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Integration of various types (Substitution)

Type I: \displaystyle \int { \sin ^{ m }{ x } .\cos ^{ n }{ x } } dx

1. Where m, n belongs to natural number.
2. If one of them is odd, then substitute for term of even power.
3. If both are odd, substitute either of them
4. If both are even, use trigonometric identities only.
5. If m and n are rational numbers and \displaystyle \left( \frac { m+n-2 }{ 2 } \right) is a negative integer, then substitute cot x=p or tan x=p which so ever is found suitable

Example

\displaystyle \int { \sin ^{ 3 }{ x } .\cos ^{ 5 }{ x } } dx

Solution: 

\displaystyle I=\int { \sin ^{ 3 }{ x } .\cos ^{ 5 }{ x } } dx\\ Let\quad \cos { x } =t\quad \Rightarrow -\sin { x } dx=dt\\ I=-\int { \left( 1-{ t }^{ 2 } \right) } .{ t }^{ 5 }dt\\ I=\int { { t }^{ 7 } } dt-\int { { t }^{ 5 }dt } =\frac { { t }^{ 8 } }{ 8 } -\frac { { t }^{ 6 } }{ 6 } +c\\ I=\frac { \cos ^{ 8 }{ x } }{ 8 } -\frac { \cos ^{ 6 }{ x } }{ 6 } +c

 

Type II: \displaystyle \int { \frac { 1 }{ a{ x }^{ 2 }+bx+c } } dx

To evaluate this type of integrals we express \displaystyle { a{ x }^{ 2 }+bx+c } as the sum or difference of two squares by using the following algorithm.

1. Make the coefficient of \displaystyle { x }^{ 2 } unity, if it is not, by multiplying and dividing by it.
2. Add and subtract the square of the half of coefficient of x to express \displaystyle { a{ x }^{ 2 }+bx+c } in the form \displaystyle a\left[ { \left( x+\frac { b }{ 2a } \right) }^{ 2 }+\frac { 4ac-{ b }^{ 2 } }{ 4{ a }^{ 2 } } \right]
3. Use the suitable formula from the following formulas :
\displaystyle \int { \frac { 1 }{ { a }^{ 2 }+{ x }^{ 2 } } } dx=\frac { 1 }{ a } \tan ^{ -1 }{ \left( \frac { x }{ a } \right) } +c\\ \int { \frac { 1 }{ { a }^{ 2 }+{ x }^{ 2 } } } dx=\frac { 1 }{ 2a } \log { \left| \frac { a+x }{ a-x } \right| } +c\\ \int { \frac { 1 }{ { x }^{ 2 }-{ a }^{ 2 } } } dx=\frac { 1 }{ 2a } \log { \left| \frac { x-a }{ x+a } \right| } +c

Example

\displaystyle \frac { 1 }{ { x }^{ 2 }-x+1 } dx

Solution: 

\displaystyle \frac { 1 }{ { x }^{ 2 }-x+1 } dx\\ =\int { \frac { 1 }{ { x }^{ 2 }-x+\frac { 1 }{ 4 } -\frac { 1 }{ 4 } +1 } } dx\\ =\int { \frac { 1 }{ { \left( x-\frac { 1 }{ 2 } \right) }^{ 2 }+\frac { 3 }{ 4 } } dx } \\ =\int { \frac { 1 }{ { \left( x-\frac { 1 }{ 2 } \right) }^{ 2 }+{ \left( \frac { \sqrt { 3 } }{ 2 } \right) }^{ 2 } } } dx\\ =\frac { 1 }{ \frac { \sqrt { 3 } }{ 2 } } \tan ^{ -1 }{ \left( \frac { x-\frac { 1 }{ 2 } }{ \frac { \sqrt { 3 } }{ 2 } } \right) } +c\\ =\frac { 2 }{ \sqrt { 3 } } \tan ^{ -1 }{ \left( \frac { 2x-1 }{ \sqrt { 3 } } \right) } +c

Type III: Integrals of the type \displaystyle \int { \frac { 1 }{ \sqrt { a{ x }^{ 2 }+bx+c } } } dx

