Integration of Functions

Integration is the process of finding the antiderivative. The integration of g′(x) with respect to dx is given by

∫ g′(x) dx = g(x) + C, where C is the constant of integration.

The two types of integrals include:

  • Definite Integral: An integral with limits namely upper and lower limit without the constant of integration.
  • Indefinite integral: An integral without limits and with an arbitrary constant.

This article covers standard integrals, properties of integration, important formulas and examples on integration which helps students to have a deep knowledge on the topic.

Standard Integrals

Integrals of Rational and Irrational Functions

xndx=xn+1n+1+C,n11xdx=lnx+Ccdx=cx+Cxdx=x22+Cx2dx=x33+C1x2dx=1x+C\int x^n dx = \frac{x^{n+1}}{n+1} + C , n \ne 1\\ \int \frac{1}{x} dx = \ln|x| + C \\ \int c \, dx = c \cdot x + C \\ \int x \, dx = \frac{x^2}{2} + C \\ \int x^2 \, dx = \frac{x^3}{3} + C \\ \int \frac{1}{x^2} dx = -\frac{1}{x} + C \\ xdx=2xx3+C11+x2dx=arctanx+C11x2dx=arcsinx+C\int \sqrt{x} \, dx = \frac{2\cdot x \cdot \sqrt{x} }{3} + C\\ \int \frac{1}{1+x^2} dx = \arctan x + C \\ \int \frac{1}{\sqrt{1-x^2}} dx = \arcsin x + C \\

Integrals of Trigonometric Functions

sinxdx=cosx+Ccosxdx=sinx+Ctanxdx=lnsecx+Csecxdx=lntanx+secx+C\int \sin x\,dx = -\cos x + C \\ \int \cos x\,dx = \sin x + C\\ \int \tan x\,dx = \ln|\sec x| + C \\ \int \sec x\,dx = \ln|\tan x + \sec x | + C \\ sin2xdx=12(xsinxcosx)+Ccos2xdx=12(x+sinxcosx)+Ctan2xdx=tanxx+Csec2xdx=tanx+C\int \sin^2x\,dx = \frac{1}{2}(x-\sin x \cdot \cos x) + C\\ \int \cos^2x\,dx = \frac{1}{2}(x + \sin x \cdot \cos x) + C \\ \int \tan^2x\,dx = \tan x – x + C \\ \int \sec^2x\,dx = \tan x + C \\

Integrals of Exponential and Logarithmic Functions

lnxdx=xlnxx+Cxnlnxdx=xn+1n+1lnxxn+1(n+1)2+Cexdx=ex+Caxdx=axlna+C\int \ln x \,dx =x \cdot \ln x -x + C \\ \int x^n \cdot \ln x \,dx =\frac{x^{n+1}}{n+1} \ln x – \frac{x^{n+1}}{(n+1)^2} + C \\ \int e^x\,dx = e^x + C \\ \int a^x\,dx = \frac{a^x}{\ln a} + C\\

Properties of  Integration

Property 1: aaf(x)dx=0\int\limits_{a}^{a}{f(x)\,dx=0}

Property 2: abf(x)=baf(x)dx\int\limits_{a}^{b}{f(x)=}-\int\limits_{b}^{a}{f(x)dx}

Property 3: abf(x)=abf(t)\int\limits_{a}^{b}{f(x)=}\int\limits_{a}^{b}{f(t)}

Property 4: abf(x)dx=acf(x)dx+abf(x)dx\int\limits_{a}^{b}{f(x)dx=}\int\limits_{a}^{c}{f(x)dx+\int\limits_{a}^{b}{f(x)dx}}

Property 5: (i) abf(x)dx=abf(a+bx)dx\int\limits_{a}^{b}{f(x)dx=}\int\limits_{a}^{b}{f(a+b-x)dx}

(ii) 0af(x)dx=0af(ax)dx\int\limits_{0}^{a}{f(x)dx=}\int\limits_{0}^{a}{f(a-x)dx}

(iii) f(x)={x2+x1x23x2x3f(x)=\left\{ \begin{matrix} {{x}^{2}}+x & 1\le x\le 2 \\ 3x & 2\le x\le 3 \\ \end{matrix} \right. 13f(x)=343\int\limits_{1}^{3}{f(x)=\frac{34}{3}}

