2}}\right) d x = e^x \tan^{-1} x + C$
View full question & answer→Question 1241 Mark
Find the integral: $\int\left(x^{\frac{2}{3}}+1\right) d x$
AnswerWe have
$\int\left(x^{\frac{2}{3}}+1\right) d x=\int x^{\frac{2}{3}} d x+\int 1d x$
= $\frac{x^{\frac{2}{3}+1}}{\frac{2}{3}+1}+x+\mathrm{C}=\frac{3}{5} x^{\frac{5}{3}}+x+\mathrm{C}$
View full question & answer→Question 1251 Mark
Find $\int e^{x} \sin x d x$
AnswerTake $e^x$ as the first function and sin x as second function.
Using Integrating by parts, we have
$\mathrm{I}=\int e^{x} \sin x d x=e^{x}(-\cos x)+\int e^{x} \cos x d x$
$= -e^x \cos x + I_1$ ......(i)
Taking $e^x$ and cos x as the first and second functions, respectively, in $I_1$, we get
$I_{1}=e^{x} \sin x-\int e^{x} \sin x d x$
Substituting the value of $I_1$ in (i), we get
$I = -e^x \cos x + e^x sin x - I$
$\Rightarrow~2I = e^x$ (sin x - cos x)
Hence, $I=\int e^{x} \sin x d x=\frac{e^{x}}{2}(\sin x-\cos x)+C$
View full question & answer→Question 1261 Mark
Find the integral: $\int \frac{x^{3}-1}{x^{2}} d x$
AnswerWe have
$\int \frac{x^{3}-1}{x^{2}} d x=\int x d x-\int x^{-2} d x$
= $\left(\frac{x^{1+1}}{1+1}+C_{1}\right)-\left(\frac{x^{-2+1}}{-2+1}+C_{2}\right); C_1, C_2$ are constants of integration
= $\frac{x^{2}}{2}+C_{1}-\frac{x^{-1}}{-1}-C_{2}$ = $\frac{x^{2}}{2}+\frac{1}{x}+C_{1}-C_{2}$
= $\frac{x^{2}}{2}+\frac{1}{x}+C$ where $C = C_1 - C_2$ is another constant of integration
View full question & answer→Question 1271 Mark
Find $\int \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} d x$
AnswerLet first function be $\sin^{-1} x$ and second function be $\frac{x}{\sqrt{1-x^{2}}}$.
First we find the integral of the second function, i.e., $\int \frac{x d x}{\sqrt{1-x^{2}}}$
Put $t =1 - x^2$. Then dt = -2x dx
Therefore, $\int \frac{x d x}{\sqrt{1-x^{2}}}=-\frac{1}{2} \int \frac{d t}{\sqrt{t}}=-\sqrt{t}=-\sqrt{1-x^{2}}$
Hence, $\int \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} d x$ = $(\sin ^{-1} x)$$\int \frac{x d x}{\sqrt{1-x^{2}}}$ $-\int [\frac d{dx} sin^{-1}x.\int \frac{x d x}{\sqrt{1-x^{2}}} ] d x$
= $\left(\sin ^{-1} x\right)(-\sqrt{1-x^{2}})-\int \frac{1}{\sqrt{1-x^{2}}}(-\sqrt{1-x^{2}}) d x$
= $-\sqrt{1-x^{2}} \sin ^{-1} x+x+C$
= $x-\sqrt{1-x^{2}} \sin ^{-1} x+\mathrm{C}$
View full question & answer→Question 1281 Mark
Find $\int x e^{x} d x$
AnswerTake first function as x and second function as $e^x$. The integral of the second function is $e^x$
Therefore, $\int x e^{x} d x=x e^{x}-\int 1 \cdot e^{x} d x=x e^{x}-e^{x}+C$
View full question & answer→Question 1291 Mark
Find $\int \log x d x$
AnswerWe take log x as the first function and the constant function 1 as the second function.
Hence, $\int(\log x ).1 d x=\log x \int 1 d x-\int\left[\frac{d}{d x}(\log x) \int 1 d x\right] d x$
= $(\log x) \cdot x-\int \frac{1}{x} x d x=x \log x-x+C$
View full question & answer→Question 1301 Mark
Find $\int x \cos x d x$
AnswerPut f (x) = x (first function) and g (x) = cos x (second function).
