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Integration Of Power Of Sin And Cos Functions And Reduction Formula

Learn method for evaluating Integrals by Inverse Trigonometric Function, reduction formula & form, practice reduction formulae to find the value of definite integral.

Reduction Formulae to Find the Value of Definite Integral

Finding Integrals Using Given Reduction Formula

Consider,

\(I_n=\int (cosx)^n\;dx\), where \( n\) is a natural number

\(\Rightarrow I_n=\int \underbrace {(cosx)}_{g'(x)}× \underbrace {(cosx)^{n-1}}_{f(x)}\;dx\)

\(=(cos\,x)^{n-1}\;sin\;x-\int[(n-1)(cosx)^{n-2}×(-sinx)]×sinx\;dx\)

\(=(cosx)^{n-1}sinx+(n-1)\int (cosx)^{n-2}sin^2x\;dx\)

\(=(cosx)^{n-1}sinx+(n-1)\int (cosx)^{n-2}×(1-cos^2x)\;dx\)

\(=(cosx)^{n-1}sinx+(n-1)\Bigg[\displaystyle\int (cosx)^{n-2}-\int(cosx)^n\;dx\Bigg]\)

\(\therefore \;I_n=(cosx)^{n-1}sinx+(n-1)I_{n-2}-(n-1)I_n\)

\(\therefore \;I_n(1+n-1)=sinx×(cosx)^{n-1}+(n-1)I_{n-2}\)

\(\Rightarrow I_n=\dfrac {(sinx)(cosx)^{n-1}}{n}+\dfrac {n-1}{n}I_{n-2}\) ...(1)

Such an equation is called reduction formula; we can express integral of any power of \(cos x\) in terms of lower power of \(cos x\), and by successive reduction of \(n\) ends up with either 

\(\int cosx\;dx\) or \(\int (cos\;x)^0\;dx\)

 

 

Illustration Questions

Using the reduction formula \(\int\limits _{0}^{\pi/2}(cos\,x)^n\;dx=\dfrac {n-1}{n} \int\limits _{0}^{\pi/2}(cos\,x)^{n-2}\;dx\) Find the value of \(\int\limits _{0}^{\pi/2}(cos\,x)^4\;dx\)

A \(\dfrac {3\,\pi}{4}\)

B \(\dfrac {3\,\pi}{16}\)

C \(\dfrac {1}{4}\)

D 2

×

\(\int\limits _{0}^{\pi/2}(cos\,x)^n\;dx=\dfrac {n-1}{n} \int\limits _{0}^{\pi/2}(cos\,x)^{n-2}\;dx\)

\(\Rightarrow I_n=\dfrac {n-1}{n}\,I_{n-2} \)

where  \( I_n=\int\limits_0^{\pi/2}(cosx)^n\;dx\)

We need the value of 

\(I_4=\dfrac {4-1}{4}×I_2\rightarrow\) put \(n = 4\)

\(\Rightarrow I_4=\dfrac {3}{4}I_2\)

\(I_2=\dfrac {2-1}{2}×I_0\rightarrow\) put \(n = 2\)

\(\Rightarrow I_2=\dfrac {1}{2}I_0\)

 

 

\(I_0=\int\limits _0^{\pi/2} (cos\,x)^0\;= \int\limits _0^{\pi/2} 1\;dx=\dfrac {\pi}{2}\)

\(\therefore\;I_4=\dfrac {3}{4}×\dfrac {1}{2}×\dfrac {\pi}{2}=\dfrac {3\pi}{16}\)

 

Using the reduction formula \(\int\limits _{0}^{\pi/2}(cos\,x)^n\;dx=\dfrac {n-1}{n} \int\limits _{0}^{\pi/2}(cos\,x)^{n-2}\;dx\) Find the value of \(\int\limits _{0}^{\pi/2}(cos\,x)^4\;dx\)

A

\(\dfrac {3\,\pi}{4}\)

.

