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Examples of calculus

August 16, 2020
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Examples :–

1 Evaluate $\int\frac{e^x}{1+e^{\mathrm{2x}}} \ dx$ .

Soln. Put e^x= t then e^x dx = dt

$\therefore\int\frac{e^x}{1+e^{\mathrm{2x}}}dx$ = $\int\frac{\mathrm{dt}}{1+t^2}$

= tan^-1t + c

= tan^-1 e^x + c

2 Evaluate dx

Soln. Put sec x + tan x = t, then ( sec x tan x + sec² x ) dx = dt

sec x ( tan x + sec x ) dx = dt $\large \Rightarrow$ sec x dx = $\frac{dt}{t}$

$\therefore\int s e c{x} \ dx$ = $\int\frac{dt}{t}$

= log |t| + c

= log_e $\left|sec{x}\ +\mathrm{\ tan\ x} \right|$ + c

3 Evaluate $\int_{0}^{1}\frac{e^x}{\mathbf{1}+e^{\mathrm{2x}}}\ \ dx$

Soln. Put e^x= t then e^xdx = dt

Also, x = 0 ⇒ t = 1 and x = 1 ⇒ t = e

∴ $\int_{0}^{1}\frac{e^x}{1+e^{\mathrm{2x}}}dx$ = $\int_{1}^{e}\frac{\mathrm{dt}}{1+t^2}$ = $\left[{tan}^{-1}\right]_1^e$ = tan^-1 e – tan^-1 1 = tan^-1 e – $\frac{\pi}{4}$

4 Evaluate $\int\frac{\mathrm{dx}}{\mathbf{1}+\mathrm{tan\ x}}$ = $\int\frac{cos{x}}{\mathrm{cos\ x\ }+\mathrm{\ sin\ x}}\ dx$

Soln. I = $\frac{1}{2}dx$ = $\frac{1}{2} + \int\frac{\mathrm{cos\ x\ -\ sin\ x}}{cos{x}\ +\mathrm{\ sin\ x} }dx$

= $\frac{1}{2}x + loge\ \left|cos{x}\ +\mathrm{\ sin\ x} \right| + c$ [ $\because \int{\frac{f^\prime\ (x)}{f\ (x)}\ dx}$ = |log f(x)| + c ]

5 Evaluate $\int\frac{\mathrm{dx}}{x^\mathbf{2}+\mathrm{x\ } +\mathrm{\ 1} }$

Soln. We have x² + x + 1 = $\left(\mathrm{x\ }+\ \frac{1}{2}\right)^2 + \frac{3}{4}$

$\therefore\int\frac{\mathrm{dx}}{x^2+\mathrm{x\ } +\mathrm{\ 1} }$ = $\int\frac{\mathrm{dx}}{\left(\mathrm{x\ }+\ \frac{1}{2}\right)^2+\left(\frac{\sqrt3}{2}\right)^2}$

= $\frac{1}{\frac{\sqrt3}{2}}tan-1\ \frac{\mathrm{x\ }+\ \frac{1}{2}}{\frac{\sqrt3}{2}} + c$

= $\frac{2}{\sqrt3} tan-1\ \frac{\ 2\mathrm{x\ }+\mathrm{\ 1}}{\ \sqrt3} + c$

6 Integration $\frac{1}{\sqrt{3-4\mathrm{x\ -\ 2}\mathrm{x}^\mathbf{2}}}$

Soln. We have 3 – 4x – 2x² = – $2\left[x^2+2\mathrm{x\ -\ } \frac{3}{2}\right]$ = – 2 $\left[(\mathrm{x\ }+\mathrm{\ 1})^2-\frac{5}{2}\right]$

$\therefore \int\frac{\mathrm{dx}}{\sqrt{\mathrm{3-4x-2}\mathrm{x}^2}}$ = $\int\frac{\mathrm{dx}}{\sqrt2\sqrt{\left(\sqrt{\frac{5}{2}}\right)^2-(\mathrm{x\ } +\mathrm{\ 1\ } )^2}}$

= $\frac{1}{\sqrt2}$ sin^-1 + c

= $\frac{1}{\sqrt2}$ sin^-1 $\frac{\sqrt2(\mathrm{x\ } +\mathrm{\ 1} )}{\sqrt5} + c$

7 If $\int\frac{3\mathrm{x\ }+\mathrm{\ 2}}{\mathrm{4}\mathrm{x}^\mathbf{2}+4\mathrm{x\ } +\mathrm{\ 5} }dx$ is A log ( 4x+ 4x + 5 ) + B tan^-1 $\left(x+\frac{\mathbf{1}}{\mathbf{2}}\right)$ + c. Then A and B are

(a) $\frac{3}{2}, \frac{1}{2}$ (b) $\frac{3}{8}, \frac{1}{8}$ (c) $\frac{3}{2}, \frac{1}{8}$ (d) $\frac{3}{8}, \frac{1}{2}$

