It is interesting.Gamma functions often occur as part of definite integral

 Consider this integral of parameter 'a'

 

                 

₀   X^a-1 e^-x dx

 

That is integral of X to the power a-1 multiplied with e to the power -X within limits 0 and infinity.

This integral is known as Gamma (a)

If you Evaluate it,you get (a-1)!

replace a by 'n'.

 

Thus While Integrating if you are able to reduce any integral in form of the above mentioned integral you can easily put the Value of the integral.You can need it while evaluating certain Remanian(Sorry if i Have misspelled) integral.

 

Try out integrating This.You will understand the utility of gamma function yourself.

 

       

                   1

I have a better solution for the second Question.

tanx dx

take half common and multiply by 2

1/22tanx dx

now write it as

    

                    (tanx +cotx)dx + (tanx -cotx)dx

                         i1                          i2

 

break tan and cot into sin and cos

 

            i1=  (sinx+cosx)/sinxcosx

multiply numerator and denominator by 4

  

then the denominator becomes

 

                              2sinxcosx

write it as (1-(sinx-cosx)²

tak e sinx-cosx as Z  then                               

i1 will reduce to standard form                       dZ/1-Z²

 

similarlly for i2 you will first break into sinx and cosx multiply by 4 in numerator and denominator and write the denominator as

                    

                        (sinx+cosx)²-1   

consider sinx +cosx as V 

  it will reduce to standard form                                 

 

                              dV/V²-1

 

now you can integrate i1 and i2.

in both cases i arrange the numerator as the derivative of the denominator.

Have a good day

 

 

one can never integrate this

this is from fundamental of calculas .

integrate this

 

 

^[3]tanx dx

 

third root of tanx

Please integrate this

 

 

^[3]tanx dx

 

third root of tanx.

Please integrate this

 

 

^{[ 3]}tanx dx

 

Third root of 'tanx'.

X²-Y²=435678902

 

Find the total number of integral solution.

A number when devided by 13 gives remender 11,when devided by 11 gives remender 7,when devided by 7 gives remender 5,when devided by 5 gives remender 3,when devided by 3 gives remender 2,and when devided by 2 gives remender 1.Find the Number.

mod(x),mod(x-1),mod(x+1)

They are the first three terms in an A.P, find the sum upto first 10 terms.

a> 45

b>25

c>35

d>15

sorry for,i could not give the mod sign