B

\(\dfrac {3\,\pi}{16}\)

C

\(\dfrac {1}{4}\)

D

2

Option B is Correct

Integration of Inverse Trigonometric Function

  • To find

\(I=\int\,sin^{-1}x\;dx\rightarrow\)    take \(u=sin^{-1}x\) and \(dv=dx\)     in

 \(\int u\,dv=uv-\int v\,du\)

  • \(\displaystyle \therefore I=\int (sin^{-1}x)×1\;dx=x\,sin^{-1}x-\int\dfrac {1}{\sqrt {1-x^2}}×x\;dx\)

\(=x\,sin^{-1}x-\displaystyle\int\dfrac {x}{\sqrt {1-x^2}}\;dx\)           

 Put  \(1-x^2=t \Rightarrow-2x\;dx=dt\)

     \(\Rightarrow x\;dx=-\dfrac {1}{2}dt\)

\(\therefore I=x\,sin^{-1}x+\dfrac {1}{2}\displaystyle\int\dfrac {dt}{\sqrt t}\)

\(=x\,sin^{-1}x+\sqrt t+C\)

\(=x\,sin^{-1}x+\sqrt {1-x^2}+C\)

  • The integral of all other Inverse Trigonometric Function can be evaluated in a similar manner.

Illustration Questions

Evaluate \(I=\int tan^{-1}x\;dx\)

A \(x\, tan^{-1}x-\dfrac {1}{2}\ell n (1+x^2)+C\)

B \(tan\,x^2+C\)

C \(x\,tan\,x^{-1}+x^2+C\)

D \(x\,e^{x}+C\)

×

\(I=\int tan^{-1}x\;dx\)

\(=\int (tan^{-1}x)\;×1\;dx\)

Take 

\(u=tan^{-1}x\),  \(dv=dx\Rightarrow v=x\)

\(I=x\,tan^{-1}x-\displaystyle\int\dfrac {1}{1+x^2}×x\;dx\)

Put \(1+x^2=t\Rightarrow 2x\;dx=dt\)

\(\Rightarrow x\;dx=\dfrac {1}{2}dt\)

\(\therefore I=x\,tan^{-1}x-\dfrac {1}{2}\displaystyle\int\dfrac {dt}{t}=x\,tan^{-1}x-\dfrac {1}{2}\,\ell n\,t\)

\(=x\,tan^{-1}x-\dfrac {1}{2}\ell n\,(1+x^2)+C\)

Evaluate \(I=\int tan^{-1}x\;dx\)

A

\(x\, tan^{-1}x-\dfrac {1}{2}\ell n (1+x^2)+C\)

.

B

\(tan\,x^2+C\)

C

\(x\,tan\,x^{-1}+x^2+C\)

D

\(x\,e^{x}+C\)

Option A is Correct

Integrals involving Powers of Sine and Odd Powers of Cosine

Method for Evaluating Integrals of the Form:

\(I=\int\limits(sin\,x)^m\;(cos\,x)^n\;dx\)   or 

\(I=\int\limits(sin^m\,x)\;(cos^n\,x)\;dx\)

where, \( m, n\) are integers and \(m, n\geq 0\)

  • If power of cosine is odd \((i.e. \;n = 2 k + 1 )\) we substitute \(sin\;x = t\)

Example:

\(I=\int\limits sin^2\,x\;cos^3\,x\;dx\rightarrow\) put \(sin\;x = t\)

This substitution makes use of the fact that even powers of \(sin\,x\) and \(cos\,x\) can be expressed in  terms of each other without radical signs. 

