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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