(iv) 11e(x)dx=2(e1)\int\limits_{-1}^{1}{{{e}^{(x)}}}dx=2(e-1)

(v) 23xdx=132\int\limits_{-2}^{3}{\left| \left. x \right| \right.dx=\frac{13}{2}} 11x3xdx=0\int\limits_{-1}^{1}{{{x}^{3}}\left| x \right|dx=0}

(vi) 28x5dx=9\int\limits_{2}^{8}{\left| x-5 \right|dx=9} 1/412x1dx=516\int\limits_{{}^{1}/{}_{4}}^{1}{\left| 2x-1 \right|dx=\frac{5}{16}}

(vii) π/2π/2sinxdx=2\int\limits_{{}^{\pi }/{}_{2}}^{{}^{\pi }/{}_{2}}{\left| \sin x \right|dx=2} 22x+x1dx=9\int\limits_{-2}^{2}{\left| x \right|+\left| x-1 \right|dx=9}

⇒ Also Read Definite and Indefinite Integration

Useful Formulas

  • eaxsinbx=eaxa2+b2[asinbxbcosbx]\int{{{e}^{ax}}\sin bx=\frac{{{e}^{ax}}}{{{a}^{2}}+{{b}^{2}}}\left[ a\sin bx-b\cos bx \right]}
  • eaxcosbx=eaxa2+b2[acosbx+bsinbx]\int{{{e}^{ax}}\cos bx=\frac{{{e}^{ax}}}{{{a}^{2}}+{{b}^{2}}}\left[ a\cos bx+b\sin bx \right]}

Illustration:

  • e2xsinxcos3xdx\int{{{e}^{-2x}}\sin x\cos 3xdx}
  • =e2x(sin4xsin2x)dx=\int{{{e}^{-2x}}\left( \sin 4x-\sin 2x \right)dx}
  • =e2x[(sin2x+cos2x8)sin4x+2cos4x20]={{e}^{-2x}}\left[ \left( \frac{\sin 2x+\cos 2x}{8} \right)-\frac{\sin 4x+2\cos 4x}{20} \right]
  • ex(f(x)+f(x))=exf(x)\int{{{e}^{x}}\left( f(x)+f'(x) \right)}={{e}^{x}}f(x)

Illustration:

ex(1x+1x2)dx=exx+c\int{{{e}^{x}}\left( \frac{1}{x}+\frac{1}{{{x}^{2}}} \right)dx}={{\frac{e}{x}}^{x}}+c ex(sinx+cosx)dx=exsinx+c\int{{{e}^{x}}(\sin x+\cos x)dx={{e}^{x}}\sin x+c} ex(lnx+1x)dx=exlnx+c\int{{{e}^{x}}(lnx+\frac{1}{x})dx={{e}^{x}}lnx+c}

Integration of Trigonometric Functions

Type 1: I=sinmxcosnxdxI=\int{{{\sin }^{m}}x{{\cos }^{n}}xdx}

1. If m –odd put cosx=t\cos x=t

2. If n odd put sinx=t\sin x=t

3. If m, n rationales then put tanx=t\tan x=t

4. If both even then use reduction method

Qcos3xsin6xdx=1t2t6dtQ\int{\frac{{{\cos }^{3}}x}{{{\sin }^{6}}x}dx=\int{\frac{1-{{t}^{2}}}{{{t}^{6}}}dt}}

Where t=sinxt=\sin x =t6t4dt=\int{{{t}^{-6}}-{{t}^{-4}}dt} =15sin5x+13sin3x+c=-\frac{1}{5si{{n}^{5}}x}+\frac{1}{3{{\sin }^{3}}x}+c

Type 2: dxacosx+bsinx+c\int{\frac{dx}{a\cos x+b\sin x+c}}

Put t=tan(x2)t=\tan \left( \frac{x}{2} \right)