One power of \(cos\,x\) will go with \(dx\) and the remaining even power of \(cos\,x\) will be expressed in terms of \(sin\,x\)

\(\Rightarrow cos\,x\;dx=dt\)

\(I=\int sin^2\,x×cos^2\;x×(cos\,x\;dx)\)       (separate one power of \(cos\,x\))

\(=\int t^2(1-t^2)\;\;dt\)           \(\rightarrow\,cos^2\;x=1-sin^2x\,=1-t^2\)

\(=\int (t^2-t^4)\;dt\)

\(=\dfrac {t^3}{3}-\dfrac {t^5}{5}+C\)\(=\dfrac {sin^3\,x}{3}-\dfrac {sin^5\,x}{5}+C\)

Illustration Questions

Evaluate \(I=\int sin^4\,x\;cos^3\,x\;dx\)

A \(\dfrac {cos^5\,x}{5}-\dfrac {cos^7\,x}{7}+C\)

B \(\dfrac {cos^3\,x}{3}-\dfrac {sin^5\,x}{5}+C\)

C \(\dfrac {sin^5\,x}{5}-\dfrac {sin^7\,x}{7}+C\)

D \(x\,e^x+sin\,x+C\)

×

\(I=\int (sin^4\,x\;cos^3\,x)\;dx\)

Since the power of \(cos\,x\) is odd,  

put \(sin\,x=t\Rightarrow cos\,x\;dx=dt\)

\(I=\int sin^4x×cos^2\,x×(cos\,x)\;dx\) (separate one power of \(cos\,x\) )

\(=\int t^4(1-t^2)\;dt\rightarrow\)\(cos^2x=1-sin^2\,x=1-t^2\)

\(=\int (t^4-t^6)\;dt \Rightarrow\dfrac {t^5}{5}-\dfrac {t^7}{7}+C\)

Put  \(t=sin\,x\)  again 

\(\Rightarrow I=\dfrac {sin^5\,x}{5}-\dfrac {sin^7\,x}{7}+C\)

Evaluate \(I=\int sin^4\,x\;cos^3\,x\;dx\)

A

\(\dfrac {cos^5\,x}{5}-\dfrac {cos^7\,x}{7}+C\)

.

B

\(\dfrac {cos^3\,x}{3}-\dfrac {sin^5\,x}{5}+C\)

C

\(\dfrac {sin^5\,x}{5}-\dfrac {sin^7\,x}{7}+C\)

D

\(x\,e^x+sin\,x+C\)

Option C is Correct

Integrals involving Powers of Cosine and Odd Powers of Sine

Method of Evaluating Integrals of the form:

\(I=\int\limits(sin\,x)^m\;(cos\,x)^n\;dx\)   or 

\(I=\int\limits(sin^m\,x)\;(cos^n\,x)\;dx\)

where, \( m, n\) are integers and \(m, n\geq 0\)

  • If power of sine is odd (i.e. \(m = 2 k + 1\) ) we substitute \(cos\;x = t\)

This substitution makes use of the fact that even powers of \(sin\,x\) and \(cos\,x\) can be expressed in  terms of each other without radical signs. 

One power of \(sin\,x\) will go with \(dx\) and the remaining even power will be expressed in terms of \(cos\,x\).

Example:

\(I=\int\limits cos^2\,x\;sin^3\,x\;dx\)     put \(cost\;x = t\)

\(\Rightarrow -sin\,x\;dx=dt\) \(\Rightarrow sin\,x\;dx=-dt\) 

\(\therefore \;I=\int cos^2\,x×sin^2\;x×(sin\,x\;dx)\) (separate one power of \(sin\,x\))

\(=\int\:t^2 (1-t^2)\;\;\times(-dt)\)        \(\rightarrow\,sin^2\;x=1-cos^2x\,=1-t^2\)

\(=\int (t^4-t^2)\;dt\)

\(=\dfrac {t^5}{5}-\dfrac {t^3}{3}+C\)\(=\dfrac {cos^5\,x}{5}-\dfrac {cos^3\,x}{3}+C\)

Illustration Questions

Evaluate \(I=\int cos^4\, x \;sin ^3\, x \;dx\)

