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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.

Standard Integrals

Integrals of Rational and Irrational Functions

∫xndx=xn+1n+1+C,n≠1∫1xdx=ln⁡∣x∣+C∫c dx=c⋅x+C∫x dx=x22+C∫x2 dx=x33+C∫1x2dx=−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 \\ ∫x dx=2⋅x⋅x3+C∫11+x2dx=arctan⁡x+C∫11−x2dx=arcsin⁡x+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

∫sin⁡x dx=−cos⁡x+C∫cos⁡x dx=sin⁡x+C∫tan⁡x dx=ln⁡∣sec⁡x∣+C∫sec⁡x dx=ln⁡∣tan⁡x+sec⁡x∣+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 \\ ∫sin⁡2x dx=12(x−sin⁡x⋅cos⁡x)+C∫cos⁡2x dx=12(x+sin⁡x⋅cos⁡x)+C∫tan⁡2x dx=tan⁡x–x+C∫sec⁡2x dx=tan⁡x+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

∫ln⁡x dx=x⋅ln⁡x−x+C∫xn⋅ln⁡x dx=xn+1n+1ln⁡x–xn+1(n+1)2+C∫ex dx=ex+C∫ax dx=axln⁡a+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+b−x)dx\int\limits_{a}^{b}{f(x)dx=}\int\limits_{a}^{b}{f(a+b-x)dx}

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

Q f(x)={x2+x1≤x≤23x2≤x≤3f(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}}

Q ∫−11e(x)dx=2(e−1)\int\limits_{-1}^{1}{{{e}^{(x)}}}dx=2(e-1)

Q ∫−23∣x∣dx=132\int\limits_{-2}^{3}{\left| \left. x \right| \right.dx=\frac{13}{2}} ∫−11x3∣x∣dx=0\int\limits_{-1}^{1}{{{x}^{3}}\left| x \right|dx=0}

Q ∫28∣x−5∣dx=9\int\limits_{2}^{8}{\left| x-5 \right|dx=9} ∫1/41∣2x−1∣dx=516\int\limits_{{}^{1}/{}_{4}}^{1}{\left| 2x-1 \right|dx=\frac{5}{16}}

Q ∫π/2π/2∣sin⁡x∣dx=2\int\limits_{{}^{\pi }/{}_{2}}^{{}^{\pi }/{}_{2}}{\left| \sin x \right|dx=2} ∫−22∣x∣+∣x−1∣dx=9\int\limits_{-2}^{2}{\left| x \right|+\left| x-1 \right|dx=9}

⇒ Also Read Definite and Indefinite Integration

Useful Formulas

  • ∫eaxsin⁡bx=eaxa2+b2[asin⁡bx−bcos⁡bx]\int{{{e}^{ax}}\sin bx=\frac{{{e}^{ax}}}{{{a}^{2}}+{{b}^{2}}}\left[ a\sin bx-b\cos bx \right]}
  • ∫eaxcos⁡bx=eaxa2+b2[acos⁡bx+bsin⁡bx]\int{{{e}^{ax}}\cos bx=\frac{{{e}^{ax}}}{{{a}^{2}}+{{b}^{2}}}\left[ a\cos bx+b\sin bx \right]}

Illustration:

  • ∫e−2xsin⁡xcos⁡3xdx\int{{{e}^{-2x}}\sin x\cos 3xdx}
  • =∫e−2x(sin⁡4x−sin⁡2x)dx=\int{{{e}^{-2x}}\left( \sin 4x-\sin 2x \right)dx}
  • =e−2x[(sin⁡2x+cos⁡2x8)−sin⁡4x+2cos⁡4x20]={{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(sin⁡x+cos⁡x)dx=exsin⁡x+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=∫sin⁡mxcos⁡nxdxI=\int{{{\sin }^{m}}x{{\cos }^{n}}xdx}

1. If m –odd put cos⁡x=t\cos x=t

2. If n odd put sin⁡x=t\sin x=t

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

4. If both even then use reduction method

Q∫cos⁡3xsin⁡6xdx=∫1−t2t6dtQ\int{\frac{{{\cos }^{3}}x}{{{\sin }^{6}}x}dx=\int{\frac{1-{{t}^{2}}}{{{t}^{6}}}dt}}