Illustration

dx2+sinx\int{\frac{dx}{2+\sin x}} t=tan(x2)\Rightarrow t=\tan \left( \frac{x}{2} \right) dx=2dt1+t2dx=\frac{2dt}{1+{{t}^{2}}} =2dt1+t22+2t1+t2=\int{\frac{\frac{2dt}{1+{{t}^{2}}}}{2+\frac{2t}{1+{{t}^{2}}}}} dtt2+t+1\Rightarrow \int{\frac{dt}{{{t}^{2}}+t+1}} =23tan1(2t+13)=\frac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{2t+1}{\sqrt{3}} \right) =23tan1(2tanx2+13)+c=\frac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{2\tan \frac{x}{2}+1}{\sqrt{3}} \right)+c

Some useful Substitutions for Irrational Functions

  • Form 1 : linearQuadralisdx\int{linear\sqrt{Quadralis}\,dx}

Substitute liner=mQudratic+nliner=mQudratic’+n

  • Form 2: dxlinlin1,linlin1dx,lin1lindx\int{\frac{dx}{lin\sqrt{li{{n}_{1}}}},}\,\int{\frac{lin}{\sqrt{li{{n}_{1}}}}}dx,\,\int{\frac{\sqrt{li{{n}_{1}}}}{lin}dx}

Substitute lin1=t2li{{n}_{1}}={{t}^{2}}

  • Form 3: 1linQuadx\int{\frac{1}{lin\sqrt{Qua}}dx}

Substitute lin=1tlin=\frac{1}{t}

  • Form 4: dx(ax2+b)(x2+d)\int{\frac{dx}{\left( a{{x}^{2}}+b \right)\sqrt{\left( {{x}^{2}}+d \right)}}}

Substitute x=1tx=\frac{1}{t}& then, u2{{u}^{2}}for at2+ba{{t}^{2}}+b

Integration Formulas

  1. abf(x)dx=abf(t)dt\int\limits_{a}^{b}{f(x)dx=}\int\limits_{a}^{b}{f(t)dt}
  2. abf(x)dx=baf(x)dx\int\limits_{a}^{b}{f(x)dx=}-\int\limits_{b}^{a}{f(x)dx}
  3. abf(x)dx=acf(x)dx+cbf(x)dx\int\limits_{a}^{b}{f(x)dx=}\int\limits_{a}^{c}{f(x)dx}+\int\limits_{c}^{b}{f(x)dx}
  4. abf(x)dx=abf(a+bx)dx\int\limits_{a}^{b}{f(x)dx=}\int\limits_{a}^{b}{f(a+b-x)dx}
  5. 02af(x)dx=0af(x)dx+0af(2ax)dx\int\limits_{0}^{2a}{f(x)dx}=\int\limits_{0}^{a}{f(x)dx}+\int\limits_{0}^{a}{f(2a-x)dx}=0iff(2ax)=f(x)=0\,\,if\,\,f(2a-x)=-f(x) and =20af(x)iff(2ax)=f(x)=2\int\limits_{0}^{a}{f(x)}\,\,if\,\,f(2a-x)=f(x)
  6. aaf(x)dx={0iff(x)isodd20af(x)dxiff(x)isodd\int\limits_{-a}^{a}{f(x)dx=\left\{ \begin{matrix} 0 & if\,\,f\left( x \right)\,is\,odd \\ 2\int\limits_{0}^{a}{f\left( x \right)dx} & if\,\,f\left( x \right)\,is\,odd \\ \end{matrix} \right.\,\,\,\,\,\,}

Problems on Integration

Illustration:

02x2[x]dx=01x2[x]dx+12x2[x]dx\int\limits_{0}^{2}{{{x}^{2}}\left[ x \right]dx=}\int\limits_{0}^{1}{{{x}^{2}}\left[ x \right]dx}+\int\limits_{1}^{2}{{{x}^{2}}\left[ x \right]dx} =01x2.0dx+12x2[1]dx=\int\limits_{0}^{1}{{{x}^{2}}.0\,dx}+\int\limits_{1}^{2}{{{x}^{2}}\left[ 1 \right]dx} =0+x3312=0+\left. \frac{{{x}^{3}}}{3} \right|_{1}^{2} =813=73=\frac{8-1}{3}\,=\frac{7}{3}

Illustration:

π/6  π/3  sinxsinx+cosxdx=cosxsinx+cosxdx\int\limits_{{\pi }/{6}\;}^{{\pi }/{3}\;}{\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx=}\int{\frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx} (by xπ2xx\to \frac{\pi }{2}-x)