Algorithm

1. Make the coefficient of \displaystyle { x }^{ 2 } unity, if it is not
2. Find half of the coefficient of x
3.Add and subtract \displaystyle { \left( \frac { 1 }{ 2 } \quad coeff.\quad of\quad x \right) }^{ 2 } inside the square root to express the quantity inside the square root in the form \displaystyle { \left( x+\frac { b }{ 2a } \right) }^{ 2 }+\frac { 4ac-{ b }^{ 2 } }{ 4{ a }^{ 2 } } \quad or\quad \frac { 4ac-{ b }^{ 2 } }{ 4{ a }^{ 2 } } -{ \left( x+\frac { b }{ 2a } \right) }^{ 2 }
4. Use the suitable formula from the following formulas:
\displaystyle \int { \frac { 1 }{ \sqrt { { a }^{ 2 }+{ x }^{ 2 } } } } dx=\log { \left| x+\sqrt { { a }^{ 2 }+{ x }^{ 2 } } \right| } +c\\ \int { \frac { 1 }{ \sqrt { { x }^{ 2 }-{ a }^{ 2 } } } dx } =\log { \left| x+\sqrt { { x }^{ 2 }-{ a }^{ 2 } } \right| } +c \\ \int { \frac { 1 }{ \sqrt { { a }^{ 2 }-{ x }^{ 2 } } } } dx=\sin ^{ -1 }{ \left( \frac { x }{ a } \right) } +c

Example

\displaystyle \int { \frac { 1 }{ \sqrt { \left( x-1 \right) \left( x-2 \right) } } } dx

Solution:

\displaystyle Let\quad I=\int { \frac { 1 }{ \sqrt { \left( x-1 \right) \left( x-2 \right) } } } dx\\ =\int { \frac { 1 }{ \sqrt { { x }^{ 2 }-3x+\frac { 9 }{ 4 } -\frac { 9 }{ 4 } +2 } } } dx\\ =\int { \frac { 1 }{ \sqrt { { \left( x-\frac { 3 }{ 2 } \right) }^{ 2 }-{ \left( \frac { 1 }{ 2 } \right) }^{ 2 } } } } dx\\ =\log { \left| \left( x-\frac { 3 }{ 2 } \right) +\sqrt { { \left( x-\frac { 3 }{ 2 } \right) }^{ 2 }-{ \left( \frac { 1 }{ 2 } \right) }^{ 2 } } \right| } +c\\ =\log { \left| \left( x-\frac { 3 }{ 2 } \right) +\sqrt { { x }^{ 2 }-3x+2 } \right| } +c

 

Type IV: Integrals Reducible to the form \displaystyle \int { \frac { 1 }{ \sqrt { a{ x }^{ 2 }+bx+c } } } dx

Evaluate:

\displaystyle \int { \frac { { x }^{ 2 } }{ \sqrt { 1-{ x }^{ 6 } } } } dx

Solution:

\displaystyle Let\quad I=\int { \frac { { x }^{ 2 } }{ \sqrt { 1-{ x }^{ 6 } } } } dx\\ Put\quad { x }^{ 3 }=t\quad so\quad that\quad 3{ x }^{ 2 }dx\quad =dt\\ \Rightarrow dx=\frac { dt }{ 3{ x }^{ 2 } } \\ \therefore I=\frac { 1 }{ 3 } \int { \frac { dt }{ \sqrt { 1-{ t }^{ 2 } } } } \\ =\frac { 1 }{ 3 } \sin ^{ -1 }{ \left( t \right) } +c\\ =\frac { 1 }{ 3 } \sin ^{ -1 }{ \left( { x }^{ 3 } \right) } +c

 

Type V: Integrals of the form \displaystyle \int { \frac { px+q }{ a{ x }^{ 2 }+bx+c } } dx

To evaluate this type of integrals, we use the following algorithm:

Algorithm

1. Write the numerator px+q in the following form:
\displaystyle px+q=\lambda \left\{ \frac { d }{ dx } \left( a{ x }^{ 2 }+bx+c \right) \right\} +\mu \\ i.e.\quad px+q=\lambda \left( 2ax+b \right) +\mu
2. Obtain the values of \displaystyle \lambda and \displaystyle \mu by equating the coefficients of like powers of x on both sides.
3. Replace \displaystyle px+q=\lambda \left( 2ax+b \right) +\mu in the given integral to get:
\displaystyle \int { \frac { px+q }{ a{ x }^{ 2 }+bx+c } } dx\\ =\lambda \int { \frac { 2ax+b }{ a{ x }^{ 2 }+bx+c } } dx+\mu \int { \frac { 1 }{ a{ x }^{ 2 }+bx+c } } dx
4. Integrate RHS in step III and put the values of \displaystyle \lambda and \displaystyle \mu obtained in step II.

