IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

joyfrancis

Evaluate :

1/root(sin³x.sin(x+a))

sboosy

dx/

(sin³xsin(x+a))

multiply and divide by sinx

sinx/ sin²x(

sinxsin(x+a))dx

sin(x+a-a) / sin²x(

sinxsin(x+a))dx

[sin(x+a)cosa - cos(x+a)sina]/ sin²x(

sinxsin(x+a)) dx

splitting into 2 integrals

[sin(x+a)cosa/ sin²x(

sinxsin(x+a)) dx .....1st integral

cosa

cosec²x

(sin(x+a)/sinx) dx

-cosa/sina

-sina cosec²x

(cosa+cotx sina) dx

put cosa+cotx sina = t

-sina cosec²x dx = dt

so we get

-cota

t dt

so integrating we get

-2/3 cota [cosa+cotxsina]^3/2 .....X

similarly in the second integral ..

we have

-sina

cosec²x

(cot(x+a) [cotx cosa - sina]) dx

tana

-cosec²a cosa

(cot(x+a) [cotx cosa - sina]) dx

put cotx cosa - sina = t

-cosec²x cosa dx = dt

tana

t/

((t+sina)tana+cosa) dt

now apply parts

tana

t d(2

((t+sina)tana+cosa)))

tana [ t * (2

((t+sina)tana+cosa))) -

(2

((t+sina)tana+cosa))) dt ]

tana [cotx cosa - sina] (2

((cotx cosa - sina+sina)tana+cosa))) - 4/3 ((cotx cosa - sina+sina)tana+cosa))) ^3/2 .............Y

answer is X+Y+c

varshavallig

a simpler method:

I=1/rootsin^3 sin(x+a)

multiply and divide by root sinx

I=

sinx / sin^2x

sin(x+a)______(1)

take (sin(x+a)/sinx) = t

dt=[sinx cos(x+a)-cosxsin(x+a)]/sin^2x

=-sin(x+a-x)/sin^2x

=-sina/sin^2x

substituting in (1),

I=(-1/sina)

1/

t dt

integrating,

I =(-2/sina)

(sin(x+a)/sinx)

hope this helps

!!!!!!!!!!!!!!!!!!

sandeepramesh

this soln has been posted elsewhere so see properly first

ill post the link wait

sandeepramesh

this is the link

http://www.goiit.com/posts/list/integration-simple-one-46920.htm

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