2I=π/6  π/3  sinx+cosxsinx+cosx=π/6  π/3  1dx=π62I=\int\limits_{{\pi }/{6}\;}^{{\pi }/{3}\;}{\frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}}=\int\limits_{{\pi }/{6}\;}^{{\pi }/{3}\;}{1\,dx}=\frac{\pi }{6} I=π12I=\frac{\pi }{12}

Illustration:

I=sin100xcos99xI=\int{{{\sin }^{100}}x{{\cos }^{99}}x}

Here f(2πx)=f(x)f(2\pi -x)=f(x)

Or I=20πsin100xcos99xI=2\int\limits_{0}^{\pi }{{{\sin }^{100}}x{{\cos }^{99}}x} =20πsin100(πx)cos99(πx)=2\int\limits_{0}^{\pi }{{{\sin }^{100}}\left( \pi -x \right){{\cos }^{99}}\left( \pi -x \right)}

I = -I

I = 0

Illustration:

55x3=0\int\limits_{-5}^{5}{{{x}^{3}}=0} as f(x)f(x) is odd

Area

Problems on integration area 1

A=abf(x)dx=abydxA=\int\limits_{a}^{b}{f(x)dx=\int\limits_{a}^{b}{y\,dx}}

Problems on integration example 2

A=ab(f(x)g(x))dxA=\int\limits_{a}^{b}{\left( f(x)-g(x) \right)}\,dx

Illustration:

Find area between y2=12x{{y}^{2}}=12xand x2=12y{{x}^{2}}=12y

Problems on integration example 3

A=01212xx212dxA=\int\limits_{0}^{12}{\sqrt{12x}-\frac{{{x}^{2}}}{12}dx} =12x3/2  3/2  x212dx=\frac{12{{x}^{{3}/{2}\;}}}{{3}/{2}\;}-\frac{{{x}^{2}}}{12}dx =8x3/2  012x336012=\left. 8{{x}^{{3}/{2}\;}} \right|_{0}^{12}-\left. \frac{{{x}^{3}}}{36} \right|_{0}^{12} =8.123/2  1443=8\,\,.\,\,{{12}^{{3}/{2}\;}}-\frac{144}{3}

Linear Differential Equation

Equation of form dydx+py=θ\frac{dy}{dx}+py=\theta.

Ar solve by multiplying integrating function

If epdx{{e}^{\int{pdx}}}

To get final solution of

y(if)=Q(IF)+cy(if)=\int{Q(IF)+c}

Illustration

Solve x2dydx+y+1=1{{x}^{2}}\frac{dy}{dx+y+1}=1

Ans : Given equation is

dydx+1x2y=1xx2\frac{dy}{dx}+\frac{1}{{{x}^{2}}}y=\frac{1x}{{{x}^{2}}}which is linear

p=1x2,q=1x2p=\frac{1}{{{x}^{2}}},q=\frac{1}{{{x}^{2}}}

IF=e1xxdx=e1x={{e}^{\int{\frac{1}{{{x}^{x}}}dx}}}=e\frac{1}{x}

Ans:

ye1x=p1x2(1x2)dx+cy\,e\,\frac{-1}{x}=\int{{{p}^{\frac{-1}{{{x}^{2}}}}}}(\frac{1}{{{x}^{2}}})dx+c y=1+ce1xy=1+c{{e}^{\frac{1}{x}}}

General Form

We use combinations of different terms to get united term

Following terms are useful

1. xdy+ydx=d(xy)xdy+ydx=d(xy)

2. xdyydxx2=d(yx)\frac{xdy-ydx}{{{x}^{2}}}=d\left( \frac{y}{x} \right).