Where t=sin⁡xt=\sin x =∫t−6−t−4dt=\int{{{t}^{-6}}-{{t}^{-4}}dt} =−15sin5x+13sin⁡3x+c=-\frac{1}{5si{{n}^{5}}x}+\frac{1}{3{{\sin }^{3}}x}+c

Type 2: ∫dxacos⁡x+bsin⁡x+c\int{\frac{dx}{a\cos x+b\sin x+c}}

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

Illustration

∫dx2+sin⁡x\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}} =23tan⁡−1(2t+13)=\frac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{2t+1}{\sqrt{3}} \right) =23tan⁡−1(2tan⁡x2+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 : ∫linearQuadralis dx\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+b−x)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(2a−x)dx\int\limits_{0}^{2a}{f(x)dx}=\int\limits_{0}^{a}{f(x)dx}+\int\limits_{0}^{a}{f(2a-x)dx}=0  if  f(2a−x)=−f(x)=0\,\,if\,\,f(2a-x)=-f(x) and =2∫0af(x)  if  f(2a−x)=f(x)=2\int\limits_{0}^{a}{f(x)}\,\,if\,\,f(2a-x)=f(x)
  6. ∫−aaf(x)dx={0if  f(x) is odd2∫0af(x)dxif  f(x) is odd      \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} =∫01x2.0 dx+∫12x2[1]dx=\int\limits_{0}^{1}{{{x}^{2}}.0\,dx}+\int\limits_{1}^{2}{{{x}^{2}}\left[ 1 \right]dx} =0+x33∣12=0+\left. \frac{{{x}^{3}}}{3} \right|_{1}^{2} =8−13 =73=\frac{8-1}{3}\,=\frac{7}{3}

Illustration:

∫π/6  π/3  sin⁡xsin⁡x+cos⁡xdx=∫cos⁡xsin⁡x+cos⁡xdx\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 x→π2−xx\to \frac{\pi }{2}-x)

2I=∫π/6  π/3  sin⁡x+cos⁡xsin⁡x+cos⁡x=∫π/6  π/3  1 dx=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=∫sin⁡100xcos⁡99xI=\int{{{\sin }^{100}}x{{\cos }^{99}}x}

Here f(2π−x)=f(x)f(2\pi -x)=f(x)

Or I=2∫0πsin⁡100xcos⁡99xI=2\int\limits_{0}^{\pi }{{{\sin }^{100}}x{{\cos }^{99}}x} =2∫0πsin⁡100(π−x)cos⁡99(π−x)=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

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

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

A=∫01212x−x212dxA=\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  ∣012−x336∣012=\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 e∫pdx{{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=e∫1xxdx=e1x={{e}^{\int{\frac{1}{{{x}^{x}}}dx}}}=e\frac{1}{x}

Ans:

y e −1x=∫p−1x2(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. xdy−ydxx2=d(yx)\frac{xdy-ydx}{{{x}^{2}}}=d\left( \frac{y}{x} \right).

3. ydx−xdyy2=d(xy)\frac{ydx-xdy}{{{y}^{2}}}=d\left( \frac{x}{y} \right)

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

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

6. xdy−ydxx2+y2=d(tan⁡−1(yx))\frac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}}=d\left( {{\tan }^{-1}}\left( \frac{y}{x} \right) \right)

Illustration

Solve xdx+ydy=xdy−ydxx2+y2xdx+ydy=\frac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}}

Ans:

12d(x2+y2)=d(tan⁡−1(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)=tan⁡−1(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.

ddx∫u(x)v(x)f(t)dt=f(v(x))dv(x)dx−f(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. ∫x2x31log⁡tdt=y\int\limits_{{{x}^{2}}}^{{{x}^{3}}}{\frac{1}{\log t}dt=y} find dydx=x(x−1)(log⁡x)−1\frac{dy}{dx}=x\left( x-1 \right){{\left( \log x \right)}^{-1}}

Q. If ∫sin⁡x1t2f(t)dt=1−sin⁡x.\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 lim⁡x→2 ∫6f(x)4t3x−2dt=18\underset{x\to 2}{\mathop{\lim }}\,\int\limits_{6}^{f\left( x \right)}{\frac{4{{t}^{3}}}{x-2}}dt=18

Q. lim⁡x→∞ ∫0xex2dx∫0xe2x2dx=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=∫18sin⁡2xsin⁡−1t dt+∫18cos⁡2xcos⁡−1t dt.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) x∫0x(1−t)f(t)dt=∫0xt f(t)dt      x∈R+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(1−1x)f\left( x \right).\,\,\,\,\,\,\,f\left( x \right)=\frac{1}{{{x}^{3}}}{{e}^{\left( 1-\frac{1}{x} \right)}}