Then, integration by parts gives
$\int x \cos x d x=x \int \cos x d x-\int\left[\frac{d}{d x}(x) \int \cos x d x\right] d x$
= $x \sin x-\int \sin x d x=x \sin x+\cos x+\mathrm{C}$
View full question & answer→Question 1311 Mark
Find: $\int \frac{x^{2}+x+1}{(x+2)\left(x^{2}+1\right)} d x$
AnswerLet, $I=\int \frac{x^{2}+x+1}{(x+2)\left(x^{2}+1\right)} d x$
By partial fractions
$\frac{x^{2}+x+1}{(x+2)\left(x^{2}+1\right)}=\frac{A}{x+2}+\frac{B x+C}{x^{2}+1}$
$x^2 + x + 1 = A(x^2+ 1) + Bx + C(x + 2)$
$=Ax^2+ A + Bx^2+ 2Bx + Cx + 2Cx^2+ x + 1$
$= x^2(A + B) + x(2B + C) + A + 2C$
On comparing coefficients of equation $(i),$ we get
$1 = A + B$ and $1 = 2B + C$
On solving above three equations, we get
$A=\frac{3}{5}$, $ B=\frac{2}{5} $ and $C=\frac{1}{5}$
Hence, $\frac{x^{2}+x+1}{(x+2)\left(x^{2}+1\right)}=\frac{3}{5(x+2)}+\frac{\frac{2}{5} x+\frac{1}{5}}{x^{2}+1}$
Therefore, $\int \frac{\left(x^{2}+x+1\right) d x}{(x+2)\left(x^{2}+1\right)}=\frac{3}{5} \int \frac{1}{x+2} d x+\frac{2}{5} \int \frac{x}{x^{2}+1} d x +\frac{1}{5} \int \frac{1}{\left(x^{2}+1\right)} d x$
= $\frac{3}{5} \log |x+2|+\frac{1}{5} \int \frac{2 x}{x^{2}+1}+\frac{1}{5} \int \frac{d x}{x^{2}+1}$
= $\frac{3}{5} \log |x+2|+\frac{1}{5} \log \left|x^{2}+1\right|+\frac{1}{5} \tan ^{-1} x+c$
View full question & answer→Question 1321 Mark
Find: $\int \frac{(3 \sin \phi-2) \cos \phi}{5-\cos ^{2} \phi-4 \sin \phi} d \phi$
AnswerLet $y=\sin \phi$ $\Rightarrow d y=\cos \phi d \phi$
Therefore, $\int \frac{(3 \sin \phi-2) \cos \phi}{5-\cos ^{2} \phi-4 \sin \phi} d \phi$ = $\int \frac{(3 y-2) d y}{5-\left(1-y^{2}\right)-4 y}$
= $\int \frac{3 y-2}{y^{2}-4 y+4} d y$
Now, we write $\frac{3 y-2}{(y-2)^{2}}=\frac{A}{y-2}+\frac{B}{(y-2)^{2}}$
Therefore, 3y - 2 = A (y - 2) + B
Comparing the coefficients of y and constant term,
we get A = 3 and B = 4. for y = 2, y= 0
Therefore, the required integral is given by
$I=\int\left[\frac{3}{y-2}+\frac{4}{(y-2)^{2}}\right] d y$ = $3 \int \frac{d y}{y-2}+4 \int \frac{d y}{(y-2)^{2}}$
= $3 \log |y-2|+4\left(-\frac{1}{y-2}\right)+C$
= $3 \log |\sin \phi-2|+\frac{4}{2-\sin \phi}+C$
= $3 \log (2-\sin \phi)+\frac{4}{2-\sin \phi}+C$ (since, sin $\phi $ $\in$[-1,1], sin $\phi $< 2, 2 - sin $\phi $ is always positive)
View full question & answer→Question 1331 Mark
Find: $\int \frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x$
AnswerConsider $\frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)}$ and put $x^2 = y$
Then, $\frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{y}{(y+1)(y+4)}$
Write, $\frac{y}{(y+1)(y+4)}=\frac{A}{y+1}+\frac{B}{y+4}$
So that, y = A (y + 4) + B (y + 1)
Comparing coefficients of y and constant terms on both sides,
we get A + B = 1 and 4A + B = 0, which give
$A=-\frac{1}{3}$ and $B=\frac{4}{3}$
Thus, $\frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=-\frac{1}{3\left(x^{2}+1\right)}+\frac{4}{3\left(x^{2}+4\right)}$
= $-\frac{1}{3} \tan ^{-1} x+\frac{4}{3} \times \frac{1}{2} \tan ^{-1} \frac{x}{2}+C$
= $-\frac{1}{3} \tan ^{-1} x+\frac{2}{3} \tan ^{-1} \frac{x}{2}+C$
View full question & answer→Question 1341 Mark
Find: $\int \frac{3 x-2}{(x+1)^{2}(x+3)} d x$
AnswerWe write
$\frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{x+3}$