3. ydxxdyy2=d(xy)\frac{ydx-xdy}{{{y}^{2}}}=d\left( \frac{x}{y} \right)

4. xdy+ydxxy=d(laxy)\frac{xdy+ydx}{xy}=d(la\,xy)

5. xdyydxxy=d(1(yx))\frac{xdy-ydx}{xy}=d\left( 1\left( \frac{y}{x} \right) \right)

6. xdyydxx2+y2=d(tan1(yx))\frac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}}=d\left( {{\tan }^{-1}}\left( \frac{y}{x} \right) \right)

Illustration

Solve xdx+ydy=xdyydxx2+y2xdx+ydy=\frac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}}

Ans:

12d(x2+y2)=d(tan1(yx))\frac{1}{2}d({{x}^{2}}+{{y}^{2}})=d\left( {{\tan }^{-1}}\left( \frac{y}{x} \right) \right)

Integrating we get

12(x2+y2)=tan1(yx)+c\frac{1}{2}({{x}^{2}}+{{y}^{2}})={{\tan }^{-1}}\left( \frac{y}{x} \right)+c

Leibnitz’s rule

f cont [a, b] & u(x),v(x)u\left( x \right),v\left( x \right) different function.

ddxu(x)v(x)f(t)dt=f(v(x))dv(x)dxf(u(x))u(x)\frac{d}{dx}\int\limits_{u\left( x \right)}^{v\left( x \right)}{f\left( t \right)dt=f\left( v\left( x \right) \right)}\frac{dv\left( x \right)}{dx}-f\left( u\left( x \right) \right)u’\left( x \right)

Proof:

ddxF(x)=f(x)\frac{d}{dx}F\left( x \right)=f\left( x \right)

u(x)v(x)f(t)dt=F(v(x))F(u(x))\int\limits_{u\left( x \right)}^{v\left( x \right)}{f\left( t \right)dt=F\left( v\left( x \right) \right)}-F\left( u\left( x \right) \right)

Practice Problems

Q. x2x31logtdt=y\int\limits_{{{x}^{2}}}^{{{x}^{3}}}{\frac{1}{\log t}dt=y} find dydx=x(x1)(logx)1\frac{dy}{dx}=x\left( x-1 \right){{\left( \log x \right)}^{-1}}

Q. If sinx1t2f(t)dt=1sinx.\int\limits_{\sin x}^{1}{{{t}^{2}}f\left( t \right)dt=1-\sin x.} where x(0,π2)x\in \left( 0,\frac{\pi }{2} \right) . find f(13)=3f\left( \frac{1}{\sqrt{3}} \right)=3

Q. f(2)=6,f(2)=148f\left( 2 \right)=6,f’\left( 2 \right)=\frac{1}{48}. Find limx26f(x)4t3x2dt=18\underset{x\to 2}{\mathop{\lim }}\,\int\limits_{6}^{f\left( x \right)}{\frac{4{{t}^{3}}}{x-2}}dt=18

Q. limx0xex2dx0xe2x2dx=0\underset{x\to \infty }{\mathop{\lim }}\,\frac{\int\limits_{0}^{x}{{{e}^{{{x}^{2}}}}dx}}{\int\limits_{0}^{x}{{{e}^{2{{x}^{2}}}}dx}}=0

Q. y=18sin2xsin1tdt+18cos2xcos1tdt.y=\int\limits_{\frac{1}{8}}^{{{\sin }^{2}}x}{{{\sin }^{-1}}\sqrt{t}\,dt}+\int\limits_{\frac{1}{8}}^{{{\cos }^{2}}x}{{{\cos }^{-1}}\sqrt{t}\,dt}. x[0,π2]x\in \left[ 0,\frac{\pi }{2} \right] is equation of straight line. Find it

(dydx=0)y=const\left( \frac{dy}{dx}=0 \right)y=const put x=π4y=3π16x=\frac{\pi }{4}y=\frac{3\pi }{16}

Q. f:(0,)(0,)f:\left( 0,\infty \right)\to \left( 0,\infty \right) x0x(1t)f(t)dt=0xtf(t)dtxR+x\int\limits_{0}^{x}{\left( 1-t \right)f\left( t \right)dt=\int\limits_{0}^{x}{t\,f\left( t \right)dt}}\,\,\,\,\,\,x\in {{R}^{+}}

Given f(1)=1.f\left( 1 \right)=1. Find f(x).f(x)=1x3e(11x)f\left( x \right).\,\,\,\,\,\,\,f\left( x \right)=\frac{1}{{{x}^{3}}}{{e}^{\left( 1-\frac{1}{x} \right)}}