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

Q. lim⁡x→4 ∫4x4t−f(t)x−4dt=16−f(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. lim⁡x→0 ∫0xcos⁡t2dtx=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(t−1)(t−2)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

∫acos⁡x+bsin⁡x+pccos⁡x+dsin⁡x+q dx,∫aex+be−x+cdex+fe−x+h dx\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+mDr′Nr=nDr+mDr'

Illustration:

∫2cos⁡x+3sin⁡x4cos⁡x+5sin⁡x dx\int{\frac{2\cos x+3\sin x}{4\cos x+5\sin x}}\,dx

Let 2cos⁡x+3sin⁡x=a(4cos⁡x+5sin⁡x)+b(−4sin⁡x+5cos⁡x)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=2341      b=−241a=\frac{23}{41}\,\,\,\,\,\,b=\frac{-2}{41}

∴ I=∫2341−241(−4sin⁡x+5cos⁡x4cos⁡x+5sin⁡x)dxI=\int{\frac{23}{41}-\frac{2}{41}\left( \frac{-4\sin x+5\cos x}{4\cos x+5\sin x} \right)dx} =2341x−241ℓn∣4cos⁡x+5sin⁡x∣+c=\frac{23}{41}x-\frac{2}{41}\ell n\left| 4\cos x+5\sin x \right|+c

By Parts Integration

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

Illustration:

Q. ∫ℓn x=∫ℓn  x.1  dx\int{\ell n\,x}=\int{\ell n\,\,x.1\,\,dx} =x ℓn x−∫1x. x dx=x\,\ell n\,x-\int{\frac{1}{x}.\,x}\,dx =x ℓn x−x=x\,\ell n\,x-x

Q. ∫x exdx=x∫exdx−∫(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}} =xex−∫exdx=x{{e}^{x}}-\int{{{e}^{x}}dx} =xex−ex=x{{e}^{x}}-{{e}^{x}}

Q. ∫(x3+3x)sin⁡3x dx=(x3+3x)(−cos⁡3x3)−(3x2+3)(−sin⁡3x32)+6x(cos⁡3x33)−6(sin⁡3x34)+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(x−3)x+1dx\int{\frac{x}{\left( x-3 \right)\sqrt{x+1}}dx}

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

I=∫(t2−1)2t dt(t2−4)t=2∫t2−1t2−4dtI=\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} =2∫1+3t2−4dt=2\int{1}+\frac{3}{{{t}^{2}}-4}dt =2t+32ℓn∣t−2t+2∣+c=2t+\frac{3}{2}\ell n\left| \frac{t-2}{t+2} \right|+c =2x+1+32ℓn∣x+1−2x+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(2a−x)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(2a−x)=−f(x)f\left( 2a-x \right)=-f\left( x \right) =2∫0af(x)=2\int\limits_{0}^{a}{f\left( x \right)} if f(2a−x)=f(x)f\left( 2a-x \right)=f\left( x \right)

Illustration:

Q. I=∫sin⁡100xcos⁡99xI=\int{{{\sin }^{100}}x{{\cos }^{99}}x}

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

Or I=2∫0πsin⁡100xcos⁡99xI=2\int\limits_{0}^{\pi }{{{\sin }^{100}}x{{\cos }^{99}}x} =2∫0πsin⁡100(π−x)cos⁡99(π−x)=2\int\limits_{0}^{\pi }{{{\sin }^{100}}\left( \pi -x \right)}{{\cos }^{99}}\left( \pi -x \right)

I = -I

I = 0

⇒ ∫−aaf(x)dx={0if  f(x)  is  odd2∫0af(x)dxif  f(x)  is  even\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 y2≤4x,x2+y2≥2x{{y}^{2}}\le 4x,{{x}^{2}}+{{y}^{2}}\ge 2x and x≤y+2.x\le y+2. in first quadrant.

Answer:

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

=4[2x3/23]0(3+1)2−π2−12(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)33−π2=\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 α3−4α2+8.{{\alpha }^{3}}-4{{\alpha }^{2}}+8. find f(x).f\left( x \right).

Answer:

α3−4α2+8=∫2α(x2+2−f(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

3α2−8α=α2+2−f(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