A \(\dfrac {cos^7\,x}{7}-\dfrac {cos^5\,x}{5}+C\)

B \(\dfrac {sin^7\,x}{7}-\dfrac {sin^5\,x}{5}+C\)

C \(\dfrac {cos^5\,x}{5}-\dfrac {sin^3\,x}{3}+C\)

D \(x^3\,e^x+tan\;x+C\)

×

\(I=\int cos^4\, x \;sin ^3\, x \;dx\)

Since the power of \(sin\;x\) is odd, put \(cos\; x = t\)

\(\Rightarrow -sin\;x\;dx=dt \Rightarrow sin\;x\;dx=-dt\)

\(\therefore \;I=\int cos^4\,x×sin^2\;x×(sin\,x\;dx)\)    (separate one power of \(sin\,x\))

\(=\int t^4(1-t^2)\;×(-dt)\)        \((sin^2x=1-cos^2x\Rightarrow 1-t^2)\)

\(=\int (t^6-t^4)\;dt\)

\(=\dfrac {t^7}{7}-\dfrac {t^5}{5}+C\)

\(=\dfrac {cos^7\,x}{7}-\dfrac {cos^5\,x}{5}+C\rightarrow \) put back \(t = cos\;x\)

Evaluate \(I=\int cos^4\, x \;sin ^3\, x \;dx\)

A

\(\dfrac {cos^7\,x}{7}-\dfrac {cos^5\,x}{5}+C\)

.

B

\(\dfrac {sin^7\,x}{7}-\dfrac {sin^5\,x}{5}+C\)

C

\(\dfrac {cos^5\,x}{5}-\dfrac {sin^3\,x}{3}+C\)

D

\(x^3\,e^x+tan\;x+C\)

Option A is Correct

Integrating Even Powers of Sine and Cosine Functions by Substitution

Method of Evaluating Integrals of the Form:

\(I=\int\limits(sin\,x)^m\;(cos\,x)^n\;dx\) or 

\(I=\int\limits(sin^m\,x)\;(cos^n\,x)\;dx\)

where, \(m, n\) are integers and \(m, n\geq 0\)

  • If powers of both \(sin\,x\) and \(cos\,x\) are even then we use the trigonometric identities

\(sin^2\,x=\dfrac {1-cos\,2x}{2}\)   or   \(cos^2\,x=\dfrac {1+cos\,2x}{2}\)  or  \(sin\,x\;cos\,x=\dfrac {sin\,2x}{2}\)

to simplify the integrand, then integrate.

  • For powers of \(sin^\,x\) and \(cos^\,x\) which are even and higher than 2, can also be reduced to simpler forms.
  • \((sin\,x)^4=(sin^2x)^2=\left ( \dfrac {1-cos\,2x}{2}\right)^2\)

\(=\dfrac {1}{4}(1+cos^2\,2x-2cos\,2x)\)                            ?

or   \((cos\,x)^4=(cos^2x)^2=\left ( \dfrac {1+cos\,2x}{2}\right)^2\)

\(=\dfrac {1}{4}(1+cos^2\,2x+2\,cos\,2x)\)                          ?

Illustration Questions

Evaluate \(I=\int\limits sin^4\,x\;dx\)

A \(\dfrac {3x}{8}+\dfrac {sin\,4x}{32}-\dfrac {sin\,2x}{4}+C\)

B \(cos^2\,x+sin^3x+C\)

C \(2x+sin\,4x-\dfrac {sin\,2x}{3}\)

D \(\dfrac {sin\,8x}{8}+\dfrac {cos\,4x}{4}+C\)

×

\(I=\int\limits sin^4\,x\;dx\) \(= \int\limits sin^4\,x\;cos^0\,x\;dx\)

Power of \(sin\,x\) is 4, \(cos\,x\) is 0, therefore use identity

\(sin^2x = \dfrac {1-cos\,2x}{2}\)

 