Hint differentiate twice. fxf(x)=13xx2\int{\frac{f’x}{f\left( x \right)}}=\int{\frac{1-3x}{{{x}^{2}}}}

Q. limx44x4tf(t)x4dt=16f(4)\underset{x\to 4}{\mathop{\lim }}\,\int\limits_{4}^{x}{\frac{4t-f\left( t \right)}{x-4}dt=16-f\left( 4 \right)}

Q. limx00xcost2dtx=1\underset{x\to 0}{\mathop{\lim }}\,\frac{\int\limits_{0}^{x}{\cos {{t}^{2}}dt}}{x}=1

Q. Find points of minima for f(x)=0xt(t1)(t2)dtf\left( x \right)=\int\limits_{0}^{x}{t\left( t-1 \right)\left( t-2 \right)dt} Ans n=0,2n=0,2

Type

acosx+bsinx+pccosx+dsinx+qdx,aex+bex+cdex+fex+hdx\int{\frac{a\cos x+b\sin x+p}{c\cos x+d\sin x+q}}\,dx,\int{\frac{a{{e}^{x}}+b{{e}^{-x}}+c}{d{{e}^{x}}+f{{e}^{-x}}+h}}\,dx

This is solved by Nr=nDr+mDrNr=nDr+mDr'

Illustration:

2cosx+3sinx4cosx+5sinxdx\int{\frac{2\cos x+3\sin x}{4\cos x+5\sin x}}\,dx

Let 2cosx+3sinx=a(4cosx+5sinx)+b(4sinx+5cosx)2\cos x+3\sin x=a\left( 4\cos x+5\sin x \right)+b\left( -4\sin x+5\cos x \right)

Solving by comparing we get

a=2341b=241a=\frac{23}{41}\,\,\,\,\,\,b=\frac{-2}{41}

I=2341241(4sinx+5cosx4cosx+5sinx)dxI=\int{\frac{23}{41}-\frac{2}{41}\left( \frac{-4\sin x+5\cos x}{4\cos x+5\sin x} \right)dx} =2341x241n4cosx+5sinx+c=\frac{23}{41}x-\frac{2}{41}\ell n\left| 4\cos x+5\sin x \right|+c

By Parts Integration

uvdx=uvdxu(vdx)dx\int{uv\,dx}=u\int{vdx}-\int{u’\left( \int{vdx} \right)}\,dx

Illustration:

Q. nx=nx.1dx\int{\ell n\,x}=\int{\ell n\,\,x.1\,\,dx} =xnx1x.xdx=x\,\ell n\,x-\int{\frac{1}{x}.\,x}\,dx =xnxx=x\,\ell n\,x-x

Q. xexdx=xexdx(x)1(exdx)dx\int{x\,{{e}^{x}}dx=x\int{{{e}^{x}}dx-\int{{{\left( x \right)}^{1}}\left( \int{{{e}^{x}}dx} \right)}dx}} =xexexdx=x{{e}^{x}}-\int{{{e}^{x}}dx} =xexex=x{{e}^{x}}-{{e}^{x}}

Q. (x3+3x)sin3xdx=(x3+3x)(cos3x3)(3x2+3)(sin3x32)+6x(cos3x33)6(sin3x34)+c\int{\left( {{x}^{3}}+3x \right)\sin 3x\,dx=\left( {{x}^{3}}+3x \right)\left( \frac{-\cos 3x}{3} \right)}-\left( 3{{x}^{2}}+3 \right)\left( \frac{-{{\sin }^{3}}x}{{{3}^{2}}} \right)+6x\left( \frac{\cos 3x}{{{3}^{3}}} \right)-6\left( \frac{{{\sin }^{3}}x}{{{3}^{4}}} \right)+c

