Informative line

### 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 $$\int (cos\;x)^0\;dx$$

#### 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.

#### 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

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

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

#### 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.

#### 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

# 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 $$cos\,x$$)

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

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

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

#### 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 $$sin\,x$$)

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

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

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

#### 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