So that $3x - 2 = A (x + 1) (x + 3) + B (x + 3) + C (x + 1)^2$
$= A (x^2 + 4x + 3) + B (x + 3) + C (x^2 + 2x + 1)$
Comparing coefficient of $x^2, x$ and constant term on both sides,
we get $A + C = 0, 4A + B + 2C = 3$ and $3A + 3B + C = -2$. Solving these equations, we get
$A=\frac{11}{4}, B=\frac{-5}{2}$ and $C=\frac{-11}{4}$. Thus the integrand is given by
$\frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{11}{4(x+1)}-\frac{5}{2(x+1)^{2}}-\frac{11}{4(x+3)}$
Therefore, $\int \frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{11}{4} \int \frac{d x}{x+1}-\frac{5}{2} \int \frac{d x}{(x+1)^{2}}-\frac{11}{4} \int \frac{d x}{x+3}$
= $\frac{11}{4} \log |x+1|+\frac{5}{2(x+1)}-\frac{11}{4} \log |x+3|+\mathrm{C}$
= $\frac{11}{4} \log \left|\frac{x+1}{x+3}\right|+\frac{5}{2(x+1)}+C$
View full question & answer→Question 1351 Mark
Write an anti derivative of function using the method of inspection: $\frac{1}{x}, x \neq 0$
AnswerWe know that
$\frac{d}{d x}(\log x)=\frac{1}{x}, x>0$ and $\frac{d}{d x}[\log (-x)]=\frac{1}{-x}(-1)=\frac{1}{x}, x<0$
Combining above, we get $\frac{d}{d x}(\log |x|)=\frac{1}{x}, x \neq 0$
Therefore, $\int \frac{1}{x} d x=\log |x|$ is one of the anti derivatives of $\frac{1}{x}$
View full question & answer→Question 1361 Mark
Find $\int \frac{x^{2}+1}{x^{2}-5 x+6} d x$
AnswerHere the integrand $\frac{x^{2}+1}{x^{2}-5 x+6}$ is not proper rational function, so we divide $x^2 + 1$ by $x^2 - 5x + 6$ and find that
$\frac{x^{2}+1}{x^{2}-5 x+6}=1+\frac{5 x-5}{x^{2}-5 x+6}=1+\frac{5 x-5}{(x-2)(x-3)}$
Let $\frac{5 x-5}{(x-2)(x-3)}=\frac{A}{x-2}+\frac{B}{x-3}$
So that 5x - 5 = A (x - 3) + B (x - 2)
Equating the coefficients of x and constant terms on both sides,
we get A + B = 5 and 3A + 2B = 5.
Solving these equations, we get A = -5 and B = 10
Thus, $\frac{x^{2}+1}{x^{2}-5 x+6}=1-\frac{5}{x-2}+\frac{10}{x-3}$
Therefore, $\int \frac{x^{2}+1}{x^{2}-5 x+6} d x=\int d x-5 \int \frac{1}{x-2} d x+10 \int \frac{d x}{x-3}$
= x - 5 log | x - 2| + 10 log | x - 3| + C.
View full question & answer→Question 1371 Mark
Write an anti derivative of function using the method of inspection: $3x^2 + 4x^3$
AnswerA function whose anti derivative is $3x^2 + 4x^3$.
$\frac{d}{d x}\left(x^{3}+x^{4}\right)=3 x^{2}+4 x^{3}$
Therefore, an anti derivative of $3x^2 + 4x^3$ is $x^3 + x^4$
View full question & answer→Question 1381 Mark
Find $\int \frac{d x}{(x+1)(x+2)}$
AnswerThe integral is a proper rational function.
Therefore, by using the form of partial fraction, we write
$\frac{1}{(x+1)(x+2)}=\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B}}{x+2}$ ......(i)
where, real numbers A and B are to be determined suitably. This gives
1 = A (x + 2) + B (x + 1).
Equating the coefficients of x and the constant term, we get
A + B = 0 and 2A + B = 1
Solving these equations, we get A = 1 and B = -1.
Thus, the integral is given by
$\frac{1}{(x+1)(x+2)}=\frac{1}{x+1}+\frac{-1}{x+2}$
Therefore, $\int \frac{d x}{(x+1)(x+2)}=\int \frac{d x}{x+1}-\int \frac{d x}{x+2}$
= log |x + 1| - log |x + 2| + C
= $\log \left|\frac{x+1}{x+2}\right|+C$
View full question & answer→Question 1391 Mark
Write an anti derivative of function cos 2x using the method of inspection.
AnswerA function whose derivative is cos 2x.