Example

\displaystyle \int { \frac { 4x+1 }{ { x }^{ 2 }+3x+2 } } dx

Solution:

\displaystyle Let\quad 4x+1\quad =\lambda .\frac { d }{ dx } \left( { x }^{ 2 }+3x+2 \right) +\mu =1\\ \Rightarrow \lambda =2\quad and\quad \mu =-5\\ \therefore \int { \frac { 4x+1 }{ { x }^{ 2 }+3x+2 } } dx=\int { \frac { 2\left( 2x+3 \right) -5 }{ { x }^{ 2 }+3x+2 } } dx\\ =2\int { \frac { \left( 2x+3 \right) -5 }{ { x }^{ 2 }+3x+2 } } dx-5\int { \frac { 1 }{ { x }^{ 2 }+3x+2 } } dx\\ =2\log { \left| { x }^{ 2 }+3x+2 \right| } -5\int { \frac { 1 }{ { x }^{ 2 }+3x+\frac { 9 }{ 4 } -\frac { 9 }{ 4 } 2 } } dx\\ =2\log { \left| { x }^{ 2 }+3x+2 \right| } -5\int { \frac { 1 }{ { \left( x+\frac { 3 }{ 2 } \right) }^{ 2 }-{ \left( \frac { 1 }{ 2 } \right) }^{ 2 } } } dx\\ =2\log { \left| { x }^{ 2 }+3x+2 \right| } -5.\frac { 1 }{ 2\left( \frac { 1 }{ 2 } \right) } \log { \left| \frac { x+\frac { 3 }{ 2 } -\frac { 1 }{ 2 } }{ x+\frac { 3 }{ 2 } +\frac { 1 }{ 2 } } \right| } +c\\ =2\log { \left| { x }^{ 2 }+3x+2 \right| } -5\log { \left| \frac { x+\frac { 3 }{ 2 } -\frac { 1 }{ 2 } }{ x+\frac { 3 }{ 2 } +\frac { 1 }{ 2 } } \right| } +c

 

Type: VI Integrals of the form \displaystyle \int { \frac { P\left( x \right) }{ a{ x }^{ 2 }+bx+c } } dx   

Where p(x) is a polynomial of degree greater than or equal to 2

To evaluate this type of integrals we divide the numerator by the denominator and express the integral as

\displaystyle Q\left( x \right) +\frac { R\left( x \right) }{ a{ x }^{ 2 }+bx+c }  

Where R(x) is a linear function of x.

\displaystyle \therefore \int { \frac { P\left( x \right) }{ a{ x }^{ 2 }+bx+c } } dx = Q\left( x \right) +\frac { R\left( x \right) }{ a{ x }^{ 2 }+bx+c }  

Now to evaluate the second integral on RHS apply the method discussed earlier.

Example

\displaystyle \int { \frac { { x }^{ 2 }-x+1 }{ { x }^{ 2 }+1 } } dx

Solution:

\displaystyle we\quad have\quad I=\int { \frac { { x }^{ 2 }-x+1 }{ { x }^{ 2 }-1 } } dx\\ =\int { x } +\frac { 2x+1 }{ { x }^{ 2 }-1 } dx\\ =\int { x } dx+\int { \frac { 2x+1 }{ { x }^{ 2 }-1 } dx } \int { \frac { 1 }{ { x }^{ 2 }-1 } } dx\\ =\frac { { x }^{ 2 } }{ 2 } +\log { \left| { x }^{ 2 }-1 \right| } +\frac { 1 }{ 2 } \log { \left| \frac { x-1 }{ x+1 } \right| } +c

 

Type: VII Integrals of the form \displaystyle \int { \frac { px+q }{ \sqrt { a{ x }^{ 2 }+bx+c } } } dx