Soln. Let 3x + 2 = m $\frac{d}{\mathrm{dx}} ( 4x+ 4x + 5 )$ + n i.e., 3x + 2 = m( 8x + 4 ) + n

Equating coefficients of x from both the sides we get 3 = $8m \Rightarrow m = \frac{3}{8}$

Equating constant term from both sides, we get 2 = 4m + n

$\large \Rightarrow$ n = 2 – 4 $\times\frac{3}{8}$ = 2 – $\frac{3}{2} = \frac{1}{2}$

Thus, $\int\frac{3\mathrm{x\ }+\mathrm{\ 2}}{\mathrm{4}\mathrm{x}^2+4\mathrm{x\ } +\mathrm{\ 5} }dx$ = $\int\frac{\frac{3}{8}(8\mathrm{x\ } +\mathrm{\ 4\ } )\ +\ \frac{1}{2}}{4x^2+4\mathrm{x\ } +\mathrm{\ 5} }\ dx$

= $\frac{3}{8}dx + \frac{1}{2}\int\frac{\mathrm{dx\ }}{\mathrm{4}\mathrm{x}^2+4\mathrm{x\ } +\mathrm{\ 5} }$

= $\frac{3}{8}log ( 4x2+ 4x + 5 ) +$ $\frac{1}{8}\int\frac{\mathrm{dx}}{\left(x+\frac{1}{2}\right)^2+1}$

= $\frac{3}{8}$ log ( 4x² + 4x + 5 ) + $\frac{1}{8}$ tan^-1 $\left(x+\frac{1}{2}\right)$ + c

Clearly A = $\frac{3}{8}$ and B = $\frac{1}{8}.$

8 If $\int\frac{x ^\mathbf{2}+\mathrm{x\ } +\mathrm{\ 1} }{x^\mathbf{2}-x}$ = Ax + B log $\frac{\left|\mathrm{x-1}\right|^3}{\left|x\right|}$ + c. Then,

(a) A = $\frac{1}{2}$ , B = $\frac{1}{2}$ (b) A = 1, B = 1 (c) A = – 1, B = $\frac{1}{2}$ (d) A = – $\frac{1}{2}$ , B = 1

Soln. Carry out the division $\frac{x^2+x +1 }{x^2-x}$ , we get

$\frac{x^2+x +1 }{x^2-x}$ = 1 + $\frac{x^2+x +1 }{x^2-x}$

Now, $\int\frac{x^2+x +1 }{x^2-x}$ = $\int\left[1+\frac{2\mathrm{x\ }+\mathrm{\ 1}}{x^2-x}\right]dx$

= $\int\mathrm{dx} + \int\frac{2\mathrm{x\ -\ 1}}{x^2-x}dx + 2\int\frac{\mathrm{dx}}{x^2-x}$

[ As in above example, ( 2x + 1 ) = m $\frac{d}{\mathrm{dx}}$ ( x² – x ) + n, etc. ]

= x + log $\left|x^2-x\right|$ + 2 $\int\frac{\mathrm{dx}}{\left(\mathrm{x\ -\ }\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2}$

= x + log $\left|x^2-x\right|$ + $2.\frac{1}{2\cdot\frac{1}{2}}log \left|\frac{\mathrm{x-}\frac{1}{2}-\frac{1}{2}}{\mathrm{x\ -\ }\frac{1}{2}+\frac{1}{2}}\right|$ + C

= x + log $\left[\left|\mathrm{\ x\ }(\mathrm{x\ -\ 1\ })\right|\left(\frac{\mathrm{x\ -\ 1}}{x}\right)^2\right]$ + C

= x + $log \left[\frac{\left|\mathrm{x\ -\ 1}\right|^3}{\left|x\right|}\right]$ + C

Clearly A = 1 and B = 1.

9 Evaluate $\int\frac{\mathrm{dx}}{\mathbf{2}+\mathrm{3co}\mathrm{s}^\mathbf{2}\mathrm{\ x} }$

Soln. We first divide the expression by cos²x, we get

I = $\int\frac{\mathrm{dx}}{2+\mathrm{3co}\mathrm{s}^2\mathrm{\ x} }$ = $\int\frac{{sec}^2{\mathrm{x\ dx}}}{\mathrm{2se}\mathrm{c}^2\mathrm{x\ } +\mathrm{\ 3} }$

We replace sec²x by 1 + tanx in the denominator

I = $\int\frac{{sec}^2{\mathrm{x\ dx}}}{\mathrm{5\ }+\mathrm{\ 2ta}\mathrm{n}^2x}$ = $\frac{1}{2}$ $\int\frac{{sec}^2{\mathrm{x\ dx}}}{\mathrm{ta}\mathrm{n}^2\mathrm{x\ } +\frac{5}{2}}$