$\frac{d}{d x} \sin 2 x=2 \cos 2 x$
or $\cos 2 x=\frac{1}{2} \frac{d}{d x}(\sin 2 x)=\frac{d}{d x}\left(\frac{1}{2} \sin 2 x\right)$
Therefore, an anti derivative of cos 2x is $\frac{1}{2} \sin 2 x$
View full question & answer→Question 1401 Mark
Find the integral: $\int \frac{x+3}{\sqrt{5-4 x-x^{2}}} d x$
AnswerGiven integral is:$\int \frac{x+3}{\sqrt{5-4 x-x^{2}}} d x$
Let us express
$x+3=A \frac{d}{d x}\left(5-4 x-x^{2}\right)+B$ = A (- 4 - 2x) + B
Equating the coefficients of x and the constant terms from both sides, we get
- 2A = 1 and -4 A + B = 3, i.e., $A=-\frac{1}{2}$ and B = 1
Therefore, $\int \frac{x+3}{\sqrt{5-4 x-x^{2}}} d x=-\frac{1}{2} \int \frac{(-4-2 x) d x}{\sqrt{5-4 x-x^{2}}}+\int \frac{d x}{\sqrt{5-4 x-x^{2}}}$
= $-\frac{1}{2} I_{1}+I_{2}$ ........(i)
In $I_1$, put $5 - 4x - x^2 = t$, so that (- 4 - 2x) dx = dt.
Therefore, $I_{1}=\int \frac{(-4-2 x) d x}{\sqrt{5-4 x-x^{2}}}=\int \frac{d t}{\sqrt{t}}=2 \sqrt{t}+C_{1}$
= $2 \sqrt{5-4 x-x^{2}}+\mathrm{C}_{1}$ .......(ii)
Now consider $I_{2}=\int \frac{d x}{\sqrt{5-4 x-x^{2}}}=\int \frac{d x}{\sqrt{9-(x+2)^{2}}}$
Put x + 2 = t, so that dx = dt.
Therefore, $I_{2}=\int \frac{d t}{\sqrt{3^{2}-t^{2}}}=\sin ^{-1} \frac{t}{3}+C_{2}$
= $\sin ^{-1} \frac{x+2}{3}+C_{2}$ .......(iii)
Substituting (ii) and (iii) in (i), we obtain
$\int \frac{x+3}{\sqrt{5-4 x-x^{2}}}=-\sqrt{5-4 x-x^{2}}+\sin ^{-1} \frac{x+2}{3}+\mathrm{C}, $ where $\mathrm{C}=\mathrm{C}_{2}-\frac{\mathrm{C}_{1}}{2}$
View full question & answer→Question 1411 Mark
Find the integral: $\int \frac{x+2}{2 x^{2}+6 x+5} d x$
AnswerLet,$ I = ∫\frac{(x+2)}{(2x^2+6x+5)}dx$
we express
$x+2=\mathrm{A} \frac{d}{d x}\left(2 x^{2}+6 x+5\right)+\mathrm{B} = A(4x + 6) + B$
Equating the coefficients of x and the constant terms from both sides, we get
$4A = 1$ and $6A + B = 2$ or $A=\frac{1}{4}$ and $B=\frac{1}{2}$
Therefore, $\int \frac{x+2}{2 x^{2}+6 x+5}=\frac{1}{4} \int \frac{4 x+6}{2 x^{2}+6 x+5} d x+\frac{1}{2} \int \frac{d x}{2 x^{2}+6 x+5}$
$= \frac{1}{4} \mathrm{I}_{1}+\frac{1}{2} \mathrm{I}_{2}$ .....(i)
In $I_1,$ put $2x^2 + 6x + 5 = t,$ so that $(4x + 6) dx = dt$
Therefore, $I_{1}=\int \frac{d t}{t}=\log |t|+C_{1}$
$= \log |2x^2 + 6x + 5| + C .....(ii)$
and $I_{2}=\int \frac{d x}{2 x^{2}+6 x+5}=\frac{1}{2} \int \frac{d x}{x^{2}+3 x+\frac{5}{2}}$
$= \frac{1}{2} \int \frac{d x}{\left(x+\frac{3}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}}$
Put $x+\frac{3}{2}=t$, so that $dx = dt,$ we get
$I_{2}=\frac{1}{2} \int \frac{d t}{t^{2}+\left(\frac{1}{2}\right)^{2}}=\frac{1}{2 \times \frac{1}{2}} \tan ^{-1} 2 t+C_{2}$
$\tan ^{-1} 2\left(x+\frac{3}{2}\right)+\mathrm{C}_{2}=\tan ^{-1}(2 x+3)+\mathrm{C}_{2}......(iii)$
Using (ii) and (iii) in (i), we get
$\int \frac{x+2}{2 x^{2}+6 x+5} d x=\frac{1}{4} \log \left|2 x^{2}+6 x+5\right|+\frac{1}{2} \tan ^{-1}(2 x+3)+\mathrm{C}$
where, $C=\frac{C_{1}}{4}+\frac{C_{2}}{2}$
View full question & answer→