Algorithm

1. Write the numerator px+q in the following form:
\displaystyle px+q=\lambda \left\{ \frac { d }{ dx } \left( a{ x }^{ 2 }+bx+c \right) \right\} +\mu \\ i.e.\quad px+q=\lambda \left( 2ax+b \right) \mu
2. Obtain the values of \displaystyle \lambda and \displaystyle \mu by equating the coefficients of like powers of x on both sides
3. Replace latex \displaystyle px+q=\lambda \left( 2ax+b \right) \mu &s=1 $ in the given integral to get
\displaystyle \int { \frac { px+q }{ \sqrt { a{ x }^{ 2 }+bx+c } } dx } =\lambda \int { \frac { 2ax+b }{ \sqrt { a{ x }^{ 2 }+bx+c } } } dx+\mu \int { \frac { 1 }{ \sqrt { { ax }^{ 2 }+bx+c } } } dx
4. Integrate rhS in step III and put the values of \displaystyle \lambda and \displaystyle \mu obtained in step II.

Example

\displaystyle \int { \frac { x+2 }{ \sqrt { { x }^{ 2 }+5x+6 } } }  dx

Solution:

\displaystyle I=\int { \frac { x+2 }{ \sqrt { { x }^{ 2 }+5x+6 } } } dx\\ =\int { \frac { \frac { 1 }{ 2 } \left( 2x+5 \right) -\frac { 1 }{ 2 } }{ \sqrt { { x }^{ 2 }+5x+6 } } } dx\\ =\frac { 1 }{ 2 } \int { \frac { \left( 2x+5 \right) }{ \sqrt { { x }^{ 2 }+5x+6 } } } dx-\frac { 1 }{ 2 } \int { \frac { 1 }{ \sqrt { { x }^{ 2 }+5x+6 } } } dx\\ =\frac { 1 }{ 2 } \int { \frac { dt }{ \sqrt { t } } } -\frac { 1 }{ 2 } \int { \frac { 1 }{ \sqrt { \left( x+\frac { 5 }{ 2 } \right) -{ \left( \frac { 1 }{ 2 } \right) }^{ 2 } } } } dx\\ where\quad t={ x }^{ 2 }+5x+6\\ =\sqrt { t } -\frac { 1 }{ 2 } \log { \left| \left( x+\frac { 5 }{ 2 } \right) +\sqrt { { x }^{ 2 }+5x+6 } \right| } +c\\ =\sqrt { { x }^{ 2 }+5x+6 } -\frac { 1 }{ 2 } \log { \left| \left( x+\frac { 5 }{ 2 } \right) +\sqrt { { x }^{ 2 }+5x+6 } \right| } +c

 

Type: VIII Integrals of the form

\displaystyle \int { \frac { 1 }{ a\sin ^{ 2 }{ x } +b\cos ^{ 2 }{ x } } dx } ,\\ \int { \frac { 1 }{ a+b\sin ^{ 2 }{ x } } } dx,\\ \int { \frac { 1 }{ a+b\cos ^{ 2 }{ x } } } dx,\\ \int { \frac { 1 }{ { \left( { a\sin { x } +b\cos { x } } \right) }^{ 2 } } } dx,\\ \int { \frac { 1 }{ a+b\sin ^{ 2 }{ x } +c\cos ^{ 2 }{ x } } }dx

Algorithm

1. Divide numerator and denominator both by \displaystyle \cos ^{ 2 }{ x }
2. Replace \displaystyle \sec ^{ 2 }{ x } , if any, in denominator by \displaystyle 1+\tan ^{ 2 }{ x }
3. Put \displaystyle \tan { x } =t so that \displaystyle \sec ^{ 2 }{ x } dx=dt
This substitution reduces the integral in the form \displaystyle \int { \frac { 1 }{ a{ t }^{ 2 }+bt+c } } dt
Evaluate the integral obtained in step III by using the methods discussed earlier

Example

\displaystyle \frac { \sin { x } }{ \sin { 3x } } dx

Solution:

\displaystyle Let\quad I=\frac { \sin { x } }{ \sin { 3x } } dx\\ =\int { \frac { \sin { x } }{ 3\sin { x } -4\sin ^{ 3 }{ x } } } dx\\ =\int { \frac { 1 }{ 3-4\sin ^{ 2 }{ x } } } dx

Dividing numerator and denominator by   \displaystyle \cos ^{ 2 }{ x }

\displaystyle =\int { \frac { \sec ^{ 2 }{ x } }{ 3\sec ^{ 2 }{ x } -4\tan ^{ 2 }{ x } } } dx