Put tan x = t ⇒ sec² x dx = dt, we get

I = $\frac{1}{2}\int\frac{\mathrm{\ dt}}{t^2+\frac{5}{2}}$

= $\frac{1}{2}\times\frac{1}{\sqrt{\frac{5}{2}}}$ tan^-1 $\frac{t}{\sqrt{\frac{5}{2}}}$

= $\frac{1}{\sqrt{10}}$ tan^-1 $\left(\sqrt{\frac{2}{5}}tan{x}\right) + C.$

10 Integration of $\int\frac{dx}{(x-1)\ (x+2)(2x+3)}$ is

Soln. Let $\frac{1}{(\mathrm{x\ -\ 1\ })\ (\mathrm{\ x\ }+\mathrm{\ 2\ })\ (\mathrm{\ 2x\ }+\mathrm{\ 3\ })\ }$ = $\frac{A_1}{\mathrm{x\ -\ 1}} + \frac{A_2}{\mathrm{x\ }+2} + \frac{A_3}{\mathrm{2x\ }+3}$

A₁ = $\frac{1}{(1+2)(2\times 1+3)} = \frac{1}{15}$

A₂ = $\frac{1}{(-2-1)(-2\cdot2+3)} = \frac{1}{3}$

A₃ = $\frac{1}{\left(-\frac{3}{2}-1\right)\left(-\frac{3}{2}+2\right)}$ = $\frac{1}{-\frac{5}{4}}$ = – $\frac{4}{5}$

Now I = $\int\frac{dx}{(x-1)\ (x+2)(2x+3)}$

= $\int\left[\frac{1}{15(\mathrm{x\ -\ 1})}+\frac{1}{3(\mathrm{x\ }+\mathrm{\ 2\ })}-\frac{4}{5(2\mathrm{x\ }+\mathrm{\ 3\ })}\right]dx$

= $\frac{1}{15} log |x-1|$ $+ \frac{1}{3}log ( x + 2 )$ $- \frac{2}{5} log \left|2\mathrm{x\ }+\mathrm{\ 3}\right| + c$

11 Evaluate : ${\int{e^x\left(\frac{1-x}{\mathbf{1}+x^\mathbf{2}}\right)}}^2dx$

Soln. I = ${\int{e^x\left(\frac{1-x}{1+x^2}\right)}}^2dx$

= $\int{e^x\left(\frac{1-2\mathrm{x\ }+\mathrm{\ }\mathrm{x}^2}{(1+x^2)^2}\right)}dx$

= $\int{e^x\left\{\frac{1}{1+x^2}-\frac{2x}{(\mathrm{\ 1\ }+\mathrm{\ }\mathrm{x}^2)^2}\right\}}dx$

Let f ( x ) = $\frac{1}{1+x^2} then f^\prime ( x )$ = $\frac{-2x}{(\mathrm{\ 1\ }+\mathrm{\ }\mathrm{x}^2)^2}$

$\therefore$ I = $\int{e^x\left\{\frac{1}{1+x^2}-\frac{2x}{(\mathrm{\ 1\ }+\mathrm{\ }\mathrm{x}^2)^2}\right\}}dx$

= $\frac{e^x}{1+x^2} + c$

12 If $\int{sin(log x\ )dx = \frac{x}{\sqrt2} f ( x ) + c$ Then f ( x ) = ………..

Soln. Let I = $\int{sin{(}log{x})} dx$

Put log x = t ⇒ $\frac{1}{x}dx$ = dt ⇒ dx = x dt ⇒ dx = e^t dt

$\therefore I = \int{(sin{t})e^t}dt$ = sin t ^. e – $\int{(cos{t})e^t}dt$

( $\because \int{u\ \ v\ \ dx}$ = $u \int{v\ \ \ dx}$ – $\int\left[\frac{du}{dx}\ \ \int{v\ \ dx}\right] + c$ )

= e^t sin t – $\left\lfloor c o s{t}\cdot e^t-\int{(-sin{t})e^t\mathrm{dt} }\right\rfloor$

= e^t sin t – e^t cos t – I

$\therefore I$ = $\frac{1}{2}e ( sin t - cos t )$ + c

= $\frac{x}{2}\left[sin{(}log{x})\mathrm{\ -\ cos\ } (\mathrm{\ log\ x\ } )\right]$ + c

= $\frac{x}{\sqrt2}\left[\frac{1}{\sqrt2}sin{(}log{x})\mathrm{\ -\ } \frac{1}{\sqrt2}cos{(}log{x})\right]$ + c

= $\frac{x}{\sqrt2}sin \left(log{x}\mathrm{\ -\ } \frac{\pi}{4}\right) + c$

$\therefore f$ ( x ) = sin $\left(log{x}-\frac{\pi}{4}\right)$

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Examples of calculus

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