\(I=\int\limits (sin^2\,x)^2\;dx\)

\(=\displaystyle \int\limits \left ( \dfrac {1-cos\,2x}{2} \right)^2\;dx\)

\(=\displaystyle\dfrac {1}{4}\int(1+cos^2\,2x-2cos\,2x)\;dx\)       

\(=\displaystyle\dfrac {1}{4}\int\left(1+\dfrac {1+cos4\,x}{2}-2\,cos\,2x\right)\;dx\)         \(\left( \because cos^2\,2x= \dfrac {1+cos\,4x}{2}\right)\)

\(=\displaystyle\dfrac {1}{4}\int\left(1+\dfrac {1}{2}+\dfrac {cos\,4x}{2}-2\,cos\,2x\right)\;dx\)

\(=\displaystyle\dfrac {1}{4}\left[x+\dfrac {1}{2}x+\dfrac {sin\,4x}{8}-\,sin\,2x\right]+C\)

\(=\dfrac {3x}{8}+\dfrac {sin\,4x}{32}-\dfrac {sin\,2x}{4}+C\)

Evaluate \(I=\int\limits sin^4\,x\;dx\)

A

\(\dfrac {3x}{8}+\dfrac {sin\,4x}{32}-\dfrac {sin\,2x}{4}+C\)

.

B

\(cos^2\,x+sin^3x+C\)

C

\(2x+sin\,4x-\dfrac {sin\,2x}{3}\)

D

\(\dfrac {sin\,8x}{8}+\dfrac {cos\,4x}{4}+C\)

Option A is Correct

Evaluation of Integrals

Evaluation of Integrals of the Form:

  1. \(\int sin\,mx\;cos\,nx\;dx\)
  2. \(\int cos\,mx\;cos\,nx\;dx\)
  3. \(\int sin\,mx\;sin\,nx\;dx\)

where \(m, n\) are any real numbers.

  • Use 

\(2\,sin\,mx\;cos\,nx=sin(m+n)\,x+sin(m-n)\,x\)

\(2\,cos\,mx\;cos\,nx=cos(m+n)\,x+cos(m-n)\,x\)

\(2\,sin\,mx\;sin\,nx=cos(m-n)\,x-cos(m+n)\,x\)

and then integrate.

 

Illustration Questions

Evaluate \(I=\int\,sin\,3x\;cos7x\;dx\)

A \(\dfrac {-cos\,10x}{20}+\dfrac {cos\,4x}{8}+C\)

B \(\dfrac {sin\,10x}{20}+\dfrac {sin\,4x}{8}+C\)

C \(cos^2\,5x-sin\,x+C\)

D \(x\,e^x\,+ log_2\,x+C\)

×

\(I=\int\,sin\,3x\;cos7x\;dx\)

\(I=\dfrac {1}{2}\int\,2\,sin\,3x\;cos7x\;dx\)

\(\big[\because 2\,sinA\;cosB=sin(A+B)+sin(A-B)\big]\)

\(I=\dfrac {1}{2}\int\,(sin(3x+7x)+sin(3x-7x))\;dx\)

\(I=\dfrac {1}{2}\int\,(sin\,10x-sin\;4x)\;dx\)

\(= \dfrac {1}{2} \left [ \dfrac {-cos\,10x}{10}+\dfrac {cos\;4x}{4} \right]+C\)

\(= \dfrac {-cos\,10x}{20}+\dfrac {cos\;4x}{8} +C\)

Evaluate \(I=\int\,sin\,3x\;cos7x\;dx\)

A

\(\dfrac {-cos\,10x}{20}+\dfrac {cos\,4x}{8}+C\)

.

B

\(\dfrac {sin\,10x}{20}+\dfrac {sin\,4x}{8}+C\)

C

\(cos^2\,5x-sin\,x+C\)

D

\(x\,e^x\,+ log_2\,x+C\)

Option A is Correct

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