\displaystyle Put\quad \tan { x } =t\quad so\quad that\quad \sec ^{ 2 }{ x } dx=dt\\ \therefore I=\int { \frac { dt }{ 3\left( 1+{ t }^{ 2 } \right) -4{ t }^{ 2 } } } \\ =\int { \frac { dt }{ 3-{ t }^{ 2 } } } \\ =\int { \frac { 1 }{ { \left( \sqrt { 3 } \right) }^{ 2 }-{ t }^{ 2 } } } dt\\ =\frac { 1 }{ 2\sqrt { 3 } } \log { \left| \frac { \sqrt { 3 } +t }{ \sqrt { 3 } -t } \right| } +c\\ =\frac { 1 }{ 2\sqrt { 3 } } \log { \left| \frac { \sqrt { 3 } +\tan { x } }{ \sqrt { 3 } -\tan { x } } \right| } +c

 

Type: IX Integrals of the form

\displaystyle \int { \frac { 1 }{ 1+a\sin { x } +b\cos { x } } dx } ,\int { \frac { 1 }{ a+b\cos { x } } } dx,\int { \frac { 1 }{ a+b\sin { x } } dx } ,\int { \frac { 1 }{ a\sin { x } +b\cos { x } +c } } dx

To evaluate this type of integrals, we use the following algorithm:

Algorithm

1. Put \displaystyle \sin { x } =\frac { 2\tan { \frac { x }{ 2 } } }{ 1+\tan ^{ 2 }{ \frac { x }{ 2 } } } ,\quad \cos { x } =\frac { 1-\tan ^{ 2 }{ \frac { x }{ 2 } } }{ 1+\tan ^{ 2 }{ \frac { x }{ 2 } } }
2. Replace \displaystyle 1+\tan ^{ 2 }{ \frac { x }{ 2 } } in the numerator by \displaystyle \sec ^{ 2 }{ \frac { x }{ 2 } }
3. \displaystyle \tan { \frac { x }{ 2 } } =t so that \displaystyle \frac { 1 }{ 2 } \sec ^{ 2 }{ \frac { x }{ 2 } } dx=dt This substitution reduces the integral form \displaystyle \int { \frac { 1 }{ a{ t }^{ 2 }+bt+c } } dt
4. Evaluate the integral in step 3 by using methods discussed earlier

Example

\displaystyle \int { \frac { 1 }{ 1+\sin { x } +\cos { x } } } dx

Solution:

\displaystyle Putting\quad \sin { x } =\frac { 2\tan { \frac { x }{ 2 } } }{ 1+\tan ^{ 2 }{ \frac { x }{ 2 } } } \quad and\quad \cos { x } =\frac { 1-\tan ^{ 2 }{ \frac { x }{ 2 } } }{ 1+\tan ^{ 2 }{ \frac { x }{ 2 } } }

\displaystyle \int { \frac { 1 }{ 1+\sin { x } +\cos { x } } } dx\\ =\int { \frac { 1 }{ 1+\frac { 2\tan { \frac { x }{ 2 } } }{ 1+\tan ^{ 2 }{ \frac { x }{ 2 } } } +\frac { 1-\tan ^{ 2 }{ \frac { x }{ 2 } } }{ 1+\tan ^{ 2 }{ \frac { x }{ 2 } } } } } dx\\ =\int { \frac { 1+\tan ^{ 2 }{ \frac { x }{ 2 } } }{ 1+\tan ^{ 2 }{ \frac { x }{ 2 } } +2\tan { \frac { x }{ 2 } } +1-\tan ^{ 2 }{ \frac { x }{ 2 } } } } dx\\ =\int { \frac { \sec ^{ 2 }{ \frac { x }{ 2 } } }{ 2+2\tan { \frac { x }{ 2 } } } } dx

Put \displaystyle \tan { \frac { x }{ 2 } } =t so that \displaystyle \frac { 1 }{ 2 } \sec ^{ 2 }{ \frac { x }{ 2 } } dx=dt\quad or\quad \sec ^{ 2 }{ \frac { x }{ 2 } } dx=2dt