Integration of Irrational Algebraic Functions

Type dx(ax+b)kpx+q\int{\frac{dx}{{{\left( ax+b \right)}^{k}}\sqrt{px+q}}}

Q. x(x3)x+1dx\int{\frac{x}{\left( x-3 \right)\sqrt{x+1}}dx}

Put x+1=t2,x+1={{t}^{2}}, we get

I=(t21)2tdt(t24)t=2t21t24dtI=\int{\frac{\left( {{t}^{2}}-1 \right)2t\,dt}{\left( {{t}^{2}}-4 \right)t}}=2\int{\frac{{{t}^{2}}-1}{{{t}^{2}}-4}dt} =21+3t24dt=2\int{1}+\frac{3}{{{t}^{2}}-4}dt =2t+32nt2t+2+c=2t+\frac{3}{2}\ell n\left| \frac{t-2}{t+2} \right|+c =2x+1+32nx+12x+1+2+c=2\sqrt{x+1}+\frac{3}{2}\ell n\left| \frac{\sqrt{x+1}-2}{\sqrt{x+1}+2} \right|+c

02af(x)dx=0af(x)dx+0af(2ax)dx\int\limits_{0}^{2a}{f\left( x \right)dx=\int\limits_{0}^{a}{f\left( x \right)dx+}}\int\limits_{0}^{a}{f\left( 2a-x \right)dx}

= 0 if f(2ax)=f(x)f\left( 2a-x \right)=-f\left( x \right) =20af(x)=2\int\limits_{0}^{a}{f\left( x \right)} if f(2ax)=f(x)f\left( 2a-x \right)=f\left( x \right)

Illustration:

Q. I=sin100xcos99xI=\int{{{\sin }^{100}}x{{\cos }^{99}}x}

Here f(2πx)=f(x)f\left( 2\pi -x \right)=f\left( x \right)

Or I=20πsin100xcos99xI=2\int\limits_{0}^{\pi }{{{\sin }^{100}}x{{\cos }^{99}}x} =20πsin100(πx)cos99(πx)=2\int\limits_{0}^{\pi }{{{\sin }^{100}}\left( \pi -x \right)}{{\cos }^{99}}\left( \pi -x \right)

I = -I

I = 0

aaf(x)dx={0iff(x)isodd20af(x)dxiff(x)iseven\int\limits_{-a}^{a}{f\left( x \right)dx=\left\{ \begin{matrix} 0 & if\,\,f\left( x \right)\,\,is\,\,odd \\ 2\int\limits_{0}^{a}{f\left( x \right)dx} & if\,\,f\left( x \right)\,\,is\,\,even \\ \end{matrix} \right.}

Illustration:

Q. 55x3=0\int\limits_{-5}^{5}{{{x}^{3}}=0} as f(x)f\left( x \right) is odd

Q. Find area between y24x,x2+y22x{{y}^{2}}\le 4x,{{x}^{2}}+{{y}^{2}}\ge 2x and xy+2.x\le y+2. in first quadrant.

Answer:

Problems on integration example 4

A=0(3+1)24xdxA=\int\limits_{0}^{{{\left( \sqrt{3}+1 \right)}^{2}}}{\sqrt{4x}\,\,dx-}Area (semicircle) – Area of ΔABC

=4[2x3/23]0(3+1)2π212(3+1)22(3+1)=\sqrt{4}\left[ \frac{2{{x}^{3/2}}}{3} \right]_{0}^{{{\left( \sqrt{3}+1 \right)}^{2}}}-\frac{\pi }{2}-\frac{1}{2}{{\left( \sqrt{3}+1 \right)}^{2}}2\left( \sqrt{3}+1 \right) =(3+1)33π2=\frac{{{\left( \sqrt{3}+1 \right)}^{3}}}{3}-\frac{\pi }{2}

Optimisation of area for greatest and least values

Illustration:

If area by y=f(x)y=f\left( x \right) and y=x2+2y={{x}^{2}}+2 between abscissa x=2&x=αx=2\And x=\alpha is α34α2+8.{{\alpha }^{3}}-4{{\alpha }^{2}}+8. find f(x).f\left( x \right).

Answer:

α34α2+8=2α(x2+2f(x))dx{{\alpha }^{3}}-4{{\alpha }^{2}}+8=\int\limits_{2}^{\alpha }{\left( {{x}^{2}}+2-f\left( x \right) \right)dx}

Differentiating with Labniz equation

3α28α=α2+2f(α)3{{\alpha }^{2}}-8\alpha ={{\alpha }^{2}}+2-f\left( \alpha \right) f(x)=2x2+8x+2f\left( x \right)=-2{{x}^{2}}+8x+2