\displaystyle \therefore I=\int { \frac { 2dt }{ 2+2t } } \\ =\int { \frac { 1 }{ t+1 } } dt\\ =\log { \left| t+1 \right| } +c\\ =\log { \left| \tan { \frac { x }{ 2 } } +1 \right| } +c

 

Type: X Alternative method to evaluate integrals of the form \displaystyle \int { \frac { 1 }{ a\sin { x } +b\cos { x } } } dx

To evaluate this type of integrals we substitute \displaystyle a=r\cos { \theta } ,\quad b=r\sin { \theta } and\quad so\quad r=\sqrt { { a }^{ 2 }+{ b }^{ 2 } } ,\quad \theta =\tan ^{ -1 }{ \left( \frac { b }{ a } \right) }  

\displaystyle \therefore a\sin { x } +b\cos { x } =r\cos { \theta } \sin { x } +r\sin { \theta } \cos { x } =r\sin { \left( x+\theta \right) } \\ So,\quad \int { \frac { 1 }{ a\sin { x } +b\cos { x } } } dx\\ =\frac { 1 }{ r } \int { \frac { 1 }{ \sin { \left( x+\theta \right) } } } dx\\ =\frac { 1 }{ r } \int { cosec } \left( x+\theta \right) dx\\ =\frac { 1 }{ r } \log { \left| \tan { \left( \frac { x }{ 2 } +\frac { \theta }{ 2 } \right) } \right| } +c\\ =\frac { 1 }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \log { \left| \tan { \left( \frac { x }{ 2 } +\frac { 1 }{ 2 } \tan ^{ -1 }{ \frac { b }{ a } } \right) } \right| } +c

Example

\displaystyle \int { \frac { 1 }{ \sqrt { 3 } \sin { x } +\cos { x } } } dx 

Solution:

\displaystyle Let\sqrt { 3 } =r\sin { \theta }\quad and\quad 1=r\cos { \theta } \quad Then\\ r=\sqrt { { \left( \sqrt { 3 } \right) }^{ 2 }+{ 1 }^{ 2 } } =2\quad and\quad \tan { \theta } =\frac { \sqrt { 3 } }{ 1 } \Rightarrow \theta =\frac { \pi }{ 3 } \\ \therefore \int { \frac { 1 }{ \sqrt { 3 } \sin { x } \cos { x } } } dx=\int { \frac { 1 }{ r\sin { \theta } \sin { x } +r\cos { \theta } \cos { x } } } dx

\displaystyle =\frac { 1 }{ r } \int { \frac { 1 }{ \cos { \left( x-\theta \right) } } } dx\\ =\frac { 1 }{ r } \int { \sec { \left( x-\theta \right) } dx } \\ =\frac { 1 }{ r } \log { \left| \tan { \left( \frac { \pi }{ 4 } +\frac { x }{ 2 } -\frac { \theta }{ 2 } \right) } \right| } +c\\ =\frac { 1 }{ 2 } \log { \left| \tan { \left( \frac { \pi }{ 4 } +\frac { x }{ 2 } -\frac { \theta }{ 2 } \right) } \right| } +c\\ =\frac { 1 }{ 2 } \log { \left| \tan { \left( \frac { x }{ 2 } +\frac { \pi }{ 12 } \right) } \right| } +c

Type: XI Integrals of the form \displaystyle \int { \frac { a\sin { x } +b\cos { x } }{ c\sin { x } +d\cos { x } } } dx

To evaluate this type of integrals, we use the following algorithm:

Algorithm

1. Write Numerator=\displaystyle \lambda (Diff. of denominator) + \displaystyle \mu (Denominator)
\displaystyle i.e.\quad a\sin { x } +b\cos { x } =\lambda \left( c\cos { x } -d\sin { x } \right) +\mu \left( c\sin { x } +d\cos { x } \right)
2. Obtain the values of \displaystyle \lambda and \displaystyle \mu by equating the coefficients of \displaystyle \sin { x } \quad and\quad \cos { x }
3. Replace numerator in the integrand by \displaystyle \lambda \left( c\cos { x } -d\sin { x } \right) +\mu \left( c\sin { x } +d\cos { x } \right) to obtain
\displaystyle \int { \frac { a\sin { x } +b\cos { x } }{ c\sin { x } +d\cos { x } } } dx=\lambda \int { \frac { c\cos { x } -d\sin { x } }{ c\sin { x } +\cos { x } } } dx+\mu \int { \frac { c\sin { x } +d\cos { x } }{ c\sin { x } +d\cos { x } } } dx\\= \lambda \log { \left| c\sin { x } +d\cos { x } \right| } +\quad \mu x+c

Example

\displaystyle \int { \frac { 1 }{ 1+\tan { x } } } dx

Solution:

\displaystyle Let\quad I=\int { \frac { 1 }{ 1+\tan { x } } } dx\\ =\frac { \cos { x } }{ \cos { x } +\sin { x } } dx\\ Let\quad \cos { x } =\lambda .\frac { d }{ dx } \left( \cos { x } +\sin { x } \right) +\mu .\left( \cos { x } +\sin { x } \right) \quad then,\\ \cos { x } =\lambda \left( -\sin { x } +\cos { x } \right) \quad +\mu .\left( \cos { x } +\sin { x } \right)   

Comparing the coefficients of sin x and cos x on both sides, we get

\displaystyle -\lambda +\mu =0\quad and\quad \lambda +\mu =1\quad \Rightarrow \lambda =\mu =\frac { 1 }{ 2 } \\ \therefore I=\int { \frac { \cos { x } }{ \cos { x } +\sin { x } } } dx\\ =\int { \frac { \frac { 1 }{ 2 } \left( \cos { x } -\sin { x } \right) +\frac { 1 }{ 2 } \left( \cos { x } +\sin { x } \right) }{ \cos { x } +\sin { x } } } dx\\ =\frac { 1 }{ 2 } \int { \frac { \cos { x } -\sin { x } }{ \cos { x } +\sin { x } } } dx+\frac { 1 }{ 2 } \int { \frac { \cos { x } +\sin { x } }{ \cos { x } +\sin { x } } } dx

Let t= cos x + sin x then,

\displaystyle =\frac { 1 }{ 2 } \int { \frac { dt }{ t } } +\frac { 1 }{ 2 } \int { 1 } dx\\ =\frac { 1 }{ 2 } \log { \left| t \right| } +\frac { 1 }{ 2 } x+c\\ =\frac { 1 }{ 2 } x+\frac { 1 }{ 2 } \log { \left| \cos { x } +\sin { x } \right| } +c

 

Type: XI Integrals of the form  \displaystyle \int { \frac { a\sin { x } +b\cos { x } +c }{ p\sin { x } +q\cos { x } +r } } dx

To evaluate this type of integrals, we use the following algorithm:

Algorithm

1. Write Numerator=\displaystyle \lambda (Diff. of denominator) + \displaystyle \mu (Denominator) + v
\displaystyle i.e. a\sin { x } +b\cos { x } +c=\lambda \left( p\cos { x } -q\sin { x } \right) +\mu \left( p\sin { x } +q\cos { x } +r \right) +v
2. Obtain the values of \displaystyle \lambda and \displaystyle \mu by equating the coefficients of \displaystyle \sin { x } \quad and\quad \cos { x } and the constant term both sides
3. Replace the numerator in the integrand by \displaystyle \lambda \left( p\cos { x } -q\sin { x } \right) +\mu \left( p\sin { x } +q\cos { x } \right) to obtain
\displaystyle \int { \frac { a\sin { x } +b\cos { x } }{ p\sin { x } +q\cos { x } +r } } dx=\lambda \int { \frac { p\cos { x } -q\sin { x } }{ p\sin { x } +q\cos { x } +r } } dx+\mu \int { \frac { p\sin { x } +q\cos { x } +r }{ p\sin { x } +q\cos { x } +r } } dx+v\int { \frac { 1 }{ p\sin { x } +q\cos { x } +r } } dx
\displaystyle =\lambda \log { \left| p\sin { x } +q\cos { x } +r \right| } +\mu x+v\int { \frac { 1 }{ p\sin { x } +q\cos { x } +r } } dx
4. Evaluate the integral on RHS in step 3 by using the methods discussed earlier

Example

\displaystyle \int { \frac { 3\cos { x } +2 }{ \sin { x } +2\cos { x } +3 } } dx

Solution:

\displaystyle I=\int { \frac { 3\cos { x } +2 }{ \sin { x } +2\cos { x } +3 } } dx\\ Let\quad 3\cos { x } +2=\lambda \left( \sin { x } +2\cos { x } +3 \right) +\mu \left( \cos { x } -2\sin { x } \right) +v

Comparing the coefficients of sin x , cos x and constant term on both we get

\displaystyle \lambda -2\mu =0,\quad 2\lambda +\mu =3,\quad 3\lambda +v=2\\ \Rightarrow \lambda =\frac { 6 }{ 5 } ,\quad \mu =\frac { 3 }{ 5 } \quad and\quad v=-\frac { 8 }{ 5 }

\displaystyle \therefore I=\int { \frac { \lambda \left( \sin { x } +2\cos { x } +3 \right) +\mu \left( \cos { x } -2\sin { x } \right) +v }{ \sin { x } +2\cos { x } +3 } } dx\\ =\int { dx } +\mu \int { \frac { \cos { x } -2\sin { x } }{ \sin { x } +2\cos { x } +3 } } dx+\int { \frac { 1 }{ \sin { x } +2\cos { x } +3 } } dx\\ =\lambda x+\mu \log { \left| \sin { x } +2\cos { x } +3 \right| } +v{ I }_{ 1 }

\displaystyle where\quad { I }_{ 1 }=\int { \frac { 1 }{ \sin { x } +2\cos { x } +3 } } dx\\ Putting\quad \sin { x } =\frac { 2\tan { \frac { x }{ 2 } } }{ 1+\tan ^{ 2 }{ \frac { x }{ 2 } } } ,\quad \cos { x } =\frac { 1-\tan ^{ 2 }{ \frac { x }{ 2 } } }{ 1+\tan ^{ 2 }{ \frac { x }{ 2 } } } 

\displaystyle { I }_{ 1 }=\int { \frac { 1 }{ \frac { 2\tan { \frac { x }{ 2 } } }{ 1+\tan ^{ 2 }{ \frac { x }{ 2 } } } +\frac { 2\left( 1-\tan ^{ 2 }{ \frac { x }{ 2 } } \right) }{ 1+\tan ^{ 2 }{ \frac { x }{ 2 } } } +3 } } dx

\displaystyle =\int { \frac { 1+\tan ^{ 2 }{ \frac { x }{ 2 } } }{ 2\tan { \frac { x }{ 2 } } +2\left( 1-\tan ^{ 2 }{ \frac { x }{ 2 } } \right) +3 } } dx

\displaystyle =\int { \frac { \sec ^{ 2 }{ \frac { x }{ 2 } } }{ 2\tan { \frac { x }{ 2 } } +2-2\tan ^{ 2 }{ \frac { x }{ 2 } } +3\left( 1+\tan ^{ 2 }{ \frac { x }{ 2 } } \right) } } dx

\displaystyle =\int { \frac { \sec ^{ 2 }{ \frac { x }{ 2 } } }{ \tan ^{ 2 }{ \frac { x }{ 2 } } +2\tan { \frac { x }{ 2 } } +5 } } dx

\displaystyle Put\quad \tan { \frac { x }{ 2 } } =t\quad so\quad that\quad \frac { 1 }{ 2 } \sec ^{ 2 }{ \frac { x }{ 2 } } dx=dt\quad or

\displaystyle \sec ^{ 2 }{ \frac { x }{ 2 } } dx=2dt

\displaystyle \therefore \int { \frac { 2dt }{ { t }^{ 2 }+2t+5 } } \\ =2\int { \frac { dt }{ { \left( t+1 \right) }^{ 2 }+{ 2 }^{ 2 } } } \\ =\frac { 2 }{ 2 } \tan ^{ -1 }{ \left( \frac { t+1 }{ 2 } \right) } \\ =\tan ^{ -1 }{ \left( \frac { \tan { \frac { x }{ 2 } } +1 }{ 2 } \right) } \\ Hence\quad I=\lambda x+\mu \log { \left| \sin { x } +2\cos { x } +3 \right| } +v\tan ^{ -1 }{ \left( \frac { \tan { \frac { x }{ 2 } } +1 }{ 2 } \right) } +c\\ Where\quad \lambda =\frac { 6 }{ 5 } ,\mu =\frac { 3 }{ 5 } \quad and\quad v=-\frac { 8 }{ 5 } 

April 18, 2019

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